How do you solve the system of equations #y+ 3= 2x# and #y + 4= 3x#?

1 Answer
May 14, 2018

Answer:

#x=1,y=-1#

Explanation:

Equation 1: #y+3=2x#
Equation 2: #y+4=3x#

Isolate the variable #y# in equation 1, then plug that into equation 2.
#\color(red)(y=2x-3)#
#\color(red)(y)+4=3x\rarr(\color(red)(2x-3))+4=3x#

Now simplify (add like terms, then solve for #x#)
#2x+1=3x#
#\color(blue)(1=x)#

Plug the value of #x# back into equation 1.
#y+3=2\color(blue)(x)\rarry+3=2(\color(blue)(1))#

Simplify to solve for #y#
#y=-1#


Checking your answers
Plug the values into either one of the original equations to see if they are true.
Equation 1: #(\color(red)(-1))+3\stackrel(?)(=)2(\color(blue)(1))# becomes #2=2#, so correct
Equation 2: #(\color(red)(-1))+4\stackrel(?)(=)3(\color(blue)(1))# becomes #3=3#, so correct