Even more Algebra

To confirm this result, enter the following keystrokes,

Cs1a1pqHz1.6

Q(X)$$$Q(=)1.8665










q(SOLVE) Solve for X 1=










This result confirms the answer.




Example 6-4:

Find the value of x which satisfies the expression,


3.1 4
=
1.6 9



47

After re-arranging you should have arrive with,



1.5 4
= √
9


The keystrokes for this calculation are,

CqF1.5$4a9=









To confirm this result, enter the following keystrokes,


CaQ(X)^3.1RJ(X)^1.
6$$Q(=)4a9










q(SOLVE) Solve for X 1=









Which confirms the result you calculated.


48

Example 6-5:

Find the value of x which satisfies the expression


√1 − −1.7 = 0.8702


After re-arranging you should arrive with,



1.7 1
= √
1 − 0.8702 2



The keystrokes for this calculation are,

CqF1.7$1a1p0.

8702d









To confirm this result, enter the keystrokes,

Cs1pQ(X)^z1.7$$

Q(=)0.8702









Choose a seed well away from 1, say 4

49

q(SOLVE) Solve for X 4=









Which should confirm the result you obtained.




Example 6-6:

Find the value of x which satisfies the expression,

7
5 ( −2.3) =
5

After re-arranging you should arrive with,


7
= log ( ) + 2.3
5
5

The keystrokes for this calculation are,

Ci5$7a5$$+2.3=










To confirm the result, enter the keystrokes,



50

C5^(Q(X)p2.3)$

Q(=)7a5










q(SOLVE) Solve for X 1=










Which should confirm the result you obtained.




After these examples you should have a good

understanding how the indices work and how you can
use your fx-991ES calculator to confirm your results.















51

7 Changing the Subject of an Equation

In the previous examples you will have had to use the

techniques for changing the subject of an equation. In

this chapter we shall introduce the function f(x) and
-1
its inverse f (x) where ∈ ℝ. There should be a one-
-1
to-one mapping of f(x) to f (x), although in some
functions the domain may have to be restricted.


Once you obtain an inverse, you can use your fx-991ES

calculator to check the result. The following examples
will show you how this is achieved.





Example 7-1:
Find the inverse of the function,



1 −
( ) = √
1 +


Let = ( ), then squaring both sides,


1 −
2
=
1 +

Multiply both sides by (1 + x),



52

1 −
2
(1 + ) = ( ) (1 + )
1 +

Expand the brackets

2
2
+ = 1 −
2
Add -y to both sides,
2
2
2
2
+ − = 1 − −
Leaving,
2
2
= 1 − −
Add x to both sides,


2
2
+ = 1 − − +
Leaving

2
2
+ = 1 −
Or

2
2
( + 1) = 1 −
2
Dividing both sides by (y + 1)

1 − 2 1 − 2
= =
2
+ 1 1 + 2

Therefore

53

1 − 2
−1 ( ) =
1 + 2

To test whether you have the correct answer use a

test value. The original equation is,


1 −
( ) = √
1 +



You will observe 0 ≤ ≤ 1, a restricted domain.
Consider the value = 0.25, calculate f(0.25), the

keystrokes are,

Csa1p0.25R1+0.

25=n








-1
Place this value for x into f (x) and you should recover
the value 0.25. The inverse equation is,


1 − 2
−1 ( ) =
1 + 2

Perform this calculation with x = 0.77459, the
keystrokes are using M,


54

C0.77459d=


a1pMR1+M=








As you can see the 0.25 has be recovered which

suggests the derived inverse is correct.



Example 7-2:
Make x the subject of the following equation,


( ) = cos( 1.6 + 3.4)


After re-arranging you should arrive with,

1.6
−1 ( ) = √cos −1 ( ) − 3.4


Test this result with x = 0.8, the keystrokes are,

CqF1.6$q>0.8
)p3.4=







Enter this value into the original equation and you

should recover 0.8. The keystrokes are,

55

( ) = cos( 1.6 + 3.4)

Ck8.9724^1.6$+

3.4)=








Which confirms the result for this value of x.


Example 7-3:

Make x the subject of the equation,

4.4
( ) =
1 + −2

After re-arranging you should arrive with,

1
−1 ( ) = ln ( )
2 4.4 −

Test this result with x = 2.74, the keystrokes are,


C1a2$hQ(X)a4.4p
J(X)$)=r



X? 2.72=





56

Enter this value into the original equation and you

should recover 2.74. The keystrokes are,
4.4
( ) =
1 + −2

C4.4a1+qHz2Q(X)
r


X? 0.25057=








Which confirms the result for this value of x. Here is a

-1
plot of f(x) and f (x).


























57

Example 7-4:

Make x the subject of the equation,

1
( ) = log (1 + )
6
2

After re-arranging you should arrive with,


1
−1 ( ) = √

6 − 1

Test this result with x = 1.84, the keystrokes are,

Cs1a6^Q(X)$p1r


X? 1.84=








Enter this value into the original equation and you
should recover 1.84. The keystrokes are,

Ci6$1+1aQ(X)dr


X? 0.19601=









58

You can see from the calculation the original 1.84 has
-1
been recovered. Graphs of f(x) and f (x) are shown
here,






































You will observe the asymptote at x = 0 in both as

expected from looking at the equations.








59

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60

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61