Question

Solve the system of linear equations below, using substitution method.

Answer 1

Answer:

$x = 0$

$y=4$

Step-by-step explanation:

Given

$8x + 9y = 36$

$3x + 4y = 16$

Required

Solve using substitution method

Make x the subject in $3x + 4y = 16$

$3x = 16 - 4y$

Divide through by 3

$x = \frac{1}{3}(16 - 4y)$

Substitute $\frac{1}{3}(16 - 4y)$ for x in $8x + 9y = 36$

$8*\frac{1}{3}(16 - 4y) + 9y = 36$

$\frac{8}{3}(16 - 4y) + 9y = 36$

Open bracket

$\frac{8}{3}*16 - \frac{8}{3}*4y + 9y = 36$

$\frac{8*16}{3} - \frac{8*4y}{3} + 9y = 36$

Multiply through by 3

$3 * \frac{8*16}{3} - 3 *\frac{8*4y}{3} +3* 9y = 36*3$

$8*16 - 8*4y +3* 9y = 36*3$

$128- 32y +27y = 108$

$128-5y = 108$

Collect Like Terms

$-5y = 108-128$

$-5y = -20$

Make y the subject

$y = \frac{-20}{-5}$

$y=4$

Substitute 4 for y in  $x = \frac{1}{3}(16 - 4y)$

$x = \frac{1}{3}(16 - 4*4)$

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

$x = \frac{1}{3}(0)$

$x = 0$