Answer:
[tex]x = 0[/tex]
[tex]y=4[/tex]
Step-by-step explanation:
Given
[tex]8x + 9y = 36[/tex]
[tex]3x + 4y = 16[/tex]
Required
Solve using substitution method
Make x the subject in [tex]3x + 4y = 16[/tex]
[tex]3x = 16 - 4y[/tex]
Divide through by 3
[tex]x = \frac{1}{3}(16 - 4y)[/tex]
Substitute [tex]\frac{1}{3}(16 - 4y)[/tex] for x in [tex]8x + 9y = 36[/tex]
[tex]8*\frac{1}{3}(16 - 4y) + 9y = 36[/tex]
[tex]\frac{8}{3}(16 - 4y) + 9y = 36[/tex]
Open bracket
[tex]\frac{8}{3}*16 - \frac{8}{3}*4y + 9y = 36[/tex]
[tex]\frac{8*16}{3} - \frac{8*4y}{3} + 9y = 36[/tex]
Multiply through by 3
[tex]3 * \frac{8*16}{3} - 3 *\frac{8*4y}{3} +3* 9y = 36*3[/tex]
[tex]8*16 - 8*4y +3* 9y = 36*3[/tex]
[tex]128- 32y +27y = 108[/tex]
[tex]128-5y = 108[/tex]
Collect Like Terms
[tex]-5y = 108-128[/tex]
[tex]-5y = -20[/tex]
Make y the subject
[tex]y = \frac{-20}{-5}[/tex]
[tex]y=4[/tex]
Substitute 4 for y in [tex]x = \frac{1}{3}(16 - 4y)[/tex]
[tex]x = \frac{1}{3}(16 - 4*4)[/tex]
[tex]x = \frac{1}{3}(16 - 16)[/tex]
[tex]x = \frac{1}{3}(0)[/tex]
[tex]x = 0[/tex]
