How do you solve the system #x+y=12# and #x-y=2#?
1 Answer
Mar 15, 2017
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]}