245 Oasis School Mathematics-8 245 18.2 Some special product and formulae (Continue) Formula 4 (a + b)3 = a³ + 3a²b + 3ab² + b³ Proof: We have, (a + b)3 = (a + b) (a + b)2 = (a + b) (a² + 2ab + b²) = a(a² + 2ab + b²) + b(a² + 2ab + b²) = a³ + 2a²b + ab² + a²b + 2ab² + b³ = a³ + 3a²b + 3ab² + b³ ∴ (a + b)³ = a³ + 3a²b + 3ab² + b³ Again, (a + b)3 = a³ + 3a²b + 3ab² + b³ = a³ + 3ab (a+b) + b³ = a³+ b³+ 3ab (a + b) or, (a + b)3 – 3ab(a + b) = a³ + b³ ∴ a³ + b³ = (a + b)³ – 3ab(a + b) Formula 5 (a – b)3 = a³ – 3a²b + 3ab² –b³ Proof: We have, (a – b)3 = (a – b) (a – b) (a – b) = (a – b) (a – b)2 = (a – b) (a² – 2ab + b²) = a(a² – 2ab + b²) – b(a² – 2ab + b²) = a³ – 2a²b + ab² – a²b + 2ab² – b³ = a³ – 2a²b – a²b + 2ab² + ab² – b³ (a – b)3 = a³ – 3a²b + 3ab² – b³ ∴ (a – b)³ = a³ – 3a²b + 3ab² – b³ Cor: We have, (a – b)3 = a³ – 3a²b + 3ab² – b³ (a – b)3 = a³ – b³– 3ab (a–b) or, = (a–b)³ +3ab (a – b) = a³ – b³ ∴ a³ – b³ = (a – b)³ + 3ab(a – b) (a + b)3 = a3 + b3 + 3ab (a + b) = a3 + 3a2 b + 3ab2 + b2 (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – 3a2 b + 3ab2 – b3 Remember !
246 Oasis School Mathematics-8 Geometrical Interpretation of (a + b)3 Take a cubical object having the length of each side (a+b)cm. Cut the cube into 8 parts as shown in the figure. Here, Volume of first part = a × a × a = a3 Volume of second part = a × b × b = ab2 Volume of third part = a × b × b = ab2 Volume of fourth part = a × b × b = ab2 Volume of fifth part = a × a × b = a2 b Volume of sixth part = a × a × b = a2 b Volume of seventh part = a × a × b = a2 b Volume of eighth part = b × b × b = b3 ∴ Volume of given object = a3 + ab2 + ab2 + ab2 + a2 b + a2 b+ a2 b + b3 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 ∴ (a + b)3 = a3 + 3a2 b + 3ab2 + b3 a-b b b b b a a a
247 Oasis School Mathematics-8 247 Geometrical Interpretation of (a – b)3 Take a cubical object having the length of each side a cm. From each side, cut b cm as shown in the figure. Volume of the whole solid = a3 Here we can get 8 parts as shown in the figure. Volume of first part = (a–b)3 Volume of second part = b2 (a-b) Volume of third part = b2 (a-b) Volume of fourth part = b2 (a-b) Volume of fifth part = (a-b)2 .b Volume of sixth part = (a-b)2 .b Volume of seventh part = (a-b)2 .b Volume of eighth part = b3 Here, ∴ a3 = (a – b)3 + (a – b)b³ + (a – b)b2 + (a – b)b2 + (a – b)2 b + (a – b)2 b + (a – b)2 b + b3 a3 = (a – b)3 + 3 (a – b)b2 + 3 (a – b)2 b + b3 a3 = (a – b)3 + 3 (ab2 – b3 ) + 3(a2 – 2ab + b2 ) b + b3 a3 = (a – b)3 + 3ab2 – 3b3 + 3a2 b – 6ab2 + 3b3 + b3 a3 = (a – b)3 – 3ab2 + 3a2 b + b3 ∴ (a – b)3 = a3 – 3a2 b + 3ab2 – b3 b b a-b b a-b a a -b -b -b -b -b -b -b -b -b -b
248 Oasis School Mathematics-8 Worked Out Examples Example: 1 Find the cubes of : 2a – 1 3b Solution: We know that (a – b)3 = a³ – 3a²b + 3ab² – b³ So, the required cube of (2a – 1 3b) = (2a – 1 3b)3 = (2a)3 – 3 × (2a)2 × 1 3b + 3 × 2a × ( 1 3b) 2 – ( 1 3b)³ = 8a³ – 3 × 4a² × 1 3b + 3 × 2a × 1 9b2 – 1 27b3 = 8a³ – 4a2 b + 2a 3b2 – 1 27b3 Example: 2 If a + b = 8 and ab = 5, then find the value of a³ + b³. Solution: Now, we know that a³ + b³ = (a + b)3 – 3ab (a + b) = (8)3 – 3 × 5(8) = 512 – 120 = 392 ∴ a³ + b³ = 392 Example: 3 If a – 1 a = 7, find the value of a³ – 1 a3 . Solution: Here, a – 1 a = 7 Now, cubing on both sides, we get (a – 1 a ) 3 = (7)3 [∵ (a – b)3 = a³ – 3ab(a – b) – b³] or, (a)3 – 3a × 1 a (a – 1 a ) – 1 a3 = 343 or, a³ – 3(7) – 1 a3 = 343 [∵ a – 1 a = 7] or, a³ – 21 – 1 a3 = 343 ∴ a³ – 1 a3 = 343 + 21 = 364
249 Oasis School Mathematics-8 249 Example: 4 Simplify: (5x + 2y)3 + 3(5x + 2y)2 (3x – y) + 3(5x + 2y) (3x – y)2 + (3x – y)3 Solution: Let 5x + 2y = a and 3x – y = b, then (5x + 2y)3 + 3(5x + 2y)2 (3x – y) + 3(5x + 2y) (3x – y)2 + (3x – y)3 = a³ + 3× (a)2 × b + 3 × a × b² + (b)3 = a³ + 3a²b + 3ab² + b³ = (a + b)3 [∴a³ + 3a²b + 3ab² + b³ = (a + b)3 ] Now, replacing the value for a = 5x + 2y and b = 3x – y, we get = {(5x + 2y) + (3x – y)}3 = {5x + 2y + 3x – y}3 = (8x + y)3 Example: 5 If x + 1 x = k, prove that x3 + 1 x3 = k3 – 3k. Solution: L.H.S. = x3 + 1 x3 = (x + 1 x )3 – 3.x. 1 x (x + 1 x ) [ ∵ a³ + b³ = (a + b)³ – 3ab (a + b)] = k3 – 3(k) [∵ x + 1 x = k] = k3 – 3k = R.H.S. ∴ Hence proved. Example: 6 Evaluate: 99³ Solution: Here, 993 = (100 – 1)3 = (100)³ – (1)³ – 3 × 100× 1 (100 –1) [ ∵ (a – b)³ = a³ – b³ – 3ab (a – b)] = 1000000 – 1 – 300 × 99 = 1000000 – 1 – 29700 = 970299
250 Oasis School Mathematics-8 Exercise 18.2 1. Find the cubes of the following binomials: (a) (4a + 5b) (b) (3x – 4y) (c) y + 1 y (d) 2x – 1 3y 2. (a) If a + b = 9 and ab = 4, find the value of a³ + b³. (b) Find the value of a³ + b³, if a + b = 8 and ab = 3. (c) Find the value of x3 –y3 , if x – y = 6 and xy = 4. (d) Find the value of a³ – b³, if a – b = 3, ab = 10. (e) Find the value of x3 – 1 x3 , if x – 1 x = 4. (f) Find the value of a³ + 1 a3 , if a + 1 a = 9. 3. (a) If x + 1 x = m, prove that x3 + 1 x3 = m3 – 3m. (b) If 3x – 2y = 5, prove that 27x3 – 8y3 – 90xy = 125. (c) If a + 1 a = 3, prove that a³ + 1 a3 = 0. (d) If p – 2q = 4, prove that p3 – 8q3 –24pq = 64. (e) If x – y = 7, prove that x3 – y3 – 21xy = 343. 4. Simplify: (a) (p + q)3 – 3(p + q)2 (p – q) + 3(p + q) (p – q)2 – (p – q)3 (b) (2x + 7y)3 + 3(2x + 7y)2 (x – 3y) + 3(2x + 7y) (x – 3y)2 + (x – 3y)3 (c) (3m + 4n)3 + 3(3m + 4n)2 (m + 3n) + 3(3m + 4n) (m + 3n)2 + (m + 3n)3 5. Find the value of the following without actual multiplication. (a) (97)3 (b) (106)3 (c) (1001)3 6. Simplify the following: (a) (3a + b)3 + (3a – b)3 (b) (5m +2n)3 – (5m – 2n)3 1. (a) 64a3 + 240a2 b + 300ab2 + 125b3 (b) 27x3 -108x2 y+144xy2 -64y3 (c) y3 +3y+3/y+1/y3 (d) 8x3 -4x2 /y+2x/3y2 -1/27y3 2. (a) 621 (b) 440 (c) 288 (d) 117 (e) 76 (f) 702 4. (a) 8q3 (b) (3x+4y)3 (c) (4m+7n)3 5. (a) 912673 (b) 1191016 (c) 1003003001 6. (a) 54a3 + 18ab2 (b) 300m2 n+16n3 Answer
251 Oasis School Mathematics-8 251 19.1 Factorisation–Review Look and learn the following. 5 × 3 = 15, 5 and 3 are the factors of 15. Similarly, 6×x = 6x, 6 and x are the factors of 6x. (a+b) (a–b) = a²–b², (a+b) and (a–b) are the factors of a²–b². Thus, the product of 6 and x is 6x, the term 6x is called a multiple of 6 and x, whereas 6 and x are the factors of 6x. The process of finding two or more expressions whose product is equal to the given expression is called factorisation. For example: Factorisation of 2a + 4ab = 2a(1 + 2b) Method of factorisation of algebraic expression: Factorisation of a polynomial having a monomial factors Steps: Find by inspection, what is common throughout in the given expression, then find the quotient of the given expression by this common factor. For example: Factorise: 2x2 y + 6xy2 + 4xy Solution: By inspection, we find that each term of given expression 2x2 y + 6xy2 + 4xy is exactly divisible by 2xy. So, 2xy is a common factor. Here, 2x2 y + 6xy2 + 4xy = 2xy (x + 3y + 2) Note: If first term is '–', then it is usual to take '–' as common. When the common factor is a polynomial. In this case, we take out the common factor and use distributive property. Look and learn For example : Factorise: (a) x(a + b) + y(a + b) + z(a + b) Solution: Here, x(a + b) + y(a + b) + z(a + b)=(a + b) (x + y + z) Unit 19 Factorisation
252 Oasis School Mathematics-8 (b) ax + by + ay + bx Solution: Here, ax + by + ay + bx = ax + ay + bx + by = a(x + y) + b(x + y) = (x + y) (a + b) Worked Out Examples Example: 1 Factorise : 2a²b + 3ab² Solution: 2a²b + 3ab² = ab(2a + 3b) Example: 3 Factorise: 2x + 2y + xy + y² Solution: 2x + 2y + xy + y² = 2(x + y) + y (x + y) = (x + y) (2 + y) Exercise 19.1 1. Factorise (a) 2x + 6y (b) 15x²y + 25xy2 (c) 3a² – 15ab (d) 6x4 y – 18x3 y2 (e) – 10 x³y²z³ – 15 x²yz (f) – 21x4 a²c³ + 14 x³ac² 2. Factorise (a) 4a + 3ab + 5a² (b) 3x²y – 6xy + 9xy² (c) 4x³y7 – 2xy6 + 6y9 (d) –abx – acy – adz (e) 4x2 y – 6xy2 + 10xy 3. Factorise (a) a(x + y) + b(x + y) (b) a2 (a + 1) + 1(a + 1) (c) y(a – b) – x(a – b) (d) x(a – b) + y(b – a) (e) x(x – 2y) + y(2y – x) 4. Factorise (a) (a – 2) (x+ 4) + (a – 2) (x + 5) (b) (x – y)2 + (x – y) (c) (a – b) – (a – b)2 (d) p(a – b) + q(a – b) + r(a – b) (e) x(a + b + c) + y(a + b + c) + z(a + b + c) Example: 2 Factorise: x(a – b) + y (a – b) + z (a – b) Solution: x(a – b) + y(a – b) + z (a – b) = (a – b) (x + y + z) Example: 4 Simplify : 7 × 16 + 7 × 14 Solution: 7 × 16 + 7 × 14 = 7 (16 + 14) = 7 × 30 = 210
253 Oasis School Mathematics-8 253 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 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 Answer 19.2 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' units. Its area is a² • From the square sheet take 'b' units from each side as shown in the figure. Area of that part = b² Area of the shaded part = a² – b² .............. (i) a² – b² = (a + b) (a – b) a a a b b a
254 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, Factorisation of difference of two cube We already know that, (a – b)³ = a³ – 3a²b +3ab² – b³ 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) a³ + b³ = (a + b) (a² – ab + b²)
255 Oasis School Mathematics-8 255 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 ∴ 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: (i) (x + 3y)2 – (x – 3y)2 Solution: (i) (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). a³ – b³ = (a – b) (a² + ab + b²) 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: 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)
256 Oasis School Mathematics-8 Exercise 19.2 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 – 8 a 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 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) (2–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) (9a+b) (a+9b) 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) (9m2 +6mn+4n2 ) (g) (4a–5b) (16a2 +20ab+25b2 ) 5 1 1 1 1 1 3 4 2 2 2 2 . (a x ) ( ) ( ) x x x b a b a a b b c a b + ( ) − + − ( ) + + + ( ) − + − + + a ab b d x y x x y y e 2 2 2 2 9 12 16 3 5 9 5 3 25 ( ) ( ) 3 4 1 9 16 3 4 1 2 x x x + ( ) − + 6. (a) 2xy (x2 +3y2 ) (b) 2b (3a2 +b2 ) (c) (a–2b–2c) (a2 –4ab+4b2 +2ac–4bc+4c2 ) (d) (4–2a+3b) (16+8a–12b+4a2 –12ab+9b2 ) (e) 18y(4x2 +3y2 ) 7. (a) (a+b) (a2 –ab+b2 +1) (b) (a–b) (a2 +ab+b2 –1) Answer
257 Oasis School Mathematics-8 257 19.3 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 Worked Out Examples 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 19.3 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 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)
258 Oasis School Mathematics-8 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? 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. Answer 19.4 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) 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.
259 Oasis School Mathematics-8 259 Worked Out Examples 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) Exercise 19.4 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 Factorise (from 2 to 11): 2. (a) x2 + 3x + 2 (b) x2 + 7x + 10 (c) x2 + 10x + 9 (d) x² + 12x + 20 (e) x² + 17x + 72 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. Look! I have found 1 and 6, 3 – 2 = 1 3 × 2 = 6 Product = 42 Difference = 1 Numbers are 7 and 6 Example: 4 Factorise: 2(a+b)2 + 9(a + b) + 7 Solution: Let (a + b) = x, then 2(a + b)² +9(a + b)+7 = 2x² + 9x + 7 = 2x² + (7+2)x +7 = 2x² + 7x + 2x + 7 = x(2x+7) + 1(2x+7) = (2x + 7) (x + 1) Now, substituting for x = (a + b) we get, = {2(a+b)+7)} {(a+b)+1} = (2a+2b+7) (a+b+1)
260 Oasis School Mathematics-8 3. (a) m² – 3m + 2 (b) a² – 7a + 12 (c) m2 – 8m + 15 (d) y2 – 11y + 30 (e) b² – 9b + 20 4. (a) a² + a – 2 (b) x² + 2x – 3 (c) a² + a – 20 (d) m2 + 7m – 18 (e) x2 + 3x – 18 5. (a) x2 – x – 12 (b) x2 – 5x – 24 (c) y2 – y – 72 (d) m² – 2m – 80 (e) b² – 5b – 150 6. (a) 2a² + 7a + 5 (b) 2x2 + 7x + 3 (c) 3a² + 5a + 2 (d) 4x² + 17x + 15 7. (a) 2x2 – 3x + 1 (b) 5a² – 19a + 12 (c) 3a² – 8ab + 5b² (d) 3x² – 36x + 33 8. (a) 2m2 + m – 3 (b) 6x2 +5x – 6 (c) 5a²+11a – 12 (d) 7x² + 4x – 11 9. (a) 3a² – 2ab – 8b² (b) 3x2 – 7xy –6y2 (c) 4x2 – 16x – 9 (d) 2b² – 5b – 12 10. (a) 1 – 3x – 28x2 (b) –7x2 + 3 – 4x (c) 3x2 + 11x + 6 3 (e) (x + 1) (2x – 1) – 5 11. (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 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) (e) (x+8) (x+9) 3. (a) (m–1) (m–2) (b) (a–3) (a–4) (c) (m–5) (m–3) (d) (y–5) (y–6) (e) (b–4) (b–5) 4. (a) (a+2) (a-1) (b) (x+3) (x-1) (c) (a+5) (a-4) (d) (m+9) (m-2) (e) (x+6) (x-3) 5. (a) (a) (x-4) (x+3) (b) (x-8) (x+3) (c) (y-9) (y+8) (d) (m-10) (m+8) (e) (b-15) (b+10) 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) Answer
261 Oasis School Mathematics-8 261 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 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)
262 Oasis School Mathematics-8 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 ) (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) 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) Alternative method a4 + a²b² + b4 = (a²)² + (b²)² + a²b² = (a² + b²)² – 2a²b² + a²b² = (a² + b²)² – a²b² = (a² + b²)² – (ab)² = (a² + b² + ab) (a² + b² – ab) = (a2 + ab + b2 ) (a2 -ab+b2 )
263 Oasis School Mathematics-8 263 20.1 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. Note: 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. Let's see an example. 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 20 H.C.F. and L.C.M.
264 Oasis School Mathematics-8 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)
265 Oasis School Mathematics-8 265 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 20.1 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
266 Oasis School Mathematics-8 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) Answer 20.2 Lowest Common Multiple (L.C.M.) Let's consider any two algebraic terms a² and a³. Multiplies of a² are a², a³, a4 , a5 .............Multiplies 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)
267 Oasis School Mathematics-8 267 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) Note: 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:
268 Oasis School Mathematics-8 Fist expression = x³ + 1 = (x)³ + 1³ = (x + 1) (x² – x + 1) Second expression = x² + 3x + 2 = 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) 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) 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)
269 Oasis School Mathematics-8 269 Exercise 20.2 Find the L.C.M. of the following by factorisation: 1. (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. (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. (b) (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. (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. (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 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) Answer
270 Oasis School Mathematics-8 21.1 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. Note: 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. 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. 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] • Resolve the numerator and denominator in their factors. • Cancel the common factors. Steps Note: +x +y = x y , –x –y = x y , +x –y = – x y , – x y = – x y Unit 21 Rational Expressions
271 Oasis School Mathematics-8 271 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 21.1 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 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 (g) x³ – y³ – 3xy (x – y) x³ – y³ (h) 4 – (a + b)² (a + 2)² – b² (i) a² – 5a – 14 2a³ – 13a² – 7a
272 Oasis School Mathematics-8 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) Answer 21.2 Multiplication and Division of Rational Expressions Multiplication of Rational Expressions While multiplying the rational expressions, apply the following steps. • Multiply numerator with numerator and denominator with denominator. • Reduce it into its lowest term. Steps For example : Multiply : 3a²b 4ab × 8ac² 6bc Solution: 3a²b 4ab × 8ac² 6bc = 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:
273 Oasis School Mathematics-8 273 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] 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) Alternate 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
274 Oasis School Mathematics-8 = (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 = a2 + 7a + 6 a2 – 6a + 9 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 21.2 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 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
275 Oasis School Mathematics-8 275 (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. Simplify: (a) a³ – b³ a³ + b³ ÷ a + b a – b × (a + b)² (a – b)² (b) a² – 8a – 9 a² – 17a + 72 × a² – 25 a² – 1 ÷ a² + 4a – 5 a² – 9a + 8 (c) x² – x – 20 x² – 25 × x² +3x+2 x² +6x+8 ÷ x + 1 x² + 5x 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) a2 +ab+b2 a2 –ab+b2 (b) a-5 a-1 (c) x Answer 21.3 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. • Cancel out the factors that are common to both numerator and denominator. • Find the L.C.M. of the given denominators. • Simplify the given terms. Rule:
276 Oasis School Mathematics-8 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. (i) Add or subtract the numerators. (ii) Write the result of numerators over the common denominator of the like fractions. (iii) Reduce the resulting fraction to the lowest term. Working steps Worked Out Examples Example: 1 3a x + 2b x + 5c x Solution: 3a x + 2b x + 5c x = 3a + 2b + 5c x 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
277 Oasis School Mathematics-8 277 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) = 4x3 y + 4xy3 – 4x3 y + 4xy³ (x²)² – (y²)² = 8xy3 x4 – y4 (a – b) (a – b) (a +b) (a – b)
278 Oasis School Mathematics-8 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 21.3 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
279 Oasis School Mathematics-8 279 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 (c) x – 1 x² – 3x + 2 + x – 2 x² – 5x + 6 + x – 5 x² – 8x + 15 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)
280 Oasis School Mathematics-8 22.1 Indices (Review) Let us consider an algebraic term axb . Here, 'a' is coefficient, 'x' is the base and 'b' is the index (or power) of the base. An index of a base means how many times the base is multiplied. The plural form of index is called indices. Laws of Indices If m and n are two positive integers, then (i) am × an = am+n [Product law] (ii) am ÷ an = am–n [Quotient law] (iv) (am)n = amn [Power law] (ab)m = ambm ( a b) m = am bm (v) aº = 1 [Law of zero index] (vi) n am = a [Root law of index] Worked Out Examples Example: 1 Find the products in their exponential form. (i) 3 × 3 × 3 ×3 (ii) (–2) × (–2) × (–2) × (–2) × (–2) (iii) 2 5 × 2 5 × 2 5 Solution: (i) 3 × 3 × 3 × 3 = 34 (ii) (–2) × (–2) × (–2) × (–2) × (–2) = (–2)5 (iii) 2 5 × 2 5 × 2 5 = ( 2 5 ) 3 Note: ( am bn ) –2 = ( bn am) 2 m n Unit 22 Indices
281 Oasis School Mathematics-8 281 Example: 2 Find the value of the following: (i) 32 × 53 (ii) 36 3 2 (iii) ( 8 125) 1 3 Solution: (i) Here, 32 × 53 = 3 × 3 × 5 × 5 × 5 = 9 × 125 = 1125 (ii) Here, 36 3 2 = 62 × 3/2 = 63 = 6 × 6 × 6 = 216 (iii) Here,( 8 125) 1 3 = ( 23 53 ) 1 3 = 23 × 1 3 53 × 1 3 = 2 5 [∵ ( a b) n = an bn] Example: 3 Simplify by using laws of indices. The answer should contain only positive indices. (i) 24 ÷ 26 (ii) a6 × a4 a² (iii) ax–y × ay–z × az–x Solution: (i) Here, 24 ÷ 26 = 24 26 = 24–6 = 2–2 [∵ am an = am–n] = 1 2² = 1 4 [∵ a–m = 1 am ] (ii) Here, a6 × a4 a² = a6+4 a² [∵ am × an = am+n] = a10 a² = a10–2 = a8 . [∵ am an = am–n] (iii) Here, ax–y × ay–z × az–x = ax–y+y–z+z–x [∵ am × an = am+n] = ax–x–y+y–z+z = a0 = 1 Example: 4 Evaluate: Solution: 5 9 8 27 9 4 1 3 1 2 × ( ) ÷ ( ) − 5 9 8 27 9 4 5 9 2 3 3 2 1 3 1 2 3 1 3 2 1 2 × ( ) ÷ ( ) = × ( ) ÷ ( ) − × − ( ) x = × ÷ ( ) = × ÷ ( ) = × × = − 5 9 2 3 3 2 5 9 2 3 2 3 5 9 2 3 3 2 5 9 1 1
282 Oasis School Mathematics-8 Exercise 22.1 1. Express the product in their exponent form. (a) 6 × 6 × 6 × 6 (b) (–b) × (–b) × (–b) × (–b) × (–b) (c) ( 3 5 ) × ( 3 5 ) × ( 3 5 ) (d) x 4 × x 4 × x 4 2. Express the following exponent form into the product form. (a) (2x)3 (b) (– 2 3 ) 2 (c) ( x a ) 3 3. (a) What power of x yields unit value? (b) What index of m will yield 1 m² ? (c) What is the power of 6 so that its value becomes 1 36 ? (d) If ax = 1, then what is the value of x? (e) If x = 2, what is the value of 2x ? (f) If x = 2, what is the value of 4 2x ? 4. Simplify by using laws of indices. Answer should contain only positive indices. (a) 34 × 35 (b) a6 × a–5 (c) 45 ÷ 43 (d) c6 ÷ c3 (e) (43 )4 × (44 )2 (f) (54 )3 × (52 )3 (g) (p q) 5 × (p q) 2 (h) ( 3 5 ) 2 × ( 3 5 ) 4 (i) (– 1 3 ) 3 × (– 1 3 ) 4 × (– 1 3 ) 5 (j) a² × a3 a4 (k) (4p)3 × (4p)² (4p)4 (l) 4² × 9² × 49 2³ × 7 × 3² (m) 8p2 × 5p3 (n) 4r5 × 4r4 (o) 9x2 y2 × 7x3 y4 5. Simplify: (a) bx-y × by-z× bz-x (b) ax²–y²× ay²–z² × az²–x² (c) (–4x²y)³ × (3x–3y)² (d) (–2ab²c–1)³× (3a²b³c)–1 (e) (2x²y³z-1)²× (3x²y³z)–3 (f) (3x²y³z–1) × (4x–2y3 z2 )–2 (g) ( a³b–5 a²b–3 ) –5 6. (a) If x = 1, y = 2 and z = 3, find the value of each of the following: (i) x–2 (ii) 27xºy6 z–3 (iii) (x5 y)–4 (iv) xy yz zx (b) If a = 2, b = 0, c = 1, d = –1, find the value of (i) a–1.b–2 (ii) b³c3 d2 (iii) ca . ab .db (iv) ab .cd.da
283 Oasis School Mathematics-8 283 7. Evaluate: (a) 82/3 (b) 323/5 (c) 272/3 (d) (27)–2/3 (e) (–125)–2/3 (f) ( 243 32 ) 2/5 (g) ( 1 216 ) – 1 3 (h) ( 5 4 ) –5 (i) (625)–1/4 (j) ( 1 16 ) 0.25 (k) (40.5)2 (l) ( 9 16) 1 2 × ( 32 243) 3 5 (m) (20)³ × (30)² (40)² × (10)³ (n) 8. Simplify: (a) 3x+1 – 3x 2 × 3x (b) 2x+1 – 2x 3 × 2x (c) 5x+1 – 5x 4 × 5x (d) ax+2 – ax+1 ax+1 –ax 9. Simplify: (a) 128x6 y–3 64x–3y³ (b) (8x6 y–3) 1/3 2xy (c) (4xy)² (2x–3y–3)³ (d) 81x6 y–5z² (3x²y³z)³ 10. Simplify: (a) ax+y ax+2y × ay–z a2x+z (b) x3a+3b x2a–2b × x3a+b x2a+b (c) xa(b + c) × x–b(c+a) × x-c(a + b) (d) (ya+b)a–b × (yb–c)b+c × (yc+a)c-a (e) ( za zb ) a²+ab+b² . ( zb zc ) b²+bc+c² . ( zc za ) c²+ca+a² (f) xa²–b² x2b²–a² × xb²–2a² xb² (g) (xa+b)a–b × (xb+c)b–c (xa+c)a–c 26 × 39 × 512 23 × 33 × 125 3 Answer 1. (a) 64 (b) (-b)5 (c) (3/5)3 (d) (x/4)3 2. (a) 2x×2x×2x (b) (-2/3)×(-2/3)×(-2/3) (c) x/a×x/a×x/a 3. (a) 0 (b) (-2) (c) -2 (d) 0 (e) 1 (f) 1 4. (a) 39 (b) a (c) 42 (d) c3 (e) 420 (f) 518 (g) (p/q)7 (h) (3/5)6 (i) (-1/3)12 (j) a (k) 4p (l) 126 (m) 40p5 (n) 16r9 (o) 63x5 y6 5. (a) 1 (b) 1 (c) -192y5 (d) -8/3 ab3 /c4 (e) 4/27x2 y3 z5 (f) 3x6 /16y3 z5 (g) b10/a5 6. (a) (i) 1 (ii) 64 (iii) 1/16 (iv) 24 (b) (i) 0 (ii) 0 (iii) 1 (iv) 1 7 4 8 9 1 9 1 25 9 4 6 1024 3125 1 5 1 2 . ( ) ( ) ( ) ( ) ( ) ( ) ( ) (h) ( ) ( ) ( ) a b c d e f g i j k 4 2 9 9 2 750 8 1 1 3 1 9 2 2 9 6 2 (l) ( ) ( ) . ( ) ( ) / ( ) ( ) . ( ) ( ) ( ) m n a b c d a a x y b x y c x y d y z a a b x c x d e f x z a b bc 11 11 14 1 2 2 5 2 3 10 1 1 1 1 ( ) . ( ) ( ) ( ) ( ) ( ) ( ) ( ) + + x g 3b2 ( ) 1 1 a2(x+3)
284 Oasis School Mathematics-8 23.1 Linear Equations in One Variable Let us consider some statements such as • 5 added to a number x gives 12 i.e. 5 + x = 12. • 10 subtracted from a number x results 8 i.e. x – 10 = 8 • Twice the number 'x' added to 5 gives 15 i.e. 2x + 5 = 15. All these are statements where two expressions are equal for a particular value of a variable. Hence, all three above expressions are equations. A statement where two expressions are equal for a particular value of a variable is called an equation. Rules for solving Linear equations of one variable • We can add the same number on either side of the equation. i.e. x–3=4 ⇒ x – 3 + 3 = 4 + 3 • We can subtract the same number from either side of the equation. i.e. x + 5 = 12 ⇒ x + 5 – 5 = 12 – 5 • We can multiply both sides of the equation by the constant number. i.e. x 3 = 6 ⇒ x 3 × 3 = 6 × 3 • We can divide both sides of the equation by the constant number. i.e. 5x = 10 ⇒ 5x 5 = 10 5 Unit 23 Equation, Inequality and Graph
285 Oasis School Mathematics-8 285 Note: Any term of the equation can be taken to the other side by changing its sign. This process is called transposition. e.g. 2x + 5 = 8–x ⇒ 2x + x = 8–5 Worked Out Examples Example 1: Solve : 2x–3 = 5 Solution: or 2x–3 = 5 or, 2x = 5+3 or, x = 8 2 ∴ x = 4. Example: 2 Solve: 2x + 1 3x + 5 = 11 20 Solution: 2x + 1 3x + 5 = 11 20 or, 20(2x +1) = 11 (3x +5) (Cross multiplying) or, 40x +20 = 33x + 55 or, 40x – 33x = 55 – 20 or, 7x = 35 or, x = 35 7 ∴ x = 5 Example 3: Solve : 9x – 7 3x+5 = 3x – 4 x+6 Solution: 9x – 7 3x+5 = 3x – 4 x+6 or, (9x-7) (x+6) = (3x–4) (3x+5) or, 9x2 +54x–7x–42 = 9x2 +15x–12x–20 or, 9x2 +47x–9x2 –3x = –20+42 or, 44x = 22 or, x = 22 44 or, x = 1 2
286 Oasis School Mathematics-8 Exercise 23.1 Solve the following: 1. (a) x – 2 = 3 (b) y – 2 = 10 (c) z – 1 = 3 2. (a) x + 1 = 6 (b) y + 2 = 4 (c) z + 6 = 2 3. (a) 2x = 4 (b) 3x = -18 (c) 5x = 20 4. (a) x 2 = 3 (b) x 4 = -2 (c) x 4 = 3 5. (a) 2x–5 = 15 (b) 3x–6=24 (c) 4x–2 = 18 6. (a) 2x+6 = 26 (b) 3x+5 = 20 (c) 4x + 5 = 25 7. (a) x 2 = 7 2 (b) 3x 4 = 15 (c) 2x 3 = 8 3 8. (a) 4x–3 = 2x + 5 (b) 5x–3 = 4x+6 (c) 4x – 15 = 25 – 6x 9. (a) x–2 2 = x+6 4 (b) x–8 2 = x–12 3 (c) 2x – 8 x + 5 = 8 5 (d) 2x+1 3x-1 = 3 2 (e) 2x+5 3x+5 = 3 4 10. (a) x+2 4 – x–3 3 = 1 2 (b) 4x – 3 2 + 2x + 1 6 = 1 (c) 5x-4 8 – x–3 5 = x+6 4 (d) 3x-1 5 – x – 3 4 = 3 (e) x+12 6 – x 12 = 13 2 11. (a) x + x 10 = 66 (b) x–3 3 – x = 2x–5 (c) 2(4x-1) 3 – 5x-1 3 = x+4 3 (d) 2(3x+1)–3(x+3) (x+3)+(3x+1) = 1 2 12. (a) (x+2) (x+3)–(x-1)(x+4) = 14 (b) (x+1) (x+2) – (x-1) (x+3) = 7 (c) (2x+1) (2x+3) – (2x–1) (2x–3) = 16 Answer 1. (a) 5 (b) 12 (c) 4 2. (a) 5 (b) 2 (c) -4 3. (a) 2 (b) -6 (c) 4 4. (a) 6 (b) -8 (c) 12 5. (a) 10 (b) 10 (c) 5 6. (a) 10 (b) 5 (c) 5 7. (a) 7 (b) 20 (c) 4 8. (a) 4 (b) 9 (c) 4 9. (a) 10 (b) 0 (c) 40 (d) 1 (e) 5 10. (a) 12 (b) 1 (c) 8 (d) 7 (e) 54 11. (a) 60 (b) 3/2 (c) 5/2 (d) 9 12. (a) 2 (b) 2 (c) 1
287 Oasis School Mathematics-8 287 23.2 Applications of Linear Equation We can convert the verbal problems into the linear equation of one variable. While solving such problems, apply the following steps. Worked Out Examples Example: 1 A number increased by 5 is 12. Find the number. Solution: Let the number be x; From the given condition, x + 5 = 12 or, x = 12–5 or, x = 7 ∴ Required number = 7. Example: 2 When a number is multiplied by 4 and decreased by 7, the result is 65. Find the number. Solution: Let the number be x From the given condition, 4x – 7 = 65 or, 4x = 65 + 7 or, 4x = 72 or, x = 72 4 or, x = 18. ∴ Required number = 18. • Read the given problem carefully to find out what is given and what is to find. • Denote the unknown quantity by the letter x or by some other letter such as a, b, y, z, etc. • Make the equation using unknown quantity. • Solve the equation to find the value of x. • Verify the solution, if necessary. Steps
288 Oasis School Mathematics-8 Example: 3 If 12 is added to three times of a number, the result is 42, find the number. Solution: Let the required number be x. From the given condition, 12 + 3x = 42 or, 3x = 42–12 or, 3x = 30 or, x = 30 3 ∴ Required number = 10. Example: 4 A is now 12 years old and B is 2 years old. In how many years, will A be three time as old as B? Solution: Let's suppose after x years, A will be three times as old as B. After x years, A's age = (12+x) years, B's age = (2 +x ) years From the given condition, (12 + x) = 3(2 + x) or, 12 + x = 6 + 3x or, 3x–x =12 – 6 or, 2x = 6 or, x = 6 2 x = 3 ∴ After 3 years, A will be three times as old as B. Example: 5 Length of a rectangle is 5m more than its breadth. If its perimeter is 50m, find its length and breadth. Solution: Let, the breadth of the rectangle = x, then its length = x + 5 From the given condition, Perimeter = 2(length + breadth) or, 50 = 2 (x + 5 + x) or, 50 = 2(2x + 5) or, 50 = 4x + 10
289 Oasis School Mathematics-8 289 or, 50 – 10 = 4x or, 4x = 40 or, x = 40 4 m x = 10m ∴ Length = x + 5 = (10 + 5)m = 15m, breadth = x = 10m. Exercise 23.2 1. (a) If 5 is added to a number, the sum is 15. Find the number. (b) A number increased by 20 is 45. Find the number. (c) If a number is decreased by 10, the result is 25. Find the number. (d) By how much 35 should be decreased to make 21? (e) Twice the number decreased by 3 is 17. Find the number. (f) If a number is multiplied by 4 and increased by 12, the result is 48. Find the number. (g) The sum of three consecutive natural numbers is 33. Find the numbers. (h) Find three consecutive natural numbers whose sum is 63. (i) Find three consecutive even numbers whose sum is 48. (j) Find three consecutive odd numbers whose sum is 27. 2. (a) In 16 years, Anil's age will be twice his age 12 years ago. What is his present age ? (b) A father's age is three times that of his son and in 10 years hence it will be twice as greater as his son's. Find their ages. (c) Ages of two brothers are in the ratio 2:3. 5 years hence the ratio of their ages will be 5:7. Find their present ages. (d) Ages of a son and his father are in the ratio 1:3. 10 years ago, ratio of their ages was 1:5, find their present ages. 3. Find the value of x from the given figures. C A B B A D C Perimeter = 31cm (a) (b) (c) (5x+1)cm (10x-2)cm (4x+10)cm (3x+1)cm (6x+1)cm 3x+2 Perimeter = 36cm x+4 4. (a) The length of a rectangular field is 10 m longer than its breadth. If its perimeter is 40 m, find its length and breadth. (b) Two equal sides of a triangle are 3 cm more than three times the third side. Find the sides of the triangle if its perimeter is 21 cm.
290 Oasis School Mathematics-8 23.3 Graph of the Linear Equation The equation of the form ax + by + c = 0 is called a linear equation. To draw the graph of a linear equation, let's take an equation. 2x – y + 1 = 0 which is in the form of ax + by + c = 0. It can be written as, – y = –2x–1 or, y = 2x +1 Let's fill the given table with suitable variables. x 0 1 2 y 1 3 5 Plot the points (0, 1), (1, 3) and (2, 5) on the graph and join them. Example: Draw the graph of the line 3x +2y = 6, Solution: Given equation of a straight line 3x +2y = 6 or, 2y = 6 – 3x or, y = 6 –3x 2 x 0 2 4 y 3 0 -3 When x = 0, y = 2 × 0 + 1 = 1 When, x = 1, y = 2 × 1 + 1 = 3 When, x = 2, y = 2 × 2+ 1 = 5 X' Y' (1, 3) (2, 5) X Y y=2x+1 O (0, 1) X' Y' O X Y 3x+2y=6 (0,3) (2,0) (4,-3) Answer 1. (a) 10 (b) 25 (c) 35 (d) 14 (e) 10 (f) 9 (g) 10, 11, 12 (h) 20, 21, 22 (i) 14,16,18 (j) 7, 9, 11 2. (a) 40 yrs (b) 10 yrs, 30 yrs. (c) 20 yrs, 30 yrs, (d) 20 yrs, 60 yrs. 3. (a) 2 (b) 2 (c) 3 4. (a) 15m, 5m (b) 8cm, 8cm, 5cm.
291 Oasis School Mathematics-8 291 23.4 Slope of a Straight Line The slope is usually measured as a ratio of vertical rise to the horizontal distance. In the given figure, slope of line PR is Slope (m) = Vertical rise (QR) Horizontal distance (PQ) In the graph, a straight line PQ is passing through the points P(6, 6) and Q(10, 11). Now, draw a line (PR) through P parallel to x–axis and the line (QR) through Q parallel to y–axis which intersect at R(10, 6). Here, PR = 10–6=4 QR = 11-6 = 5 ∴ From P to Q, horizontal displacement = change in x-coordinates = 4 Vertical displacement = Change in y-coordinates = 5 ∴ Slope of a line (AB) = m = change in y-coordinate change in x-coordinate = 5 4 Hence, the slope of straight line passing through A(x1 , y1 ) and B(x2 , y2 ), is Slope of AB = m = y2 –y1 x2 –x1 Standard Equation of Straight Line A linear equation is in the form of ax+by+c = 0 It can be written as by = –ax–c or, y = -ax–c b or, y = (– a b) x + (– c a ) Where, (– a b) represents the slope of straight line and (– c a ) represents the Y– intercept. Hence this equation can be written as y = mx + c. R Q θ P X' X Y' Y Q(10, 11) R(10,6) P(6,6) O A(x1 , y1 ) B(x2 , y2 ) Y Y' X' X Remember! y = mx + c. slope Y–intercept
292 Oasis School Mathematics-8 Worked Out Examples Example: 1 Find the slope of a line joining the points (2, 1) and (1, 6). Solution: Here, (2, 1) = (x1 , y1 ) and (1, 6) = (x2 , y2 ). Now, the required slope (m) = y2 –y1 x2 –x1 = 6–1 1–2 = 5 -1 = –5 Example: 2 Find the value of k if the slope of the line passing through (3, k) and (–2, 4) is – 3 5 . Solution: Here, (3, k) = (x1 , y1 ) and (–2, 4) = (x2 , y2 ) and slope (m) = – 3 5 . We have, slope (m) = y2 –y1 x2 –x1 Note : The slope of a straight line may be positive, negative, zero or undefined. X' X Positive slope Y' Y O X' X No slope Y' Y O X' X' X Negative slope Y' Y' Y Y O O Slope is undefined
293 Oasis School Mathematics-8 293 or, -3 5 = 4 – k -2–3 or, -3 5 = 4 – k –5 or, –3 × –5 = 5(4 – k) or, 15 = 20 – 5k or, 5k = 20 – 15 ∴ k = 5 5 = 1 Example: 3 Find the slope and y–intercept of the line whose equation is 3x + 2y = 6. Solution: Here, 3x + 2y = 6 or, 2y = –3x + 6 or, y = -3x 2 + 6 2 or, y = -3 2 x + 3 Now, comparing with y = mx + c, we get, Slope (m) = -3 2 and y–intercept (c) = 3. Example: 4 Draw a straight line passing through the points A(3, 2) and B(5, 4). (i) Find the difference of y–coordinate and x–coordinate for the line AB. (ii) Find the slope of AB. Solution: Now, plotting the points A(3, 2) and B(5, 4) on the graph and joining them, line AB is obtained. (i) Here, (3, 2) = (x1 , y1 ) and (5, 4) = (x2 , y2 ) difference of y–coordinates = y2 – y1 = 4 – 2 = 2 difference of x–coordinates = x2 – x1 = 5 – 3 = 2 (ii) Slope of AB = y2 –y1 x2 –x1 = 2 2 = 1 Example: 5 Draw the graph of the equation y = 3x – 2 and find its slope taking any two points. Solution: Here, y = 3x – 2 X' X Y' Y 1 2 A(3,2) B(5, 4) 1 2 3 4 O 3 4 5 6
294 Oasis School Mathematics-8 Table of values: x 0 1 2 y -2 1 4 By plotting (0, –2), (1, 1), and (2, 4) a straight line PQ is obtained. Taking any two points (1, 1) and (2, 4) (1, 1) = (x1 , y1 ) and (2, 4) = (x2 , y2 ). ∴ Slope of PQ = y2 –y1 x2 –x1 = 4 – 1 2 – 1 = 3 1 = 3 Exercise 23.3 1. Find the slope of straight lines passing through the following pairs of points. (a) A(2, 3), B(6, 5) (b) P(–5, 2), Q(3, –5) (c) M(2, 5), N(–6, 9) (d) R(4, 7), S(6, 10) (e) C(–1, –4), D(5, 6) (f) K(2, 1), L(5, 7). 2. (a) Find the value of k, if the slope of the line passing through (5, k) and (9, 8) is 1 2 . (b) Find the value of a, if the slope of the line passing through (a, 7) and (6, 10) is -3 2 . 3. Find the slope and y–intercept of the lines whose equations are: (a) y = 2x+9 (b) y = 2 3 x + 5 3 (c) y = –2x + 7 (d) 2x + 3y = 5 (e) y + 2x = 3 4. A straight line is passing through the points P(1, 2) and Q(2, 4). (i) Draw the graph of the straight line. (ii) Find the difference of x-coordinates and y-coordinates for the line PQ. (iii) Find the slope of PQ. 5. Complete the following table. 6. Draw the graphs of the following equations. (a) y = 3x + 4 (b) 2y – 3x = –5 (c) x + 2y = 12 (d) 3x + 4y = 12 (e) 2x + 6y = 3 Y' O Y X' X (0,-2) (1,1) (2, 4) 7 6 5 4 3 2 1 1 2 3 4 5 6 x y = 2x – 1 (a) (b) (c) 1 2 3 x y = 3x – 1 0 1 2 x y = 1 4 -2 2x – 1 3