Question

Which of the following equations has a solution x = 4? Select all that apply.

Answer 1

Answer: 1st and 3rd

Step-by-step explanation:

Answer 2

Answer:

$-2(3x - 4) = -16$ and $20- 15x = 6 - 2(4x + 7)$

Step-by-step explanation:

Given

$-2(3x - 4) = -16$

$\frac{1}{2}(10x + 2) = 6$

$20- 15x = 6 - 2(4x + 7)$

$\frac{1}{3}(9x - 12) + 6=-10$

Required

Equations with $x = 4$ as a solution

$-2(3x - 4) = -16$

Divide both sides by -2

$3x - 4 = 8$

Add 4 to both sides

$3x - 4 +4= 8+4$

$3x =12$

Divide both sides by 3

$x = 4$

$\frac{1}{2}(10x + 2) = 6$

Multiply both sides by 2

$10x + 2 = 12$

Subtract 2 from both sides

$10x + 2 -2= 12-2$

$10x = 10$

Divide both sides by 1

$x = 1$

$20- 15x = 6 - 2(4x + 7)$

Open bracket

$20 - 15x = 6 - 8x - 14$

Collect like terms

$8x - 15x = 6 - 14 - 20$

$-7x = -28$

Divide both sides by -7

$x = 4$

$\frac{1}{3}(9x - 12) + 6=-10$

Subtract 6 from both sides

$\frac{1}{3}(9x - 12) + 6-6=-10-6$

$\frac{1}{3}(9x - 12) =-16$

Multiply both sides by 3

$9x - 12 = -48$

Add 12 to both sides

$9x - 12 +12= -48+12$

$9x = -36$

Divide both sides by 4

$x =-4$

$-2(3x - 4) = -16$ and $20- 15x = 6 - 2(4x + 7)$ have $x = 4$ has a solution