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).
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