How do you solve #y = 4x + 3# and #2x + 3y = -5# using substitution?

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Answer 1 · 2016-03-21T13:26:25

#x=y=-1#

#color(blue)(y=4x+3#

#color(blue)(2x+3y=-5#

In the first equation substitute the value of #y# to the second equation

#rarr2x+3(4x+3)=-5#

Use distributive property #color(brown)(a(b+c)=ab+ac#

#rarr2x+12x+9=-5#

#rarr14x+9=-5#

#rarr14x=-5-9#

#rarr14x=-14#

#rArrcolor(green)(x=-14/14=-1#

Substitute the value of #x# to the first equation

#rarry=4(-1)+3#

#rarry=-4+3#

#rArrcolor(green)(y=-1#

#:.color(blue)(ul bar |x=y=-1|#