How do you solve x-2y=4 and 8y+24x=5 using substitution?

1 Answer
Mar 10, 2016

{(x=3/4),(y=-13/8):}

Explanation:

Given:

{(x-2y=4, "S1"),(8y+24x=5,"S2"):} <=> {(x-2y=4, "S1"),(24x+8y=5,"S2"):}

The Substitution solving method tells us that it's possible to find the value of one of the two variables from one of the two equations, then substitute it in the other one.

In your excercise it's easy to find x value from S1 as follows:

{(x=4+2y, "S1"),(24x+8y=5,"S2"):}

Now we can substituite it in S2

{(x=4+2y, "S1"),(24(4+2y)+8y=5,"S2"):}

{(x=4+2y),(96+48y+8y=5):}<=>{(x=4+2y),(56y=5-96):}<=>{(x=4+2y),(56y=-91):}

{(x=4+2y),(y=-(91/56)=-13/8):}

Now we can substitute the y value in S1 to find x value:

{(x=4+cancel(2)(-13/cancel(8)^4)),(y=-13/8):}

{(x=4-13/4=(16-13)/4=3/4),(y=-13/8):}

:.{(x=3/4),(y=-13/8):}

In a graph the system solution is the point where S1 intercepts S2

graph{(x-2y-4)(24x+8y-5)=0 [-3.347, 6.52, -4.167, 0.766]}