How do you solve the system of equations #y= 3x + 8# and #2y = 10x - 8#?

1 Answer
Jim G.
Mar 24, 2018

#(x,y)to(6,26)#

Explanation:

#y=3x+8to(1)#

#2y=10x-8to(2)#

#"multiply the terms in equation "(1)" by 2"#

#rArr2y=6x+16to(3)#

#"equations "(2)" and "(3)" now express x in terms of 2y"#
#"so we can equate the right sides of the equations"#

#rArr10x-8=6x+16#

#"subtract 6x from both sides"#

#10x-6x-8=cancel(6x)cancel(-6x)+16#

#rArr4x-8=16#

#"add 8 to both sides"#

#4xcancel(-8)cancel(+8)=16+8#

#rArr4x=24#

#"divide both sides by 4"#

#(cancel(4) x)/cancel(4)=24/4#

#rArrx=6#

#"substitute "x=6" into equation "(1)" for y-coordinate"#

#rArry=(3xx6)+8=18+8=26#

#rArr"the solution is "(x,y)to(6,26)#

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