How do you solve 4x+3y=22 and 8x+y=58 using substitution?

1 Answer
May 28, 2018

x=19322andy=141011

Explanation:

4x+3y=22eqn1

8x+y=58eqn2

By Substitution Method..

From eqn2

8x+y=58eqn2

Making y the subject formula..

8x+y=58

Add both sides by 8x

8x+y+8x=58+8x

y=58+8xeqn3

Substituting eqn3 into eqn1

4x+3y=22eqn1

4x+3(58+8x)=22

4x+174+24x=22

4x+24x+171=22

28x+171=22

28x=22171

28x=193

x=19322

Substituting the value of x into eqn3

y=58+8xeqn3

y=58+8(19322)

y=58+154422

y=1276+154422

y=282022

y=141011

Therefore;

x=19322andy=141011