color(red)(-14)=color(green)(-20y)\quadcolor(blue)(-7x)
color(green)(10y)+color(red)(4)=color(blue)(2x)
Manipulate the equations so the like terms (same colored) are in the same place:
color(green)(-20y)\quadcolor(blue)(-7x)=color(red)(-14)
color(green)(10y)\quadcolor(blue)(-2x)=color(red)(-4)
To eliminate a variable, they have to be opposites. We can eliminate color(green)(y) here by multiplying the bottom equation by 2. The top equations stays as it is.
2(color(green)(10y)\quadcolor(blue)(-2x))=color(red)((-4))2
Simplifying:
color(green)(20y)\quadcolor(blue)(-4x)=color(red)(-8)
Now the equations can be added together:
\qquadcolor(green)(cancel(-20y))\quadcolor(blue)(-7x)=color(red)(-14)
+\qquadcolor(green)(cancel(20y))\quadcolor(blue)(-4x)=color(red)(-8)
——————————
color(blue)(-11x)=color(red)(-22
Solving for color(blue)(x:
color(blue)(x)=2
Now that we know the value of color(blue)(x), we can plug it into one of the equations to solve for color(green)(y).
color(red)(-14)=color(green)(-20y)\quadcolor(blue)(-7x)
color(red)(-14)=color(green)(-20y)\quadcolor(blue)(-14
Isolating for color(green)(y):
0=color(green)(-20y)
color(green)(y)=0
To confirm our answers plug the values of color(blue)(x=2) and color(green)(y=0) into an equation:
color(green)(10y)+color(red)(4)=color(blue)(2)
color(green)(0)+color(red)(4)=color(blue)(4)
4=4
So the answers are correct.
