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

##### 1 Answer

#### Answer:

#### Explanation:

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,

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

Notice that the first equation has

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

Use this value of *second equation* to find the value of

#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

#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

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

Substitute this into the first equation to find the value of

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

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

Multiply all the terms by

#2y = 7 + y - 10#

Rearrange to find

#y = -3#

This will once again get you

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