#4x+3y=8#

#x-y=10#

To determine the values of #x# and #y#, we must first determine from one equation a temporary value for either of the terms, and then substitute that temporary value in the other equation.

We determine a temporary value for #x# from the second equation.

#x-y=10#

Add #y# to both sides.

#x=10+y#

In the first equation, **substitute** #x# with #color(red)((10+y))#.

#4x+3y=8#

#4color(red)((10+y))+3y=8#

Open the brackets and simplify.

#40+4y+3y=8#

#40+7y=8#

Subtract #40# from both sides.

#7y=-32#

Divide both sides by #7#.

#y=-32/7=-4 4/7#

In the second equation, **substitute** #y# with #color(blue)(-32/7)#.

#x-y=10#

#x-(color(blue)(-32/7))=10#

Open the brackets and simplify. The product of two negatives is a positive.

#x+32/7=10#

Multiply all terms by #7#.

#7x+(7xx32/7)=70#

#7x+(1cancel7xx32/(1cancel7))=70#

#7x+32=70#

subtract #32# from each side.

#7x=38#

Divide both sides by #7#.

#x=38/7=5 3/7#