How do you solve by substitution #x + y = 15# and #4x + 3y = 38#?

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Answer 1 · 2015-06-06T22:11:25

Subtract #x# from both sides of the first equation to get:

#y = 15 - x#

Substitute this expression into the second equation:

#38 = 4x+3y = 4x+3(15-x)#

#=4x+45-3x=x+45#

Subtract #45# from both ends to get:

#x = 38-45 = -7#

Then

#y = 15 - x = 15 - (-7) = 15+7 = 22#

Answer 2 · 2015-06-06T22:20:40

Separate one variable in one equation and substitute it on the other one and find the solution.

Let's separate the variable that results easier in the easiest equation. We're doing this with the #x# on the first equation:

#x=15-y#

And substitute it on the second one:

#4(15-y)+3y=38 -> 60-4y+3y=38 -> -y=38-60-> y=22#

And now, with the #y# we just found, calculate the #x# we previously isolated:

#x=15-y -> x=15-22 -> x=-7#