How do you solve the system #x+y=12# and #x-y=2#?

1 Answer
Mar 15, 2017

Answer:

#(7,5)#

Explanation:

Labelling the equation.

#xcolor(red)(+y)=12to(1)#

#xcolor(red)(-y)=2to(2)#

Note that adding + y and - y will eliminate them and leave an equation we can solve.

#"Adding "(1)+(2)"term by term gives"#

#(x+x)+(color(red)(+y-y))=(12+2)#

#rArr2x=14#

divide both sides by 2

#rArrx=7#

Substitute this value into either of the 2 equation, to find the corresponding value of y

#"Substituting in "(1)#

#7+y=12rArry=12-7#

#rArry=5#

#color(blue)"As a check"#

Substitute these values into both equations and if both sides are equal then they are the solution.

#"left side of "(1)to7+5=12to"true"#

#"left side of "(2)to7-5=2to"true"#

#rArr(7,5)" is the solution"#
graph{(y+x-12)(y-x+2)=0 [-11.1, 11.09, -5.55, 5.55]}