Help with pre-calc please?

Question

1. (7x/x-2)-(2x/x+2)=9

2.x^3+x^2-2x=2

3.(1/x-2)+(3/x+3)=(4/x^2+x-6)

4. (4/x)-(3/x+1)=7

Answer 1

Please note how I've presented your

equations with correctly placed brackets,

if I have interpreted correctly.

1. 7x/(x-2) - 2x/(x+2) = 9

Multiply through by (x-2)(x+2) :

7x(x+2) - 2x(x-2) = 9(x-2)(x+2)

Multiply out :

7x^2 + 14x - 2x^2 + 4x = 9x^2 - 36

Simplify sides :

5x^2 + 18x = 9x^2 - 36

Subtract the LHS from both sides :

4x^2 - 18x - 36 = 0

Divide through by 2 :

2x^2 - 9x - 18 = 0

Factorise :

(2x + 3)(x - 6) = 0

Therefore, 2x + 3 = 0 implies x = -3/2

or, x - 6 = 0 implies x = 6.

2. x^3 + x^2 - 2x = 2

Subtract 2 from both sides :

x^3 + x^2 - 2x - 2 = 0

Factor first two terms and last two terms separately :

x^2(x + 1) - 2(x + 1) = 0

Take out the factor (x + 1) :

(x + 1)(x^2 - 2) = 0

Therefore, x + 1 = 0 implies x = -1,

or, x^2 - 2 = 0 implies x^2 = 2, so x = ± sqrt(2).

3. 1/(x-2) + 3/(x+3) = 4/(x^2+x-6)

Factor x^2+x-6 into (x-2)(x+3) :

1/(x-2) + 3/(x+3) = 4/[(x-2)(x+3)]

Multiply through by (x-2)(x+3) :

(x+3) + 3(x-2) = 4

x + 3 + 3x - 6 = 4

4x - 3 = 4

4x = 7

x = 7/4.

4. 4/x - 3/(x+1) = 7

Multiply through by x(x+1) :

4(x+1) - 3x = 7x(x+1)

4x + 4 - 3x = 7x^2 + 7x

x + 4 = 7x^2 + 7x

7x^2 + 6x - 4 = 0

Use quadratic formula as it doesn't factor :

x = {-6 ± sqrt[6^2 - 4(7)(-4)]} / (2*7)

= [-6 ± sqrt(148)] / 14

= [-3 ± sqrt(37)] / 7