Oasis School Mathematics-8 97 5. Factorise (a) ax + ay + px + py (b) ax + bx + ma + mb (c) ab – ac – b + c (d) a² – (x + y) a + xy (e) xy² – y(x – z) – z 6. Simplify using factorisation method. (a) 6×25 + 6 × 13 (b) 29 × 86 + 29 ×14 (c) 100 × 33 + 100 × 67 (d) 279 × 300 – 279 × 200 (e) 15×15 – 15 × 12 + 15 × 13 10.2 Factorisation of difference of two squares As we know that (a + b) (a – b) = a² + ab – ab – b² = a² – b² Factors of a² – b² are (a + b) and (a – b) Hence, Geometrical Interpretation of a² – b² • Take a square sheet of paper having a side 'a' unit. Its area is a² • From the square sheet take 'b' units from each side as shown in the figure. a b b a Area of that part = b² Area of the shaded part = a² – b² .............. (i) Answer 1. (a) 2(x+3y) (b) 5xy(3x+5y) (c) 3a(a–5b) (d) 6x3 y (x–3y) (e) –5x2 yz(2xyz2 +3) (f) –7x3 ac2 (3xac–2) 2. (a) a(4+3b+5a) (b) 3xy (x–2+3y) (c) 2y6 (2x3 y–x+3y3 ) (d) –a(bx+cy+dz) (e) 2xy (2x–3y+5) 3. (a) (x+y) (a+b) (b) (a+1) (a2 +1) (c) (a–b) (y–x) (d) (a–b) (x–y) (e) (x–2y) (x–y) 4. (a) (a–2) (2x+9) (b) (x–y) (x–y+1) (c) (a–b) (1–a+b) (d) (a–b) (p+q+r) (e) (a+b+c) (x+y+z) 5. (a) (x+y) (a+p) (b) (a+b) (x+m) (c) (b–c)(a–1) (d) (a–x) (a–y) (e) (y–1) (xy+z) 6. (a) 228 (b) 2900 (c) 10000 ( d) 27900 (e) 240 a² – b² = (a + b) (a – b) a a
98 Oasis School Mathematics-8 • Cut the shaded part. • Again, cut it along the dotted line. • Join these two parts as shown in the figure. Then the new sheet is a rectangle having two sides (a + b) and (a – b) Area of shaded part ∴ Factorisation of sum of two cubes We already know that, (a + b)³ = a³ + 3a²b + 3ab² + b³ or, (a + b)³ = a² + b³ + 3ab (a + b) or, (a + b)³ – 3ab (a + b) = a³ + b³ or, (a + b) [(a + b)² – 3ab] = a³ + b³ or, (a + b) (a² + 2ab + b² – 3ab) = a³ + b³ or, (a + b) (a² – ab + b²) = a³ + b³ Hence, a a b b a – b a a b b a – b a – b a a b b a – b a – b a a b b a – b a – b a² – b² = (a + b) (a – b) Factorisation of difference of two cube We already know that, (a – b)³ = a³ – 3a²b +3ab² – b³ or, (a – b)³ + 3a²b – 3ab² = a³ – b³. or, (a – b)³ + 3ab (a – b) = a³ – b³ or, (a – b) (a² – 2ab + b² + 3ab) = a³ – b³ or, (a – b) (a² + ab + b²) = a³ – b³ a³ + b³ = (a + b) (a² – ab + b²) ∴ a³ – b³ = (a – b) (a² + ab + b²)
Oasis School Mathematics-8 99 Worked Out Examples Example 1 Factorise : x² – 16y² Solution: x² – 16y² = (x)² – (4y)² = (x + 4y) (x – 4y) [∵ a2 –b2 = (a+b ) (a-b)] Example 3 Factorise: (x + 3y)2 – (x – 3y)2 Solution: (x + 3y)2 – (x – 3y)2 = {(x + 3y) + (x – 3y)} {(x + 3y) – (x – 3y)} [∵ a² – b²= (a + b) (a – b)] = (x + 3y+ x – 3y) (x + 3y – x + 3y) = 2x × 6y = 12xy Example 5 Factorise : a³ + 27 Solution: (i) Here, a³ + 27 = (a)3 + (3)3 = (a + 3) {(a)2 – a× 3 + (3)2 } = (a + 3) (a² – 3a +9). Example 7 Factorise: 8(x – y)³ – z³ Solution: 8(x – y)³ – z³ = [2(x – y)]³ – z³ = (2x – 2y)³ – z³ = {(2x – 2y) – z} {(2x – 2y)² + (2x – 2y) z + z²} = (2x – 2y – z) (4x² – 2. 2x. 2y + 4y² + 2xz – 2yz + z²) = (2x – 2y – z) (4x² – 8xy + 4y² + 2xz – 2yz + z²) Example 2 Factorise: 25x² – 64y² Solution: 25x² – 64y² = (5x)² – (8y)² = (5x + 8y) (5x – 8y) [∵ a2 –b2 = (a+b ) (a-b)] Example 4 Factorise: 4a² – 9(b – c)² Solution: 4a² – 9 (b – c)² = (2a)² – {3(b – c)}² = (2a)² – (3b – 3c)² = {2a + (3b – 3c)} {2a–(3b–3c)} = (2a + 3b – 3c) (2a – 3b + 3c) [∵ a2 –b2 = (a+b ) (a-b)] Example 6 Factorise : 27y3 – 8z3 Solution: 27y3 – 8z3 = (3y)3 – (2z)3 = (3y–2z) [(3y)2 + 3y.2z + (2z)2 ] = (3y–2z) (3y2 – 6yz + 4z2 )
100 Oasis School Mathematics-8 Exercise 10.2 1. Factorise each of the following: (a) a² – 16 (b) 4x² – 9y² (c) 9y² – 16z² (d) 25p² – 144q² (e) a² – 1 b² (f) a² 9 – 1 4b² (g) x² 25 – y² 49 2. (a) 4 – (x – y)² (b) (x + y)² – 9z² (c) (x + y)² – (x – y)² (d) 9z² – 4(x – y)² (e) (4x – y)² – 9z² (f) (2a + 3b)² – (2a – 3b)² (g) 25(a + b)² – 16(a – b)² 3. (a) x³ – 4x (b) 2x³ – 72x (c) 5 – 125y² (d) a³b³ – 16ab z² (e) 2 a ³ b ² – 8a 4. Factorise each of the following: (a) x³ + 8 (b) x³ + 27 (c) 8x³ + 125y³ (d) m³ + 27n³ (e) 8x³ – 1 (f) 27m³ – 8n³ (g) 64a³ – 125b³ 5. Factorise: (a) x³ + 1 x³ (b) a³ – 1 b³ (c) a³ 27 + b³ 64 (d) x³ 27 – 125 y³ (e) 27x³ 64 + 1 6. Factorise: (a) (x + y)3 + (x – y)³ (b) (a + b)³ – (a – b)³ (c) (a – 2b)³ – 8c³ (d) 64 – (2a – 3b)³ (e) (2x + 3y)³ – (2x – 3y)³ 7. Factorise: (a) a³ + b³ + a + b (b) a³ – b³ – a + b Answer 1. (a) (a + 4) (a – 4) (b) (2x + 3y) (2x – 3y) (c) (3y + 4z) (3y – 4z) (d) (5p + 12q) (5p – 12q) (e) (a + 1 b )(a – 1 b ) (f) ( a 3 + 1 2b)( a 3 – 1 2b) (g) ( x 5 + y 7 )( x 5 – y 7 ) 2. (a) (2 + x – y) (x – x + y) (b) (x + y + 3z) (x + y – 3z) (c) 4xy (d) (3z + 2x – 2y) (3z – 2x + 2y) (e) (4x – y + 3z) (4x – y – 3z) (f) 24ab (g) (a + 9b) (9a + b) 3. (a) x(x + 2) (x – 2) (b) 2x (x + 6) (x – 6) (c) 5 (1 + 5y) (1 – 5y) (d) ab (ab + 4 z ) (ab – 4 z ) (e) 2a (ab + 2) (ab – 2) 4. (a) (x + 2) (x² – 2x + 4) (b) (x + 3) (x² – 3x + 9) (c) (2x + 5y) (4x² – 10xy + 25y²) (d) (m + 3n) (m² – 3mn + 9n²) (e) (2x – 1) (4x² + 2x + 1) (f) (3m – 2n) (9x² + 6mn + 4x²) (g) (4a – 5b) (16a² + 20ab + 25b²) 5. (a) (x + 1 x ) (x² – 1 + 1 x²) (b) (a – 1 b ) (a² + a b + 1 b²) (c) ( a 3 + b 4 ) ( a² 9 – ab 12 + b² 16) (d) ( x 3 – 5 y ) ( x² 9 + 5x 3y + 25 y²) (e) ( 3x 1 + 1) ( 9x² 16 – 3x 4 + 1) 6. (a) 2x (x² + 3y²) (b) 2a (a² + 3b²) (c) (a – 2b – 2c) (a² – 4ab + 4b² – 2ac – 4bc + 4c²) (d) (4 – 2a + 3b) (16 + 8a – 12b + 4a² – 12ab + 9b²) (e) 8y (12x² + 9y²) 7. (a) (a + b) (a² – ab + b² + 1) (b) (a – b) (a² + ab + b² – 1)
Oasis School Mathematics-8 101 10.3 Factorisation of the trinomial of the form a2 ± 2ab + b2 We know that (a + b)² = a² + 2ab + b² i.e. factors of expression a² + 2ab + b² are (a + b) and (a + b). Similarly, (a – b)² = a² – 2ab + b² i.e. factors of expression a² – 2ab + b² are (a – b) and (a – b) Let's convert the expression 4x² + 12xy + 9y² in the form of (a + b)² 4x² + 12xy + 9y² = (2x)² + 2. 2x. 3y + (3y)² = (2x + 3y)² Example 1 Convert the trinomial 4a² – 20ab + 25b² in the form of (a – b)². Solution: 4a² – 20ab + 25b² = (2a)² – 2.2a.5b + (5b)² = (2a – 5b)² = (2a–5b ) (2a–5b) Exercise 10.3 1. Write the missing term in each of the following to make perfect squares: (a) x² + ..... + (2)² (b) 4a² + 24ab + ..... (c) ..... + 16 mn + 16n² (d) x² – ..... + 16y² (e) 25a² – 60ab + ..... 2. Convert the following in the from of (a + b)² or (a – b)². (a) x² + 10x + 25 (b) 4x² – 4xy + y² (c) 9x² + 24xy + 16y² (d) 49a² – 84ab + 36b² (e) 64m² – 112mn + 49n² (f) x² + 2 + 1 x² 3. Factorise: (a) a² – 14a + 49 (b) 25 – 60y + 36y² (c) m² – 24m + 144 (d) 3b² – 6b + 3 (e) 32x² – 48xy + 18y² (f) x² – 1 + 1 4x² 4. (a) What must be added to x² + 14x to make it a perfect square? (b) What must be added to 9a² + 4b² to make it a perfect square? Answer 1. (a) 4x (b) 36b2 (c) 4m2 (d) 8xy (e) 36b2 2. (a) a.(x+5)2 (b) (2x–y)2 (c) (3x+4y)2 (d) (7a-6b)2 (e) (8m–7n)2 (f) (x+1/x)2 3. (a) (a-7)2 (b) (5-6y)2 (c) (m-12)2 (d) 3(b-1)2 (e) 2(4x–3y)2 (f) (x–1/2x)2 4. (a) 49 (b) 12ab. Worked Out Examples Example 2 Resolve into factors : 9x² + 24xy + 16y² Solution: 9x² + 25xy + 16y² = (3x)² + 2.3x. 4y + (4y)² = (3x + 4y)² = (3x + 4y) (3x + 4y)
102 Oasis School Mathematics-8 10.4 Factorisation of the trinomial of the form ax2 + bx + c. Let's recall the multiplication of binomials, such as (x + 2) (x + 3) Now, (x + 1) (x + 3) = x² + x + 3x + 3 = x² + 4x + 3 = x² + (sum of the constants) x + product of the constant = x2 + (1 + 3)x + 1 × 3 Hence, to factorise x2 + 5x + 6 We have to find two numbers whose sum is 5 and the product is 6. i. e. x² + 5x + 6 = x² + (2 + 3) x + 6 = x² + 2x + 3x + 6 = x (x + 2) + 3(x + 2) = (x + 2) (x + 3) Example 1 Factorise: x² – 7x + 12 Solution: = x² – (3 + 4) x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4(x – 3) = (x – 3) (x – 4) Example 3 Factorise: 3a² + a – 14 Solution: 3a² + a – 14 = 3a² + (7 – 6) a – 14 = 3a² + 7a – 6a – 14 = a(3a + 7) – 2(3a + 7) = (3a + 7) (a – 2) Steps: • Arrange the terms in ascending order or descending order of power of variable. • Find the product (P) of the coefficient of first term and the last term of given expression with their sign. • Split the middle term of the given expression such that the sum (S) or difference (D) of these two terms is equal to the middle term and their product is equal to the product obtained in step 1. • By forming the suitable groups, factorise the given trinomial. Worked Out Examples Example 2 Factorise: x² – x – 6 Solution: = x² – (3 – 2) x – 6 = x² – 3x + 2x – 6 = x(x – 3) + 2(x – 3) = (x – 3) (x + 2) I have to find two numbers whose sum is 7 and product is 12. Product = 42 Difference = 1 Numbers are 7 and 6
Oasis School Mathematics-8 103 Exercise 10.4 1. Find the two numbers, whose (a) Sum = 7, product = 12 (b) Sum = 5, product = 6 (c) Sum = 8, product = 7 (d) Difference = 5, product = 24 (e) Difference = 1, product = 72 (f) Difference = 2, product = 80 2. Factorise: (a) x2 + 3x + 2 (b) x2 + 7x + 10 (c) x2 + 10x + 9 (d) x² + 12x + 20 3. Factorise: (a) m² – 3m + 2 (b) a² – 7a + 12 (c) m2 – 8m + 15 (d) y2 – 11y + 30 4. Factorise: (a) a² + a – 2 (b) x² + 2x – 3 (c) a² + a – 20 (d) m2 + 7m – 18 5. Factorise: (a) x2 – x – 12 (b) x2 – 5x – 24 (c) y2 – y – 72 (d) m² – 2m – 80 6. Factorise: (a) 2a² + 7a + 5 (b) 2x2 + 7x + 3 (c) 3a² + 5a + 2 (d) 4x² + 17x + 15 7. Factorise: (a) 2x2 – 3x + 1 (b) 5a² – 19a + 12 (c) 3a² – 8ab + 5b² (d) 3x² – 36x + 33 8. Factorise: (a) 2m2 + m – 3 (b) 6x2 +5x – 6 (c) 5a²+11a – 12 (d) 7x² + 4x – 11 9. Factorise: (a) 3a² – 2ab – 8b² (b) 3x2 – 7xy –6y2 (c) 4x2 – 16x – 9 (d) 2b² – 5b – 12 10. Factorise: (a) 1 – 3x – 28x2 (b) –7x2 + 3 – 4x (c) 3x2 + 11x + 6 3 (e) (x + 1) (2x – 1) – 5 11. Factorise: (a) (a – b)2 + 4(a – b) + 3 (b) (x + y)2 + 4(x + y) – 12 (c) 2(x + 2)² – 3 (x + 2) – 5 (d) (2x + y)2 + 4(2x + y) – 32 Project Work • Take chart paper, interpret geometrically the relation a2 – b2 = (a + b) (a – b). • Interpret geometrically the relation : x2 + 5x + 6 = (x + 2) (x + 3)
104 Oasis School Mathematics-8 Complete the given tables: (i) Sum Product Numbers (ii) Difference Product Numbers 7 9 11 21 12 10 12 14 28 108 27 21 3,4 2 3 12 9 7 11 15 28 45 70 44 180 5,3 (iii) S.N. Algebraic Expression Middle-term (Sum of the numbers) Product Numbers Factors i. x² – 5x + 6 – 5 6 – 2, – 3 (x – 2) (x – 3) ii. x² + x – 6 iii. x² – 5x – 24 iv. x² – x – 6 v. y² – y – 12 vi. y² – 11y + 24 vii. x² – 7x + 12 viii. x² + 5x + 6 Self Practice Materials Answer 1. (a) 3 and 4 (b) 2 and 3 (c) 1 and 7 (d) 3 and 8 (e) 8 and 9 (f) 8 and 10 2. (a) (x+1) (x+2) (b) (x+5) (x+2) (c) (x+1) (x+9) (d) (x+2) (x+10) 3. (a) (m–1) (m–2) (b) (a–3) (a–4) (c) (m–5) (m–3) (d) (y–5) (y–6) 4. (a) (a+2) (a-1) (b) (x+3) (x-1) (c) (a+5) (a-4) (d) (m+9) (m-2) 5. (a) (a) (x-4) (x+3) (b) (x-8) (x+3) (c) (y-9) (y+8) (d) (m-10) (m+8) 6. (a) (2a+5) (a+1) (b) (2x+1) (x+3) (c) (a+1) (3a+2) (d) (x+3) (4x+5) 7. (a) (2x-1) (x-1) (b) (a-3) (5a-4) (c) (a-b) (3a-5b) (d) (x-11) (3x-3) 8. (a) (2m+3) (m-1) (b) (2x+3) (3x-2) (c) (a+3) (5a-4) (d) (x-1) (7x+11) 9. (a) (a-2b) (3a+4b) (b) (x-3y) (3x+2y) (c) (2x+1) (2x-9) (d) (b-4) (2b+3) 10. (a) (1+4x) (1-7x) (b) -(x+1) (7x-3) (c) (x+3 3 ) ( 3 x+2) (d) (x+2) (2x-3) 11. (a) (a-b+1) (a-b+3) (b) (x+y+6) (x+y-2) (c) (2x-3) (x+3) (d) (2x+y+8) (2x+y-4)
Oasis School Mathematics-8 105 Additional Text Factorisation of the trinomial in the form of a4 +a²b² +b4 We know that, (a + b)² = a² + 2ab + b² or, (a + b)² – 2ab = a² + b² ∴ a² + b²= (a + b)² – 2ab ……………….(i) Again (a – b)² = a² – 2ab + b² or, (a – b)2 + 2ab = a² + b² ∴ a² + b² = (a – b)2 + 2ab ……………….(ii) Example Factorise: (i) a4 + a²b² + b4 Solution: (i) Here, a4 + a²b² + b4 = (a²)² + 2 × a² × b² + (b²)² – a²b² = (a² + b²)² – (ab)² = {(a² + b²) + ab}{(a² + b²) – ab} = (a² + b² + ab) (a² + b² – ab) = (a2 + ab + b2 ) (a2 –ab + b2 ) Additional Exercise Factorise each of the following: 1. x4 + x2 y2 + y4 2. x4 + 4 3. x4 + 6x2 + 25 4. a4 + 7a²b² + 16b4 5. 9a4 – 24a²b² + 4b4 6. a² – 10a + 24 + 6b – 9b² 7. 16 – x2 + 6xy – 9y2 Answer 1. (x2 –xy+y2 ) (x2 +xy+y2 ) 2. (x2 +2x+2) (x2 –2x+2) 3. (x2–2x+5) (x2+2x+5) 4. (a2 –ab+4b2 ) (a2 +ab+4b2 ) 5. (3a2 +2b2 –6ab) (3a2 – 6ab+b2 ) (3a2 + 6ab + 2b2 ) 6. (a+3b–6) (a–3b–4) 7. (4–x+3y) (4+x–3y) (ii) a4 – 6a² – 7 – 8x – x² Solution: Here, a4 – 6a² – 7 – 8x – x² = a4 – 6a² + 9 –16 – 8x – x² = (a4 – 6a² + 9) – (16 + 8x + x²) = {(a²)² – 2.a².3 + (3)²} – {(4)² + 2.4.x + (x)²} = (a² – 3)² – (4 + x)² = {(a² – 3) + (4 + x)} {(a² – 3) – (4 + x)} = (a² – 3 + 4 + x) (a² – 3 – 4 – x) = (a² + 1 + x) (a² – 7 – x) = (a2 + x+ 1) (a2 –x–7)
106 Oasis School Mathematics-8 11.1 Highest Common Factor (H.C.F.) Let's take any two algebraic terms a² and a³. Possible factors of a² are a and a². Possible factors of a³ are a, a² and a³. Common factors of a² and a³ are 'a' and 'a2 ' Among them, the highest common factor is a². Hence, H.C.F. is the greatest number or expression which divides the given number or expression exactly without remainder. • H.C.F. is also written an G.C.F. or G.C.D. Where, G.C.F. = Greatest common factor G.C.D. = Greatest common divisor H.C.F. by factorisation method To find the H.C.F. of an algebraic expression by factorisation method, apply the following steps. • Factorise the given algebraic expressions. • Find the common factors from all expressions. • Common factor or the product of all common factors. Find the H.C.F. of x2 + 2x and x2 + x – 2. Solution: 1st expression = x2 + 2x = x(x + 2) 2nd expression = x2 + x – 2 = x2 + 2x – x – 2 = x(x + 2) – 1(x + 2) = (x + 2) (x – 1) ∴ H.C.F. = (x+2) Unit 11 H.C.F. and L.C.M.
Oasis School Mathematics-8 107 Worked Out Examples Example 1 15x³y7 and 18x7 y³ Solution: First expression = 15x³y7 = 3 × 5x³y7 Second expression = 18x7 y³ = 2 × 3 × 3x7 y³ ∴ H.C.F. = 3x³y³ Example 2 Find the H.C.F. of x2 – 4 and x2 – x – 6 Solution: First expression = (x2 – 4) = (x)2 – 22 = (x + 2) (x – 2) Second expression = x2 – x – 6 = x2 – (3 – 2) x – 6 = x² – 3x + 2x – 6 = x(x – 3) + 2(x – 3) = (x – 3) (x + 2) ∴ H.C.F. = (x + 2) Example 3 Find the H.C.F. of (x + y)2 , x2 – y2 , 2x2 + xy – y2 . Solution: First expression = (x + y)² = (x + y) (x + y) Second expression = x² – y² = (x + y) (x – y) Third expression = 2x² + xy – y² = 2x² + (2 – 1)xy – y² = 2x² + 2xy – xy – y² = 2x(x + y) –y (x + y) = (x + y) (2x – y) ∴ H.C.F. = (x + y)
108 Oasis School Mathematics-8 Example 4 Find the H.C.F. of a2 + 2a – 3, a2 – 3a + 2 and a3 – 1. Solution: First expression = a2 + 2a – 3 = a² + 3a – a – 3 = a(a + 3) – 1(a + 3) = (a + 3) (a – 1) Second expression = a² – 3a + 2 = a² – 2a – a + 2 = a(a – 2) – 1(a – 2) = (a – 2) (a – 1) Third expression = a³ – 1 = (a)³ – (1)³ = (a – 1) (a² + a + 1) ∴ H.C.F = (a – 1) Exercise 11.1 Find the H.C.F. of the following by factorisation: 1. (a) a³b², a³b (b) x5 y³, x³y4 (c) 8x²y, 4xy² (d) 5x3 y2 z, 15x2 yz3 (e) 16a³b4 c, 24a²b³c² (f) 8a²b4 c2 , 12a³bc5 and 20a³b6 (g) 21m3 n2 p, 15m2 n3 p2 and 12m4 n2 p4 2. (a) x²(x + y) and xy (x + y) (b) (x – 1) (x + 2) and (x – 1) (x – 2) (c) 3ab (a + b) (a – b), 6ab (a + b) (a + b) (d) a² + ab, a²c + abc (e) x³y² +x²y³, x²y+xy² 3. (a) x² – y², (x – y)² (b) a³–b³, a²–b² (c) a³ + b³, (a + b)³ (d) 2x² – 8, (x – 2)² 4. (a) (a² – b²) and (a² – 2ab + b²) (b) (x³ – y³) and (x² + xy + y²) (c) (8x³+27y³) and (4x² – 9y²) (d) (a + b)³, a² + 2ab + b², a² – b² (e) a³ – 8b³, a² – 4b², (a – 2b)² 5. (a) x² – 9 and x² – 5x + 6 (b) x² + x – 20 and x² – 7x + 12 (c) a² – a – 2 and a² + 2a + 1 (d) x² – 36, x² – 12x + 36 (e) x³ – 25x, x² – 6x + 5 (f) 2x² – 18, x² – 4x + 3, x² – 6x + 9 (g) x³ – xy², x² + 3xy + 2y², x² + 2xy + y² (h) 2(a² – 9), 4(a – 3)² and 6(a² – 6a + 9) (i) x² – 3x + 2, x² – 5x + 6 and x² – 6x + 8
Oasis School Mathematics-8 109 Answer 1. (a) a3 b (b) x3 y3 (c) 4xy (d) 5x2 yz (e) 8a2 b3 c (f) 4a2 b (g) 3m2 n2 p 2. (a) x(x + y) (b) (x – 1) (c) 3ab(a + b) (d) a(a + b) (e) xy (x + y) 3. (a) (x–y) (b) (a–b) (c) (a+b) (d) (x–2) 4. (a) (a–b) (b) x2 + xy + y2 (c) (2x + 3y) (d) (a + b) (e) (a – 2b) 5. (a) (x–3) (b) (x – 4) (c) (a+1) (d) (x–6) (e) (x–5) (f) (x–3) (g) (x+y) (h) 2(a–3) (i) (x–2) 11.2 Lowest Common Multiple (L.C.M.) Let's consider any two algebraic terms a² and a³. Multiples of a² are a², a³, a4 , a5 .............Multiples of a³ are a³, a4 , a5 Here, the common multiples are a³, a4 , a5 ................... Among these, lowest common multiple is a³. ∴ L.C.M. = a³. L.C.M. by factorisation method To find the L.C.M. of an algebraic expression, apply the following steps. ♦ Factorise the given algebraic expressions. ♦ Find the common factors. ♦ Multiply the common factors with all the remaining factors. ∴ The product is L.C.M. Let's be clear with the help of the given example. Example: Find the L.C.M. of x² – 2x and x² – 3x + 2. Here, First expression = x² – 2x = x(x – 2) Second expression = x² –3x + 2 = x² – (2 + 1) x + 2 = x² – 2x – x + 2 = x(x – 2) – 1 (x – 2) = (x – 2) (x – 1) Here, common factor = (x – 2) Remaining factors are x and (x – 1) ∴ L.C.M. = (x – 2) × x (x – 1) = x (x – 1) (x – 2)
110 Oasis School Mathematics-8 L.C.M. of 3 expressions = common from all expressions × common from two expressions × remaining factors from all expressions. Worked Out Examples Example 1 Find the L.C.M. of (a + b) (a – b) and (a + b)². Solution: First expression = (a + b) (a – b) Second expression = (a + b)² = (a + b) (a + b) ∴ L.C.M. = Common factors × Remaining factors = (a + b) × (a – b) × (a + b) = (a + b)² (a – b) Example 2 Find the L.C.M. of x² – 4, x³ + 8 Solution: First expression = x² – 4 = (x)² – (2)² = (x + 2) (x – 2) Second expression = x² + 8 = (x)² + (2)³ = (x + 2) (x² – 2x + 4) ∴ L.C.M. = (x + 2) (x – 2) (x² – 2x + 4) Common Remaining from first expression Remaining from second expression Example 3 Find the L.C.M. of x³ + 1, x² + 3x + 2 and (x + 1)². Solution: Fist expression = x³ + 1 = (x)³ + 1³ = (x + 1) (x² – x + 1) Second expression = x² + 3x + 2 L.C.M. = (x+1) (x2 –x+1) Common from all expression Remaing in 2nd expression Remaing in 1st expression Remaing in 3rd expression (x+2) (x+1)
Oasis School Mathematics-8 111 = x² + (2 + 1) x + 2 = x² + 2x + x + 2 = x(x + 2) + 1(x + 2) = (x + 2) (x + 1) Third expression = (x + 1)² = (x + 1) (x + 1) ∴ L.C.M = (x + 1) (x² – x + 1) (x + 2) (x + 1) = (x + 1)² (x + 2) (x² – x + 1) Example 4 Find the L.C.M. of : x³ + 7x² + 12x, 2x³ + 13x² + 20x, x³ + 64 Solution: First expression = x³ + 7x² + 12x = x(x² + 7x + 12) = x [x² + (3 + 4) x + 12] = x[x² + 3x + 4x + 12] = x[x(x + 3) + 4 (x + 3)] = x(x + 3) (x + 4) Second expression = 2x³ + 13x² + 20x = x(2x² + 13x + 20] = x(2x² + (8 + 5) x + 20] = x[2x² + 8x + 5x + 20] = x[2x (x + 4) + 5(x + 4)] = x(x + 4) (2x + 5) Third expression = x³ + 64 = x³ + 4³ = (x + 4) (x² – 4x + 16) ∴ L.C.M. = (x + 4). x (x + 3) (2x + 5) (x² – 4x + 16) = x (x+3) (x+4) (2x+5) (x2 –4x+16) Common in all = (x + 4) Common in two = x Remaining = (x + 3) (2x + 5) and (x² –4x + 16) ∴ L.C.M. = (x + 4)x (x + 3) (2x+5) (x² – 4x + 16)
112 Oasis School Mathematics-8 Exercise 11.2 1. Find the L.C.M. of: (a) a²b³c and a4 b5 c4 (b) x2 y3 z and x3 y2 z4 (c) 16a²bc, 8a³b²c³ and 4ab4 c² (d) 8a5 b³c², 4a4 b5 c³ and 12a²b³c³ (e) 6x5 y6 z4 , 3x7 y²z and 9x10y4 z² 2. Find the L.C.M. of: (a) (x–y) (x–y) and (x+y) (x+y) (x–y) (b) (x + 3) (x + 4) and (x + 3) (x+1) (c) (2x + 1) (2x – 1) and (x – 1) (2x – 1) (d) (a + 2b) (a – b) (a – b) (2a – b) and (1 + b) (a – b) (e) (x + 3) (x + 1), (x + 1) (x + 2) and (x + 2) (x + 3) 3. Find the L.C.M. of: (a) (x – y)² and (x² – y²) (b) (a³ – b³) and (a – b)² (c) x² + xy and xy + y² (d) x² – y² and x³ – y³ (e) a² – 4b², (a + 2b)² and a² + 2ab 4. Find the L.C.M. of: (a) x² + 6x + 9 and x² + 5x + 6 (b) 2x² + 7x + 6 and x² + 7x + 10 (c) x² + 2xy and x³ – 4xy² (d) x² – 9 and x³ – 7x² + 12x (e) a³ + 8, a³ + 9a² + 14a 5. Find the L.C.M. of: (a) a² + 5a +6, a² + 9a + 20 and a² + 7a + 12 (b) 2a² – 8, a³ + 8 and a² – 2a – 8 (c) a4 – 8a, 2a² – 5a + 2, and 3a² – 8a + 4 (d) 2x² – 18, x² – 2x – 3 and x² + x – 12 Answer 1. (a) a4 b5 c4 (b) x3 y3 z4 (c) 16a3 b4 c3 (d) 24a5 b5 c3 (e) 18x10y6 z4 2. (a) (x+y)2 (x-y)2 (b) (x+1) (x+3) (x+4) (c) (2x-1) (2x+1) (x-1) (d) (a+2b) (2a-b) (a-b)2 (1+b) (e) (x+3) (x+2) (x+1) 3. (a) (x-y)2 (x+y) (b) (a-b)2 (a2 +ab+b2 ) (c) xy (x+y) (d) (x+y) (x-y) (x2 +xy+y2 ) (e) a(a+2b)2 (a-2b) 4. (a) (x+3)2 (x+2) (b) (x+2) (x+5) (2x+3) (c) x(x+2y) (x-2y) (d) x(x+3) (x-3) (x-4) (e) a (a+2) (a+7) (a2 –2a+4) 5. (a) (a+2) (a+3) (a+4) (a+5) (b) 2(a+2) (a-2) (a2 –2a+4) (a–4) (c) a(a-2) (2a-1) (3a-2) (a2 +2a+4) (d) 2(x-3) (x+3) (x+1) (x+4)
Oasis School Mathematics-8 113 12.1 Rational Expressions - Review In a fraction, if the numerator and the denominator consist of algebraic expressions, then the fraction is called the algebraic fraction (or rational algebraic expression). For examples : 3x 4y , 3x + 4y 3x – 4y etc. are algebraic fractions or rational algebraic expression. Reduction of rational expression into its lowest term To reduce the rational expression into its lowest term, apply the following steps. Worked Out Examples Example 1 For what value of x is the rational expression 2x – 1 x – 2 undefined? Solution: 2x – 1 x – 2 is undefined, if its denominator is 0. If the denominator of the rational expression is zero, it has no meaning. i.e. in a b if b = 0 it has no meaning. Steps • Resolve the numerator and denominator in their factors. • Cancel the common factors. Note: +x +y = x y , –x –y = x y , +x –y = – x y , – x y = – x y Unit 12 Rational Expressions You know! • Numerator and denominator of fraction • Algebraic expression • Rational expression
114 Oasis School Mathematics-8 i.e. x – 2 = 0 or, x = 2. Example 2 Reduce –10x³y4 20x²y5 to its lowest terms. Solution: Here, –10x³y4 20x²y5 = –10 × x3–2 20 × y5–4 = –x 2y [∵ am an = am–n] Example 3 Reduce x² + x – 2 x² – 4 to its lowest terms. Solution: x² + x – 2 x² – 4 = x² + (2 – 1) x – 2 (x + 2) (x – 2) = x² + 2x – x – 2 x² – 2² = x(x + 2) – 1(x + 2) (x + 2) (x – 2) = (x + 2) (x – 1) (x + 2) (x – 2) = x – 1 x – 2 Exercise 12.2 1. For what value of x are the following rational expressions undefined ? (a) 1 x – 1 (b) 2 x + 5 (c) x + 5 x – 4 (d) x + 4 3x – 2 (e) 5x+2 5x – 2 2. Reduce the following algebraic fractions to the lowest term: (a) a4 b²c a²bc³ (b) 15p³q4 – 10pq² (c) – 6x14y9 24x10y5 (d) 30x4 y5 z6 45x7 yz9 (e) 15a7 b²c7 5a9 bc4 3. Reduce to the lowest terms: (a) 2x + 4 x² – 4 (b) 2x – 2y 2(x – y)² (c) a² + ab a² – b2 (d) 2m² + 4mn m² – 4n² (e) 2x² + 8x x² –16
Oasis School Mathematics-8 115 4. Simplify: (a) x² – 9 x² – 6x + 9 (b) a³ – 8 a² – 4 (c) – 3 – x x² – x – 12 (d) x² + 5x + 6 x² + 4x + 4 (e) a² – 5a + 6 a² – 3a + 2 (f) a³ + b³ (a – b)² + 4ab 5. (a) Factorise the algebraic expression x3 – y3 – 3xy ( – y). (b) Factorise : x3 – y3 (c) What is the result, if (a) is divided by (b). 6. (a) Using the formula a2 – b2 = (a + b) (a–b), factorise 4 – (a+b)2 and (a + 2)2 – b2 . (b) Simplify: 4 – (a + b)2 (a+2)2 – b2 Answer 1. (a) 1 (b) -5 (c) 4 (d) 2/3 (e) 2/5 2. (a) a2 b c2 (b) -3p2 q2 2 (c) -x4 y4 4 (d) 2y4 3x3 z3 (e) 3bc3 a2 3. (a) 2 x-2 (b) 1 x-y (c) a a-b (d) 2m m–2n (e) 2x x-4 4. (a) x+3 x-3 (b) a2 +2a+4 a+2 (c) – 1 (x-4) (d) x+3 x+2 (e) a-3 a-1 (f) a2 –ab+b2 a+b (g) x2 –2xy+y2 x2 +xy+y2 (h) 2–a–b a+2–b (i) a+2 a(2a+1) 5. (a) (x–y)3 (b) (2–y) (x2 + xy + y2 ) (c) (x–y)2 x2 + xy + y2 6. (a) (a + a + b) (b) (2 – a – b) (c) 2 – a – b a + 2 – b 12.2 Multiplication and Division of Rational Expressions Multiplication of Rational Expressions While multiplying the rational expressions, apply the following steps. For example : Multiply : 3a²b 4ab × 8ac² 6bc Solution: 3a²b 4ab × 8ac² 6bc Steps • Multiply numerator with numerator and denominator with denominator. • Reduce it into its lowest term.
116 Oasis School Mathematics-8 = 3 × 8a²b × ac² 4 × 6 × ab × bc = a³bc² ab²c = a²c b Division of Rational Expressions Division of rational expression means multiplication of first rational expression with the reciprocal of second. For example, Simplify : 4x² – 81y² 1 – 4a² ÷ 2x – 9y a – 2a² Solution: 4x² – 81y² 1 – 4a² ÷ 2x – 9y a – 2a² = (2x)² – (9y)² (1)² – (2a)² × a – 2a2 2x – 9y = (2x + 9y) (2x – 9y) (1 – 2a) (1 + 2a) × a(1 – 2a) (2x – 9y) = a(2x + 9y) (1 + 2a) Worked Out Examples Example 1 Multiply: 3a²b4 6a²b³ by 8a²b² 20a²b Solution: Here, 3a²b4 6a²b³ by 8a²b² 20a²b = 3a2 b4 × 8a2 b² 6a²b³ × 20a²b = 3 × 8 × a2 × a2 × b4 × b² 6 × 20 × a² × a² × b³ ×b = a4 × b6 5 × a4 × b4 = 1 5 × a4–4 × b6–4 = 1 5 × b² = b2 5 [ ∵ a0 = 1] Alternative method 3a2 b4 6a²b³ × 8a2 b2 20a2 b = 3×8×a2 ×a2 ×b2 ×b2 6×4×5a2 ×a2 ×b3 ×b = b6 5b4 = b2 5 2 3
Oasis School Mathematics-8 117 Example 2 Simplify : a² + 4a – 12 a² – 5a + 6 ÷ a² – 2a – 3 a² + 2a + 1 Solution: a² + 4a – 12 a² – 5a + 6 × a² + 2a + 1 a² – 2a – 3 = a² + (6 – 2) a – 12 a² – (3 + 2) a + b × a² + 2.a.1 + (1)² a² – (3 – 1) a – 3 = a² + 6a – 2a – 12 a² – 3a – 2a + 6 × (a + 1)² a² – 3a + a – 3 = a(a + 6) –2(a + 6) a(a – 3) – 2 (a – 3) × (a + 1)² a(a – 3) + 1 (a – 3) = (a + 6) (a – 2) (a – 3) (a – 2) × (a + 1) (a + 1) (a – 3) = a + 6 a – 3 × a + 1 a – 3 = (a + 6) (a + 1) (a – 3)² Example 3 Simplify : y a² – b² × (a + b)² a – b ÷ ax + bx (a – b)² Solution: y a² – b² × (a + b)² a – b × (a – b)2 ax+bx = y a² – b² × (a + b)² a – b × (a – b)2 x(a+b) = y (a + b) (a – b) × (a + b) (a + b) (a – b) × (a –b) (a – b) x(a + b) = y x Exercise 12.2 1. Simplify: (a) a³b² a²c² × a³c b (b) 8x5 y³ 9a4 b² × 15a³b 5x4 y³ (c) x² y × y² xz × z² zx (d) 3ab³ 4ac² × abc c³d³ × 4c4 d4 3a²b (e) 18x³y4 15a5 b³ × 25x²yz 6a²bc² × 3a6 b4 c 5x4 y5
118 Oasis School Mathematics-8 2. Simplify: (a) xy ab ÷ x²y³ a²b³ (b) 6ab² 21b³c ÷ 18a³b 63a²c² (c) x²a² b4 y² ÷ x²y³ a³b² (d) 8x5 y² 9a4 b² ÷ 6x4 y² 15a³b (e) 7x²y³ 8a5 b4 × 12ab 21xy ÷ 3a²b² 4x4 y5 (f) x 2y × 3z 4c ÷ 9x 8ac 3. Simplify: (a) a + b (a – b)² × a² – b² (a + b)² (b) x x – y × x² – y² y (c) a² + 5a + 6 a² – 4 × a² + a – 6 a³ – 9a (d) x² + 5x + 6 x² – 1 × x² – 2x – 3 x² – 9 (e) 4x² – 9y² a² – b² × a²b + ab² 4x – 6y (f) x² + 3x + 2 x² + 4x + 3 × x + 3 (x – 2)² × x² – 4 x + 2 (g) a² + ab + b² ab × a + b (a – b) ÷ a³ – b³ a² – b² 4. Simplify: (a) x² + 3x + 2 x² + 4x + 3 ÷ x² – 4 x² – 9 (b) 4x² – 100 x² + 2x – 15 ÷ 4x – 20 3 – x (c) 2x² – 5x + 2 3x² – 5x – 2 ÷ 4x – 2 3x² + x (d) a² + 8a + 15 ab – a + 5b – 5 ÷ 3b + 2a + ab + 6 ab – a + 2b – 2 5. (a) Factorise x2 – x – 20 and x2 – 25. (b) Factorise x2 + 3x + 2 and x2 + 6x + 8 (c) Find the product of: x2 – x – 20 x2 – 25 × x2 + 3x + 2 x2 + 6x + 8 (d) Divide the result by x + 1 x2 + 5x 6. (a) Factorise a2 – 8a – 9 and a2 – 17a + 72 (b) Factorise a2 – 25 and a2 – 1. Answer 1. (a) a4 b c (b) 8x 3ab (c) y (d) b3 d a (e) 3xz ac 2. (a) ab2 xy2 (b) c b2 (c) a5 b2 y5 (d) 20x 9ab (e) 2x5 y7 3a6 b5 (f) za 3y 3. (a) 1 a–b (b) x(x+y) y (c) a+3 a(a-3) (d) x+2 x–1 (e) ab(2x+3y) 2(a-b) (f) x+2 x–2 (g) (a+b)2 ab(a-b) 4. (a) x-3 x-2 (b) -1 (c) x(x-2) x-1 (d) a+2 b+2 5. (a) (x – 5) ( x + 4), (x + 5) (x – 5) (b) (x + 2) and (x + 2) (x + 4) (c) x + 1 x + 5 (d) x 6. (a) (a – 9) ( a + 1) and (a – 9) ( a – 8) (b) (a + 5) (a – 5) and (a + 1) (a – 1) (c) (a+5) (a–5) (a + 1) (a–1) (d) a – 5 a – 8
Oasis School Mathematics-8 119 12.3 Addition and Subtraction of Rational Expressions (or Algebraic Fractions): (Review) Like and Unlike Fractions: If the fractions have the same denominator, then they are called like fractions and if the denominators of the fractions are not same, then they are called unlike fractions. For example : a b and c b are like fractions and d e and e ƒ are unlike fractions. To reduce fractions to their lowest common denominator. • If possible factorise the numerator and denominator. • Canceloutthefactors thatarecommontobothnumeratoranddenominator. • Find the L.C.M. of the given denominators. • Simplify the given terms. Rules: For example: Express with the lowest common denominator of a xz, b xy and c yz. Solution: Here, L.C.M. of xz, xy and yz = xyz. Now, a xz = a xz × y y = ay xyz , b xy = b xy × z z = bz xyz and c yz = c yz × x x = cx xyz When the fractions have same denominator. Worked Out Examples Example 1 3a x + 2b x + 5c x Solution: 3a x + 2b x + 5c x = 3a + 2b + 5c x Steps • Add or subtract the numerators. • Write the result of numerators over the common denominator of the like fractions. • Reduce the resulting fraction to the lowest term.
120 Oasis School Mathematics-8 Example 2 x + 3 x – 3 – x – 3 x + 3 Solution: x + 3 x – 3 – x – 3 x + 3 = (x + 3)² – (x – 3)² (x – 3) (x + 3) = (x² + 6x + 9) – (x² – 6x+9) x² – 9 = x² + 6x + 9 – x² + 6x – 9 x² – 9 = 12x x² – 9 Example 3 a a – b – b a + b – 2ab a² – b² Solution: a a – b – b a + b – 2ab a² – b² = a a – b – b a + b – 2ab a² – b² = a(a + b) – b(a – b) (a – b) (a + b) – 2ab a² – b² = a² + ab – ab + b² a² – b² – 2ab a² – b² = a² + b² a² – b² – 2ab a² – b² = a² + b² – 2ab a² – b² = (a – b)² (a + b) (a – b) = = a – b a + b Example 4 x + y x – y – x – y x + y – 4xy x² + y² Solution: x + y x – y – x – y x + y – 4xy x² + y² = (x + y)² – (x – y)² (x – y) (x + y) – 4xy x² + y² = (x² + 2xy + y²) – (x² –2xy + y²) x² – y² – 4xy x² + y² = x² + 2xy + y² – x²+2xy – y² x² – y² – 4xy x² + y² = 4xy x² – y² – 4xy x² + y² = 4xy(x² + y²) – 4xy (x² – y²) (x² – y²) (x² + y²) (a – b) (a – b) (a +b) (a – b)
Oasis School Mathematics-8 121 = 4x3 y + 4xy3 – 4x3 y + 4xy³ (x²)² – (y²)² = 8xy3 x4 – y4 Example 5 a – 1 a² – 3a + 2 + a – 2 a² – 5a + 6 + a – 5 a² – 8a + 15 Solution: a – 1 a² – 3a + 2 + a – 2 a² – 5a + 6 + a – 5 a² – 8a + 15 = a – 1 a² – (2 + 1) a + 2 + a – 2 a² – (3 + 2) a + 6 + a – 5 a² – (5 + 3) a + 15 = a – 1 a² – 2a – a + 2 + a – 2 a² – 3a – 2 a + 6 + a – 5 a² – 5a – 3a + 15 = a – 1 a(a – 2) – 1(a – 2) + a – 2 a(a – 3) – 2(a – 3) + a – 5 a(a – 5) – 3(a – 5) = a – 1 (a – 1) (a – 2) + a – 2 (a – 2) (a – 3) + a – 5 (a – 5) (a – 3) = 1 a – 2 + 1 a – 3 + 1 a – 3 = 1(a – 3) + 1 (a – 2) + 1(a – 2) (a – 2) (a – 3) = a – 3 + a – 2 + a – 2 (a – 2) (a – 3) = 3a – 7 (a – 2) (a – 3) Exercise 12.3 Simplify the following: 1. (a) 2x 5 + 3x 5 (b) 3a b + a b (c) a + b 2ab + a – b 2ab (d) 5x x – 1 – 5 x – 1 (e) x – 3y a + b – 3x – y a + b (f) a² a + 2 + 4a + 4 a + 2 (g) a² a – 4 – 16 a – 4 (h) x + 4 x² – 36 + x + 8 x² – 36 (i) 3a² a + 4 + 15a + 12 a + 4 (j) x x² + 5x + 6 + 2x + 3 x² + 5x + 6 + 3 x² + 5x + 6 (k) 2x x – y – 3x + 4y x – y + 2x + 5y x – y 2. (a) 1 x – 2 + 1 x + 2 (b) 1 2x+y + 1 2x – y (c) x + y x – y – x – y x + y (d) a a – 5 – a² a² – 25 (e) 1 a – b + 2b b² – a² (f) 1 x² – 1 + 1 1 – x (g) 1 (a – b) (c – a) – 1 (a – c) (b – c) (h) x y(x – y) – y x(x – y) (i) a²b a – b – ab² a + b
122 Oasis School Mathematics-8 3. (a) 1 a + 2 + 1 a – 2 + 4 4 – a² (b) 1 x + y + 1 x – y + 2y y² – x² (c) 1 a – 3 + 1 2(a + 3) + a a² – 9 (d) 1 1 – x + 1 1 + x + 2 1 – x² (e) 2 x + 1 + 2x x – 1 – x² + 3 x² – 1 (f) x + y x – y – x – y x + y + 2xy x² – y² (g) x – y x – x x + y – x2 y(x+y) 4. (a) 1 (a – 3) (a – 2) + 3 (a – 2) (4 – a) + 2 (a – 3) (a – 4) (b) 1 (1 –x) (x – 2) + 1 (x – 2) (x – 3) + 1 (x – 3) (x – 1) (c) x (x – y) (x – z) + y (y – z) (y – x) + z (z – x) (z – y) 5. (a) 1 a² + 8a + 15 + 1 a² + 4a + 3 + 1 a² + 6a + 5 (b) a – 2 a² – 5a + 6 + a – 4 a² – 6a + 8 + a – 3 a² – 6a + 9 6. (a) Factorise the expression x2 – 3x + 2 and x2 – 5x + 6 (b) Simplify: x – 1 x2 – 3x + 2 and x – 2 x2 – 5x + 6 (c) Simplify: x – 1 x2 – 3x + 2 – x – 2 x2 – 5x + 6 Answer 1. (a) x (b) 4a b (c) 1 b (d) 5 (e) -2(x+y) a+b (f) (a+2) (g) (a+4) (h) 2 x - 6 (i) 3(a+1) (j) 3 x+3 (k) x+y x - y 2. (a) 2x x - 4 (b) 4x 4x - y (c) 4xy x - y 2 2 2 2 2 (d) 5a a - 25 (e) 1 a+b (f) x 1 - x (g) -1 (a - b)(b - c) (h) x+y xy (i) ab(a + 2 2 2 b ) a - b 3. (a) 2 a+2 (b) 2 (x+y) (c) 5a+3 2(a - 9) (d) 4 1 - x (e) x+5 x+1 2 2 2 2 2 (f) 6xy x - y (g) - (x - xy+y ) xy 4. (a) 1 (a - 2)(a - 3)(a - 4) (b) x (x -1) 2 2 2 2 (x - 2)(x - 3) (c) 0 5. (a) 3 (a+1)(a+5) (b) 3a - 7 (a - 2)(a - 3) (c) 3x - 7 (x - 2)(x - 3) 6. (a) (x–1) (x–2) and (x–2) (x–3) (b) 1 x – 2 and 1 x – 3 (c) –1 (x–2)(x–3)
Oasis School Mathematics-8 123 Unit 13 Equation and Graph Review Observe the given equation of one variable and find out the solution. • 5x + 12 = 37 • 8x – 12 = 84 • –3x + 18 = –27 • –4x = 5x + 45 13.1 Simultaneous Equations Let us considerthe equations x + y = 8 and 2x–y = 7. These two equations contain two variables 'x' and 'y' and both are linear equations. These equations are simultaneous equations. Hence, a pair of linear equations in two variables are called a system of simultaneous linear equations or simply simultaneous equations. Method of solving simultaneous equations There are many methods of solving simultaneous equations. Here we will discuss only three methods. (i) Graphical method (ii) Elimination method (iii) Substitution method Graphical method of solving simultaneous equation The steps for solving of simultaneous equations graphically are as follows: • Draw the graph of the first equation. • Draw the graph of the second equation with the same axes and the same scale. • From the graph, read the point of intersection of the two straight lines drawn. • The required solutions are the x and y coordinates of the point which is the point of intersection of two simultaneous equations. You know! • Equation with one variable • Equation with two variable • Solution of the equation of one variable
124 Oasis School Mathematics-8 Worked Out Examples Example 1 Solve graphically: 2x – y = 5 and x – y = 1 Solution: Here, 2x – y = 5 or 2x – 5 = y ∴ y = 2x – 5 …(i) And x – y = 1 or x –1 = y y = x – 1 …(ii) x 0 1 4 y -5 -3 3 x 0 2 3 y -1 1 2 Now, plotting these points on the graph, we get two lines. From the graph, Point of intersection of equation (i) and (ii) is (4,3) ∴ x = 4, y =3. Example 2 Solve graphically: 2x + 3y = 13, y = 3. Solution: Given equations, 2x +3y = 13 ………….(i) or, 3y = 13 – 2x or, y = 13–2x 3 x 2 -1 5 y =13–2x 3 3 5 1 Again, y = 3………… (ii) x 1 2 3 y=3 3 3 3 From the graph, Point of intersection of (i) and (ii) is (2, 3). ∴ x = 2, y = 3 X' X Y' Y O (4,3) X' X Y' Y (2, 3) y = 3 2x+3y=13 O
Oasis School Mathematics-8 125 Exercise 13.1 1. Write down the solution of each pair of equations: X' X X' X Y' Y' Y Y (c) (d) O O 2. Complete the following table and plot them on the graph and find the solution. x y = 2x+1 x y = 4x–1 (a) x y = 2 x y = x+1 (b) x y = x–6 x y = 2 – x (c) 3. Solve the following equations graphically: (a) x + y = 6, x – y = 2 (b) 2x – y = 5, x + y = 4 (c) x + 2y = 1, x – y = 4 (d) x – 2y = 5, 2x –y =1 (e) 2x = 4 + y, x + 3 = –2y (f) x + 2y + 16 = 0, 2x + 4 – 3y = 0 (g) y = 3-5x 3 , y = –4 (h) x = 3, 2x = 26 – 5y (i) 3x +y = 2, 3x – 2y = 5 X' X Y' Y X' X Y' Y (a) (b) 4 3 3 2 2 1 1
126 Oasis School Mathematics-8 4. Convert the following verbal problems into the equations and solve them graphically. (a) Find the two numbers whose sum is 7 and difference is 1. (b) The sum of two numbers is 6 and their difference is 2. Find the numbers. (c) The difference of two numbers is 2. One number is one third of the other. Find the numbers. 5. (a) In the equation, x – 2y = 3, find any three points which satisfy the given equation. (b) Find any three points on the line x + y = 6. (c) Plot these points on the graph and find the solution. 6. (a) Length of a rectangularroom is 2m more than its breadth convert this statement into equation. (b) The perimeter of the room is 16m, convert this statement into equation. (c) Solve the equation of (a) and (b) to get the length and breadth of the room. Answer 1. Consult your teacher. 2. Consult your teacher. 3. (a) 4, 2 (b) 3, 1 (c) 3, -1 (d) -1, -3 (e) 1, -2 (f) -8, -4 (g) 3, -4 (h) 3, 4 (i) 1, -1 4. (a) 4,3 (b) 4, 2 (c) 1, 3) 5. (a), (b) Consult your teacher (c) 5, 1) 6. (a), (b) Consult your teacher (c) 5m, 3m 13.2 Additional Text Elimination Method In this method we make the coefficient of one of the variables equal and we add or subtract the given equations to eliminate that variable. We substitute the value of that variable in an equation to get the value of other variable. For example, solve the equations, x +y = 5 and x –y = 3 Solution: Here, x +y = 5 …………… (i) x–y = 3 …………… (ii) Adding equation (i) and (ii), x + y = 5 x – y = 3 2x = 8
Oasis School Mathematics-8 127 or, x = 8 2 = 4 Substituting the value of x in equation (i), x + y = 5 or, 4+y = 5 or, y = 5 – 4 y = 1 ∴ x = 4 y = 1. Substitution method In this method, we express one of the variables in terms of other and we substitute its value in other equation to get the values of the variable. For example solve the equations x +y = 7 and 2x –y = 5 Solution: Given equation, x +y = 7 ……………(i) 2x – y = 5 ……………(ii) From equation (i) y = 7 – x Substituting this value in equation (ii) we get, 2x –y = 5 or, 2x – (7–x) = 5 or, 2x – 7 + x = 5 or, 3x = 5 + 7 or, 3x = 12 or, x = 12 3 = 4. Again, substituting the value of x in (i) x + y = 7 or, 4 + y = 7 or, y = 7–4 = 3 ∴ x = 4 and y=3. Note: If the co-efficeints of a variable are same in both equations and their sign is opposite add the equations. If their sign is same, subtract the equations.
128 Oasis School Mathematics-8 Worked Out Examples Example 1 Solve the given equations by elimination method: 3x – 2y = 5 and 2x +y = 8 Solution: Given equation, 3x –2y = 5 ………………….. (i) 2x +y = 8 ……………………(ii) Multiplying equation (ii) by 2 and adding these equations, 3x –2y = 5 4x +2y = 16 Adding, 7x = 21 or, x = 21 7 or, x = 3. or, 2y = 4 or, y = 4 2 or, y = 2. ∴ x = 3 and y = 2 Example 2 Solve the given equations by substitution method: 2x +y =5 and 3x –2y = 4 Solution: Given equations, 2x +y = 5 ………………(i) 3x – 2y = 4 ……………..(ii) From equation(i), y = 5 – 2x Substituting this value in equation (ii) 3x –2y = 4 or, 3x – 2(5–2x) = 4 or, 3x – 10+4x = 4 or, 7x – 10 = 4 or, 7x = 4+10 or, 7x = 14 or, x = 14 7 or, x = 2 Substituting the value of x in (i) 3x – 2y = 5 or, 3×3–2y = 5 or, 9 –2y = 5 or, –2y = 5 – 9 or, –2y = –4 Cancel (–) sign from both sides. Substituting the value of x in (i) 2x +y = 5 2×2+y = 5 or, 4 + y = 5 or, y = 5 –4 or, y = 1 ∴ x = 2 and y = 1.
Oasis School Mathematics-8 129 Exercise 13.2 1. Solve the following simultaneous equations (by elimination method). (a) x + y = 3, x – y = 1 (b) 2x + y = 7 and 3x – y = 8 (c) 3x – 2y = 4, 3x + y = 7 (d) 2x + y = 8, 2x + 3y = 12 (e) 3x – y = 15, 3x + 2y = 24 (f) 2x + 3y = 15, x –2y = –3 (g) 2x + y = 1, 3x – 2y = 5 2. Solve the following simultaneous equations (by substitution method). (a) –y = –x + 4, 2x – 3y = 6 (b) x – y = 2, x + y = 6 (c) x + 2y = 7, 2x + 3y = 12 (d) 3x + 2y = 5, 3x – y = 2 (e) 2x – y = 3, 3x + 2y = 8 (f) 2x + y = 6, 2y = 2 + x (g) y = 2x, 4x + 3y = 20 Answer 1. (a) 2, 1 (b) 3,1 (c) 2,1 (d) 3, 2 (e) 6, 3 (f) 3,3 (g) 1, -1 2. (a) 6, 2 (b) 4, 2 (c) 3, 2 (d) 1,1 (e) 2, 1 (f) 2,2 (g) 2,4 13.3 Quadratic Equation Let's take an equation x² + 2x + 5 = 0. It is the second degree equation with one variable which is in the form of ax² + bx + c = 0. This type of equation is the quadratic equation where, a = 1, b = 2 and c = 5 Hence an algebraic equation having degree two is the quadratic equation. It is in the form of ax2 + bx + c = 0. where a, b, c are constants. Examples: x2 – 5x + 6 = 0 x2 – 2x = 0 x2 – 8 = 0, etc. All these are in the form of ax2 + bx + c = 0. In the first equation, a = 1, b = –5, c = 6
130 Oasis School Mathematics-8 In the second equation, a = 1, b = –2, c = 0 and in the third equation, a = 1, b = 0, c = –8 Solution of Quadratic Equations A quadratic equation can be solved by different methods. Here we are discussing the solution of quadratic equation by factorisation method. Worked Out Examples Example 1 Solve: (x+1) (x–3) = 0 Solution: Here, (x + 1) (x – 3) = 0 Either, (x + 1) = 0 or, x – 3 = 0 or, x = 3, or, x = –1 Hence, x = –1, 3. Example 2 Solve: x2 = 5x Solution: Here, x2 = 5x or, x2 – 5x = 0 or, x(x – 5) = 0 Either, x = 0 or, x – 5 = 0 x = 5 Hence, x = 0, 5 Example 3 Solve: x2 – 49 = 0 Solution: Here, x2 – 49 = 0 or (x)2 – (7)2 = 0 or, (x + 7)(x – 7) = 0 Either, x + 7 = 0 ⇒ x = –7 or, x – 7 = 0 ⇒ x = 7 Hence, x = –7, 7 Steps • Simplify the given equation and convert it in the form of ax2 +bx+c = 0 • Factorise the expression on the left hand side. • Write each factor is equal to zero to get the values of x.
Oasis School Mathematics-8 131 Example 4 Solve: 2x2 – 5x – 12 = 0 Solution: Here, 2x2 – 5x – 12 = 0 or, 2x2 – (8 – 3)x – 12 = 0 or, 2x2 – 8x +3 x – 12 = 0 or, 2x(x – 4) + 3 (x – 4) = 0 or, (x – 4) (2x + 3) = 0 Either, x–4 = 0, or, 2x + 3 = 0 or, x = 4, or , x = -3 2 Hence, x = 4, -3 2 Example 5 Solve: 3x – 7 2x – 5 = x + 1 x – 1 Solution: Here, 3x – 7 2x – 5 = x + 1 x – 1 or (3x – 7) (x – 1) = (x + 1) (2x – 5) or 3x2 – 3x – 7x + 7 = 2x2 – 5x + 2x – 5 or 3x2 – 10x + 7 = 2x2 – 3x – 5 or 3x2 – 2x2 – 10x + 3x + 7 + 5 = 0 or x2 – 7x + 12 = 0 or x2 – 4x – 3x + 12 = 0 or x(x – 4) – 3(x – 4) = 0 or (x – 4) (x – 3) = 0 Either, x – 4 = 0 or, x – 3 = 0 ∴ x = 4 x = 3 Hence, x = 4, 3. Example 6 If 5 times of a number is added to the square of the number, the result is 66. Find the numbers. Solution: Let the required number be x. Then by the given condition, x2 + 5x = 66 or, x2 + 5x – 66 = 0
132 Oasis School Mathematics-8 or, x2 + 11x – 6x – 66 = 0 or, x(x + 11) – 6(x + 11) = 0 or, (x + 11) (x – 6) = 0 Either, x + 11 = 0 or, x – 6 = 0 ∴ x = –11 ∴ x = 6 Hence, the required number = 6, –11. Exercise 13.3 Solve the following: 1. (a) (x–3) (x+2) = 0 (b) (x+6) (x–6) = 0 (c) (x+3) (x+4) = 0 (d) (x+1) (x–5) = 0 2. (a) x2 –2x=0 (b) 2x2 +5x=0 (c) 3x2 = 6x (d) x2 = 5x 3. (a) x2 – 25 = 0 (b) x2 – 81 = 0 (c) x2 – 16 =0 (d) 4x2 –81 = 0 4. (a) x2 + 5x + 6 = 0 (b) x2 + x – 20 = 0 (c) x2 – 3x – 10 = 0 (d) 2x2 – x– 6 = 0 (e) x2 –9x = 70 (f) x2 –10x = 39 (g) 3x2 = 2x + 8 (h) 2x2 = 3x – 1 5. (a) x + 1 x = 5 2 (b) x + 1 3x–7 = x – 1 2x–5 6. Solve the following problems: (a) If 6 is added to the square of a number, the result is 31. Find the number. (b) If 2 is subtracted from the half of the square of a number, the result is 30. Find the numbers. (c) If 5 times a number is added to its square, the result is 6. Find the numbers. (d) If 6 times of a number is added to the square of the number, the result is 55. Find the numbers. (e) One number is 5 less than another and their product is 36. Find the numbers. (f) If the sum of two numbers is 6 and the product is 8, find the two numbers.
Oasis School Mathematics-8 133 7. (a) What type of equation is 2x2 – 5x + 2 = 0? Give reason. (b) Factorise the expression 2x2 + 5x + 2. (c) What are the possible solution of the equation 2x2 – 5x + 2 = 0? 8. The sum of a number and 2 times its reciprocal is 9 2 . (a) Convert the above statement into equation. (b) Find the numbers. (c) Do both numbers satisfy the above equation? Answer 1. (a) 3, -2 (b) -6, 6 (c) -3, -4 (d) -1, 5 2. (a) 0, 2 (b) 0, -5/2 (c) 0, 2 (d) 0, 5 3. (a) 5, -5 (b) 9, -9 (c) 4, -4 (d) 9/2, -9/2 4. (a) -3, -2 (b) -5, 4 (c) 5, -2 (d) 2, -3/2 (e) -5, 14 (f) -3, 13 (g) 2, -4/3 (h) 1, 1/2 5. (a) 2, 1/2 (b) 3, 4 6. (a) 5, -5 (b) 8, -8 (c) -6, 1 (d) -11, 5 (e) 4, 9 and -4,-9 (f) 4, 2 7. (a) Consult your teacher, (b) (x+2) (2x+1), (c) –2, -1/2 8. (a) Consult your teacher, (b) 4, 1/2 (c) Yes Miscellaneous Exercise 1. (a) If 2m, 2n and 2p are multiplied, write the result in exponent form. (b) What is the value of (2m)n. (c) Simplify: 2m × 2n × 2p (2m)n [Ans: (a), (b) consult your teacher, (c) 2m+n+p–mn] 2. (a) If (xa )b = xab, then evaluate (xa – b )a + b , (xb + c )b – c and (xc – a)c + a. (b) Find the product of these three terms. [Ans: (a) consult your teacher (b) 1] 3. (a) If am × an = am+n, evaluate xa + 2 – xa + 1 (b) Similarly, evaluate: xa + 1 – xa . (c) Simplify : xa + 2 – xa + 1 xa + 1 – xa [Ans: (a), (b) consult you teacher, (b) (x + 1)] 4. (a) Which one of the following property of indices is not true? (i) (xm)n = xm+n (ii) ( x y ) m = xm ym (iii) x n m – x m n (b) Simplify: 8 27 3 ÷ ( 16 81 ) 1 4 [Ans: (a) consult your teacher, (b) 1]
134 Oasis School Mathematics-8 5. (a) Which one of the following relation is not true? (i) ( am an ) –2 = ( an am ) 2 (ii) ( a b ) m = bm an (iii) a n m = a n m (b) Simplify the relation: 27 216 3 . (c) Simplify: (am + n)n –m × (am – l )l –m × (al –n)n + l . [Ans: (a) consult your teacher, (b) 1 2 (c) 1 ] Factorization H.C.F and L.C.M 6. (a) Factorize the algebraic expressions a² – 25b² (b) Take common and factorize a4 + 125ab3 using the formula a³ + b³. (c) Factorise the expression: a² + 6ab + 5b² (d) Find the H.C.F. of above three expressions. [Ans: (a) (a + 5b) (a – 5b), (b) a (a + 5b) (a² – 5ab + 25b²), (c) (a + b) (a + 5b) (d) (a + 5b)] 7. (a) Convert the algebraic expression x² + 4xy + 4y² in the perfect square. What are its factors? (b) Use the formula a³ + b³ = (a + b) (a² – ab + b²) to factorize x³ + 8y³. (c) Using mid term factorization, factorize x2 + 3xy + 2y². (d) Find the common factors from all. What is it called? [Ans: (a) (x + 2y) (x + 2y), (b) (x + 2y) (x² – 2xy + 4y²), (c) (x + y) (x + 2y) (d) (x + 2y)] 8. (a) What do you mean by H.C.F? What is the H.C.F of two terms if there is no common term? (b) Factorize the expressions x³ – xy², x² + 3xy + 2y². (c) What is their L.C.M. and H.C.F? [Ans: (b) x (x + y) (x – y), (b) (x + 2y) (x + y), (c) H.C.F. = (x + y) L.C.M = x (x + y) (x – y)] 9. (a) What are the formula of a² – b², a³ + b³ and a³ – b³? (b) Using these formula factorize : 9a² – 16b². (c) Using these formula factorize: 27a³ + 64b³. (d) Take common from the expression 9a³– 12a²b _ 16ab² (e) What is the L.C.M of three expressions? [Ans: (b) (3a – 4b) (3a + 4b) (c) (3a + 4b) (9a² – 12ab + 16b²) (d) a(9a² – 12ab + 16b²) (e) a (3a + 4b) (3a – 4b) (9a² – 12ab + 16b²)
Oasis School Mathematics-8 135 Rational Expression 10. (a) Use the formula a2 – b2 = (a + b) (a–b), to factorise 4x2 – 81y2 and 1 – 4a2 (b) Find the value of : 4a2 – 81y2 1 – 4x2 (c) Divide this by 2a – 9b x – 2x2 [Ans: (a) (2x+2y) (2x–9y), (1+2x) (1-2x) (b) (2a+9b) (2a-9b) (1-2x) (1+2x) , (c) x(2a + 9b) (1 + 2x) ] 11. (a) Using mid term factorization, factorise: x2 + 8x + 15 (b) Factorise: xy – x + 5y – 5 (c) Find the value of: x2 + 8x + 15 xy–x+5y–5 (d) Factorise and simplify the expressions : xy+2x+xy+6 xy–x+2y–2 (e) Divide the result of (c) by the result of (d). [Ans: (a) (x+5) (x+3), (b) (x+5) (y-1) (c) x + 3 y – 1 , (d) (x+3) (y+2) (y-1) (x+2) , (e) y + 2 x + 2 ] 12. (a) Simplify the rational expression x – 1 x2 – 3x + 2 . (b) Simplify the expression x – 2 x2 – 5x + 6 . (c) Simplify: x – 1 x2 + 3x + 2 – x – 1 x2 – 3x + 6 . [Ans: (a) 1 x–2 , (b) 1 x – 3 , (c) -1 (x-2) (x-3) ] 13. (a) Simplify the rational expression x + y x – y + x – y x + y . (b) Add the result of (a), with 4xy x2 – y2 . (c) Simplify the result. [Ans: (a) 2x2 + 2y2 (x-y) (x+y) , (b) Consult your teacher, (c) 2x2 + 2y2 (x-y) (x+y) ] 14. (a) For what value of x, x + 3 x + 2 does not define? (b) Factorise the expression xy – 3x + 2y – 6. (c) Factorise the expression y2 – 8y + 15. (d) Simplify the result xy – 3y + 2y – 6 y2 – 8y + 15 . [Ans: (a) Consult your teacher, (b) (x+2) (y-3), (c) (y-3) (y-5), (d) x + 2 y–5 ]
136 Oasis School Mathematics-8 Objective Questions Choose the correct alternatives. 1. Which one the given relation is not true? 2 –2 –2 2 m 2m m n m 2(m+n) n 2n n m n a2 a a a (i) = (ii) = (iii) = a. aa a a a 2. The value of 1 – 8 3 27 is (i) 2 3 (ii) 3 2 (iii) 8 27 3. Which one of the following is not equal to (x + 3y)² – (x – 3y)²? (i) 12xy (ii) [(x + 3y) + (x – 3y)] [(x + 3y) – (x – 3y)] (iii) [( x + 3y) – (x – 3y)]² 4. If 4x² + mx + 36 is a perfect square, the possible value of m is (i) ±24 (ii) ±12 (iii) ± 48 5. The H.C.F. of the expression (a + 1)², a² – 1 and a² + 1 is (i) a – 1 (ii) (a + 1) (iii) (a + 1)² (a – 1) (a² – a + 1) 6. The simplest form of the expression a + b (a + b)² × a² – b² (a – b)² is (i) 1 a + b (b) a – b a + b (iii) 1 a – b 7. The simplest form of the expression a a – b – b a + b + 2ab a² – b² is (i) a + b a – b (ii) a – b a + b (iii) a² – b² a + b 8. If 6 is added to the square of a number, the result is 31; the number is (i) 5 (ii) –5 (iii) both (i) and (ii) 9. 2 times a square number is added to 2, the result is 100, the number is (i) 7 (ii) –7 (iii) both (i) and (ii) 10. If there is no common factors in two algebraic expressions then its H.C.F. is (i) 1 (ii) their product (iii) their sum.
Oasis School Mathematics-8 137 Assessment Test Paper Full mark – 32 Attempt all the questions. 1. (a) If xa and xb are multiplied, write in exponent form. (1) (b) What is the value of (xa ) b . (1) (c) Using above results evaluate (xa–b)a+b × (xb–c)b+c × (xc–a)c+a. (2) 2. (a) Factorize the algebraic expression x(x + 2) + 3(x + 2) + y(x + 2). (1) (b) Using the formula of a² – b², factorise the expression 2x² – 8. (1) (c) Find two numbers whose product is 6 and the sum is 5, using this condition factorise the expression x² – 5x + 6. (2) (d) What is the common factor in all three expressions? What is it called? (1) (e) Find the L.C.M. of these three expressions. (2) 3. (a) For what value of x, the rational expression x + 5 x – 4 does not define? (1) (b) Factorise the expression a² – 3a + 2, and a² – 4a + 3 and a² – 5a + 6. (3) (c) Using the factors of (b) find the value of: a² – 3a + 2 a² – 5a + 6 × a² – 4a + 3 a² – 5a + 6 . (2) (d) Simplify the expression 1 a² – 3a + 2 + 1 a² – 4a + 3 + 1 a² – 5a + 6 (3) 4. (a) What type of equation is 2x + y = 8? Write y in terms of x. (1) (b) Substitute the value of y obtained in (a) in the equation 2x + 3y = 12 and get the value of 'x' and 'y'. (3) (c) Find any three points on the line (a) and on the line (b). (2) (d) Plot these points on the graph sheet and find the point of intersection. (2) 5. (a) What type of equation in 2x² – x – 6 = 0? (2) (b) Factorise the L.H.S. of this equation (2) (c) Find the two values of x which satisfy the equation (a). (1)
138 Oasis School Mathematics-8 Estimated Teaching Hours 45 Contents: • Lines and Angles • Triangle, Quadrilateral and Polygons • Congruency and Similarity • Solid objects • Co-ordinates • Symmetry and Tessellation • Bearing and Scale drawing Expected learning outcomes: At the end of this unit, students will be able to develop the following competencies. • To identify the angle formed by cutting a pair of liens by a transversal. • To find the relation between the angles formed by cutting the pair of parallel lines by a transversal. • To verify the properties of triangle experimentally. • Verify the properties of plane figure lie rhombus trapezium, kite, etc. experimentally. • To construct the rectangle, square and parallelogram from the given criteria. • To identify the polygons and calculate the interior and exterior angle of regular polygons. • To make two triangles congruent using different axioms • To solve the problems related to congruency of triangles. • To identify the similar figures • To explain about triangular prism and pyramid • To prepare the net of cube, cuboid, tetrahedron, cone and cylinder. • To calculate the distance between two points • To identify regular and irregular tessellation • To reflect the triangle in graph • To translate the point, line and triangle vertically and horizontally. • To rotate a point, a line and a triangle through 90° about origin. • To use bearing and scale drawing on map Geometry
Oasis School Mathematics-8 139 Flash back • In the given figure, ∠AOC and ∠DOB are vertically opposite angles. Similarly, ∠AOD and ∠BOC are vertically opposite angles. Vertically opposite angles are equal. So, ∠AOC = ∠DOB and ∠AOD = ∠BOC. • In the given figure, ∠AOB and ∠BOC have common vertex 'O' and common arm OB. Both angles are on opposite side of common arm OB. So, ∠AOB and ∠BOC are adjacent angles. • In the given figure, ∠AOB and ∠BOC are linear pairs. Their sum is 180°. Here, ∠AOC + ∠BOC = 180°. • In the given figure, ∠AOB = 40° and ∠BOC = 50°. ∠AOB + ∠BOC = 90°. These two angles are complementary angles. Sum of two complementary angles is 90°. • In the given figure, ∠POR + ∠QOR = 180°. So, ∠POR and QOR are supplementary angles. Sum of two supplementary angles i s 180°. • In the given figure, ∠POQ, ∠QOR and ∠POR have common vertex 'O'. ∠POQ + ∠QOR + ∠POR = 360°. Angles around a point is 360°. O A C D B O A B C A B 40° 50° C O Q P R O A B C Unit 14 Lines and Angle • P Q R O You know! • Vertical opposite angles • Adjacent angles • Linear pair • Complementary angles • Supplementary angles
140 Oasis School Mathematics-8 Review 14.1 1. Find the value of x. 2. Find the value of x, y and z in the given figures. D B C O A 400 (a) x E D C O 600 400 B A F (b) z y x S P Q Y O (d) (3x+30)0 300 Q N M P 700 (c) (2x+10)0 3. Without measuring, find the value of x in the given figures. (a) A B O 800 x (b) 4x 3x 5x P Q R O (c) 1000 800 900 x O W Z X Y 4. (a) If (3x + 11)° and (2x + 4)° are complementary angles, find the value of x. (b) If (3x + 15)° and (2x + 5)° are supplementary angles, find the value of x. B B O D C 4x0 600 4x0 (c) D C B O A 8x 3x x (d) (a) 750 x A B C D (b) A O B D C 800 50 x 0 • Vertically opposite angles are equal. • Sum of adjacent angles forming a linear pair is always 180°. • Sum of complementary angles is always 90°. • Sum of supplementary angles is always 180°. • Angles around a point is 360°. Remember !
Oasis School Mathematics-8 141 Answer 1. (a) 1050 (b) 500 (c) 150 (d) 150 2. (a) 400 (b) x = 400 , y = 600 , z = 800 (c) x = 300 (d) x = 400 3. (a) 2800 (b) 300 (c) 900 4. (a) 150 (b) 320 14.2 Angles made by a Transversal with the Lines In the given figures, PQ cuts two or more than two lines. PQ is the transversal. A straight line, which cuts two or more straight lines (parallel or non–parallel) at different points is called a transversal. In each of the above figures, PQ is a transversal line. Angles formed by two lines and their transversal line In the given figure, AB and CD are any two straight lines and transversal PQ cutsAB and CD at points Rand S respectively. It makes four angles with each of the lines AB and CD. These angles are marked a, b, c, d, e, f, g, h respectively. In the figure, angles a, b, h, g are exterior angles whereas d, c, e, f are interior angles. Alternate angles: In all of the above figures, a pair of black marked angles are interior. Let's find nature of black marked angles. • both lie on the alternate sides of transversal. • both angles are non adjacent angles Such pair of angles are alternate angles. When a transversal cuts a pair of straight lines, two non–adjacent interior angles that lie on the alternate side of transversal are called alternate angles. P P P Q Q Q P B D S R Q C A Fig. (i) e a b c d f g h
142 Oasis School Mathematics-8 Again, If a transversal cuts a pair of parallel lines then alternate angles so formed are equal. In the given figure, AB//CD, so a = d and b = c. Conversely, if a transversal cuts two straight lines such that a pair of alternate angles are equal, then the two lines are parallel. In the given figure if x = y, then AB//CD are parallel. Experimental Verification: If a transversal cuts two parallel lines, then the alternate angles so formed are equal. Draw three pairs of parallel lines AB and CD. Draw a transversal EF which cuts AB and CD at G and H respectively. Now, ∠AGH and ∠GHD are a pair of alternate angles. Similarly, ∠BGH and ∠ GHC are also a pair of alternate angles. A B D Fig. (ii) C E F G H E G H B D Fig. (i) F C A To verify: ∠AGH = ∠GHD, ∠BGH = ∠GHC Observations: Measure each pair of alternate angles and tabulate the values as below. Figure ∠AGH ∠GHD ∠BGH ∠GHC Remarks i. ..................0 ..................0 ..................0 ..................0 ∠AGH = ∠GHD, ∠BGH = ∠GHC ii. ..................0 ..................0 ..................0 ..................0 ∠AGH = ∠GHD, ∠BGH = ∠GHC Conclusion: Hence, if a transversal cuts a pair of parallel lines, alternate angles so formed are equal. y x B D A C E B M N c d a b D F C A
Oasis School Mathematics-8 143 Worked Out Examples Example 1 From the adjoining figure, calculate the values of x, y and z. Solution: x = 500 [Vertically opposite angles] Here, y = x = 50° [Alternate angles] Lastly, y + z = 180° [Linear pair] or 50° + z = 180° ∴ z = 180° – 50° = 130°, Hence, x = 50°, y = 50° and z = 130° Example 2 In the given figure, AB||CD, ∠ABE=30°, ∠CDE = 40°, find the value of x. Solution: Here, AB || CD, ∠ABE = 30°, ∠CDE = 40° ∠BED = x° Through E draw PQ || AB i.e. PQ || AB || CD Here, ∠BEQ = ∠ABE = 30° [Alternate angles] Again, ∠DEQ = ∠CDE = 40° [Alternate angles] Hence, ∠BED = x° = ∠BEQ + ∠DEQ = 30° + 40°= 70° [Whole parts axiom] Example 3 In the given figure, find the two pairs of parallel lines. Solution: Here, ∠ABC = ∠BCD = 300 [Given] then AB//CD. [Being alternate angles equal] Again, ∠BCD = ∠CDE = 300 [Given] Then CB //ED. [Being alternate angles equal] Hence, AB//CD and CB//ED are two pairs of parallel lines. 500 y x z 300 400 B Q C D E P A x A B C D E 300 300 300
144 Oasis School Mathematics-8 Exercise 14.2 1. From the given figure, (i) name four interior angles and (ii) name four exterior angles. 2. Copy the following figures in your copy and shade the alternate angles of the given angle. (a) (b) (c) (d) 3. Identify in which of the following figures, alternate angles are equal. (a) (b) (c) 4. Find the value of unknown angles in the given figure. 5. Find a pair of parallel lines in the given figures and also justify your answer. 500 500 P Q S M N P O A C D B R 560 560 500 500 (a) (b) (c) a b d f e h g c (a) (e) (b) (f) (c) (g) (d) 580 x0 y0 x0 700 1100 y 1200 z y x x 300 x 250 300 400 x y Answer 1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher. 4. (a) 700 (b) 1100 (c) 300 (d) x = 1200 , y = 1200 , z = 1200 (e) x = 580 , y = 1220 (f) 550 (g) x = 400 , y = 400 5. Consult your teacher.
Oasis School Mathematics-8 145 14.3 Corresponding angles In all of the above figures, let's find the nature of black marked angles. • each pair are non adjacent angles • one is interior and another is exterior • both lie on the same side of transversal These pair of angles are corresponding angles. Again, if a transversal cuts two parallel lines, corresponding angles so formed are equal. In the given figure, a and e are corresponding angles. b and f are corresponding angles. c and h are corresponding angles. d and g are corresponding angles. Since AB||CD, a = e, b = f, c = h and d = g. In the given figure, if x = y then AB||CD. In the above figure, x and y are a pair of corresponding angles and AB‖CD. So, they are equal. Experimental Verification If a transversal intersects two parallel lines, corresponding angles so formed are equal. Draw two pairs of parallel lines AB and CD. Also, draw a transversal EF which meets AB and CD at G and H respectively. Now, ∠AGE and ∠GHC are a pair of corresponding angles. Similarly, ∠AGH and ∠CHF, ∠EGB and ∠GHD, ∠BGH and ∠DHF are other pairs of corresponding angles. b a d e f h g c x A B C E D y
146 Oasis School Mathematics-8 E G A B D H C F Fig (i) B D F A C E G H Fig (ii) To verify : ∠AGE = ∠GHC ∠AGH = ∠CHF ∠EGB =∠GHD and ∠BGH = ∠DHF Observations: Measure each angles and tabulate their values as below. Figure ∠AGH ∠CHF ∠GHD ∠EGB Remarks i. ................0 ................0 ................0 ................0 ∠AGH = ∠FHC, ∠GHD = ∠EGB ii ................0 ................0 ................0 ................0 ∠AGH = ∠FHC, ∠GHD = ∠EGB Conclusion: Hence, if a transversal intersects two parallel lines, then the corresponding angles so formed are equal. Worked Out Examples Example 1 From the given figure, find the four pairs of corresponding angles. Solution: In the given figure, four pairs of corresponding angles are a and e, b and f, c and h, & d and g. Example 2 Identify whether AB and CD are parallel or not.. Solution: In the given figure, ∠EIB = ∠IJD = 70° Since two corresponding angles are equal, then AB||CD. b f a e d g c h B C D A I E J 70º 70º F B C D A