How do you solve this system of equations: #y= - 3x + 3 and 3x - 2y = - 3#?

1 Answer
Jim G.
Dec 12, 2017

#(x,y)to(1/3,2)#

Explanation:

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

#3x-2y=-3to(2)#

#color(blue)"substitute "y=-3x+3" into equation "(2)#

#3x-2(-3x+3)=-3#

#rArr3x+6x-6=-3#

#rArr9x-6=-3#

#"add 6 to both sides"#

#9xcancel(-6)cancel(+6)=-3+6#

#rArr9x=3#

#"divide both sides by 9"#

#(cancel(9) x)/cancel(9)=3/9#

#rArrx=1/3#

#"substitute this value into equation "(1)#

#rArry=(-3xx1/3)+3=-1+3=2#

#rArr"the point of intersection "=(1/3,2)#
graph{(y+3x-3)(y-3/2x-3/2)((x-1/3)^2+(y-2)^2-0.04)=0 [-10, 10, -5, 5]}

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