#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.
