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

1 Answer
Veddesh #phi#
Jan 28, 2016

#(x,y)=(5,-4)#

Explanation:

#4x+3y=8#

#x-2y=13#

In the second equation we can find the value of #x# by transposing #2y# to the other side of the equation and we can substitute the value of #x# to the other equation

Solve for second equation:

#rarrx-2y=13#

Add #2y# both sides:

#rarrx=13+2y#

Substitute the value of #x# to the first equation:

#rarr4(13+2y)+3y=8#

#rarr(52+8y)+3y=8#

#rarr52+8y+3y=8#

#rarr52+11y=8#

#rarr11y=8-52#

#rarr11y=-44#

#rarry=-44/11=-4#

So,substitute value of y to second equation:

#rarrx-2(-4)=13#

#rarrx-(-8)=13#

#rarrx+8=13#

#rarrx=13-8=5#

So,#(x,y)=(5,-4)#

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