Question

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

Answer 1

Given:

The system of equations is

$8x+9y=36$

$3x+4y=16$

To find:

The solution of the given system of equations using the substituting method.

Solution:

We have,

$8x+9y=36$           ...(i)

$3x+4y=16$            ...(ii)

From (ii), we get

$3x=16-4y$

$x=\dfrac{16-4y}{3}$          ...(iii)

Putting this value in (i), we get

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

Multiply both sides by 3.

$8(16-4y)+27y=108$

$128-32y+27y=108$

$-5y=108-128$

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

$y=4$

Putting y=4 in (iii), we get

$x=\dfrac{16-4(4)}{3}$

$x=\dfrac{16-16}{3}$

$x=\dfrac{0}{3}$

$x=0$

Therefore, the solution of given system of equations is (0,0).