How do you solve the system of equations #2x+3y=8# and #5x+4y=2#?

Multiplying the first equation by #5# and the second one by #-2#
we obtain

#7y=36# so

#y=36/7#

plugging this in the second equation we get

#5x+4*26/7=2#

so

#5x=2-4*36/7#
note that #2-4*36/7=-130/7#

so

#x=-26/7#

Equation 1: #2x+3y=8#

Equation 2: #5x+4y=2#

I am going to use the substitution method.

Solve Equation 1 for #x#.

#2x=8-3y#

Divide both sides by #2#.

#x=8/2-(3y)/2#

Substitute the value for #x# into Equation 2 and solve for #y#.

#5(8/2-(3y)/2)+4y=2#

Expand.

#40/2-(15y)/2+4y=2#

Multiply both sides by #2#.

#40-15y+8y=4#

Subtract #40# from both sides.

#-15y+8y=4-40#

Simplify.

#-7y=-36#

Divide both sides by #-7#.

#y=(-36)/(-7)# #larr# Two negatives make a positive.

#y=36/7# or #~~5.143#

Substitute #36/7# for #y# in the first equation and solve for #x#.

#2x+3(36/7)=8#

#2x+108/7=8#

Multiply both sides by #7#.

#14x+108=56#

Subtract #108# from both sides.

#14x=56-108#

Simplify.

#14x=-52#

Divide both sides by #14#.

#x=(-52)/14#

#x=-52/14#

Reduce.

#x=-26/7# or #~~-3.71#

The solution is the point #(36/7,-26/7)# or #(~~-3.714,5.143)#.

graph{(2x+3y-8)(5x+4y-2)=0 [-13.875, 6.125, 0.2, 10.2]}