How do you solve #y=2x-5# and #4x-y=7# using substitution?

In order to solve a system of equations using substitution, you must express one variable in terms of the other in one equation, then use this value in the second equation.

Your system of two equations with two unknowns, #x# and #y#, looks like this

#{(y = 2x - 5), (4x-y = 7):}#

Notice that the first equation has #y# isolated on one side and expressed in terms of #x#

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

Use this value of #y# in the second equation to find the value of #x#

#4x - overbrace((color(blue)(2x-5)))^(color(purple)("= y")) = 7#

#4x - 2x + 5 = 7#

#2x = 2 implies x = 2/2 = 1#

Now take this value of #x# back into the first equation and find the value of #y#

#y = 2 * 1 - 5 = -3#

The solution to your system of equations is

#{(x=1), (y=-3) :}#

Notice that you can find the same values for #x# and #y# by substituting #x# instead of #y#. Isolate #x# in the second equation to get

#4x = 7 + y implies x = color(blue)((7+y)/4)#

Substitute this into the first equation to find the value of #y#

#y = 2 * overbrace((color(blue)((7+y)/4)))^(color(purple)("= x")) - 5#

#y = (7 + y)/2 - 5#

Multiply all the terms by #2#

#2y = 7 + y - 10#

Rearrange to find

#y = -3#

This will once again get you

#x = (7 + (-3))/4 = 4/4 = 1#