How do you solve the following linear system: 4x+3y=8 , x-2y=13 4x+3y=8,x−2y=13?

1 Answer
Veddesh phiϕ
Jan 28, 2016

(x,y)=(5,-4)(x,y)=(5,−4)

Explanation:

4x+3y=84x+3y=8

x-2y=13x−2y=13

In the second equation we can find the value of xx by transposing 2y2y to the other side of the equation and we can substitute the value of xx to the other equation

Solve for second equation:

rarrx-2y=13→x−2y=13

Add 2y2y both sides:

rarrx=13+2y→x=13+2y

Substitute the value of xx to the first equation:

rarr4(13+2y)+3y=8→4(13+2y)+3y=8

rarr(52+8y)+3y=8→(52+8y)+3y=8

rarr52+8y+3y=8→52+8y+3y=8

rarr52+11y=8→52+11y=8

rarr11y=8-52→11y=8−52

rarr11y=-44→11y=−44

rarry=-44/11=-4→y=−4411=−4

So,substitute value of y to second equation:

rarrx-2(-4)=13→x−2(−4)=13

rarrx-(-8)=13→x−(−8)=13

rarrx+8=13→x+8=13

rarrx=13-8=5→x=13−8=5

So,(x,y)=(5,-4)(x,y)=(5,−4)

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