#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#
