Precalculus, Chapter 7, 7.3, Section 7.3, Problem 20

EQ1: 2x-y=0
EQ2: x-y=7
To solve this system of equations, let's apply substitution method. Let's isolate the x in the second equation.
x - y=7
x=7+y
Then, plug-in this to the first equation.
2x - y=0
2(7+y)-y = 0
And solve for y.
14+2y-y=0
14+y=0
y=-14
Now that the value of y is known, solve for x. Let's plug-in y=-14 to the second equation.
x -y=7
x-(-14)=7
x+14=7
x=7-14
x=-7

To check, plug-in x=-7 and y=-14 to one of the original equations. Let's use the first equation.
2x-y=0
2(-7) - (-14)=0
-14 + 14=0
0=0 (True)

Therefore, the solution is (-7,-14).