MANIPULATING AND SOLVING EQUATIONS

Problem 1 :

If a + b = -2, then (a + b)³ = 

a)  4    b)  0   c)  -4   d)  -8

Solution :

a + b = -2

(a + b)³ = (-2)³

(a + b)³ = -8

Problem 2 :

For what value of n is (n - 4)²  = (n + 4)²

Solution :

Expand using the algebraic identities :

(n - 4)²  = (n + 4)²

n² - 8n + 16 = n² + 8n + 16

Subtracting n² and 16 on both sides, we get

- 8n = 8n

-16 n = 0

n = 0

Problem 3 :

If (1/a) x (b/c) = 1, what is the value of b - ac ?

(a)  -3    (b)  0     (c)  2    (d) it cannot be determined

Solution :

(1/a) x (b/c) = 1

b/ac = 1

b = ac

b - ac = 0

So, the answer is 0.

Problem 4 :

If 3x - 8 = -23, what is the value of 6x - 7 ?

(a)  -5  (b)  -21    (c)  -30    (d)  -37

Solution :

If 3x - 8 = -23

3x = -23 + 8

3x = -15

x = -15/3

By applying the value x = -5 in 6x - 7, we get

6x - 7 ==> 6(-5) - 7

= -30 - 7

= -37

Problem 5 :

If 4/9 = 8m/3, what is the value of m ?

(a)  1/6   (b)  2/3   (c)  5/6   (d)  6

Solution :

4/9 = 8m/3

Doing cross multiplication, we get

12 = 72m

Dividing by 72 on both sides.

12/72 = m

m = 1/6

Problem 6 :

If 3x + 1 = -8, what is the value of (x + 2)³

Solution :

3x + 1 = -8

Subtracting 1 on both sides, we get

3x = -8 - 1

3x = -9

x = -3

By applying the value of x in (x + 2)³, we get

= (-3 + 2)³

= (-1)³

= -1

Problem 7 :

If 4/(k + 2) = x/3, where  k ≠ -2, what is k in terms of x ?

(a)  (12 - 2x)/x   (b)  (12 + 2x)/x   (c)  x/(12 + 2x)   (d)  12x - 2

Solution :

4/(k + 2) = x/3

Doing cross multiplication, we get

4(3) = x(k + 2)

12 = kx + 2x

kx = 12 - 2x

Dividing by x on both sides, we get

k = (12 - 2x)/x

Problem 8 :

If (x - 3)² = 36 and  x < 0, what is the value of x²

Solution :

(x - 3)² = 36

Taking square roots on both sides 

(x - 3) = √36

(x - 3) = ±6

x - 3 = 6 and x - 3 = -6

Since x < 0, we choose x = -9.

x² = (-9)²

x² = 81

Problem 9 :

f = p[ (1+i)ⁿ - 1 ] / i

The formula above gives the future value f of an annuity based on the monthly payment p, the interest rate i, and the number of months n. Which of the following gives p in terms of f, i and n.

Solution :

f = p[ (1 + i)ⁿ - 1 ] / i

fi = p [(1 + i)ⁿ - 1]

fi = p (1 + i)ⁿ

fi/(1 + i)ⁿ = p

p = fi/(1 + i)ⁿ

Problem 10 :

If m/2n = 2, what is the value of n/2m ?

(a)  1/8   (b)  1/4   (c)  1/2  (d)  1

Solution :

m/2n = 2

Multiplying by 2 on both sides, 

m/n = 4

Taking reciprocals on both sides

n/m = 1/4

Multiplying by 1/2 on both sides.

n/2m = 1/8

Problem 11 :

If x < 0 and x² - 12 = 4, what is the value of x ?

(a)  -16  (b)  -8   (c)  -4  (d)  -2

Solution :

x² - 12 = 4

Add 12 on both sides.

x² = 4 + 12

x² = 16

x = ± 4

x = 4 or -4

Since x < 0, we choose -4.

Problem 12 :

If x² + 7 = 21, then the value of x² + 3 ?

Solution :

x² + 7 = 21

x² + 4 + 3 = 21

Subtracting 4 on both sides.

x² + 3 = 21 - 4

x² + 3 = 17

Problem 13 :

If x² + 5x - 24 = 0

If k is a solution of the equation above and k < 0, what is the value of |k| ?

Solution :

x² + 5x - 24 = 0

Factoring the above quadratic equation, we get

x² + 8x - 3x - 24 = 0

x(x + 8) - 3(x + 8) = 0

(x + 8)(x - 3) = 0

x = -8 and x = 3

Since x < 0, we choose x = -8.

k = -8 then |k| = 8

Problem 14 :

x² (x⁴ - 9) = 8x⁴

If x > 0 for what real value of x is the equation above true ?

Solution :

x² (x⁴ - 9) = 8x⁴

(x⁴ - 9) = 8x²

x⁴ - 8x² - 9 = 0

Let x² = t

(x²)² - 8x² - 9 = 0

t² - 8t - 9 = 0

(t - 9) (t + 1) = 0

t = 9 and t = -1

x² = 9

x = 3 and -3

So, the value of x is 3.

Problem 15 :

If [2 √(x + 4)] / 3 = 6 and x > 0, what is the value of x ?

Solution :

[2 √(x + 4)] / 3 = 6

2 √(x + 4) = 18

√(x + 4) = 9

Take square on both sides.

(x + 4) = 81

x = 81 - 4

x = 77