How do you evaluate this system of equations: #y-2x=-5; y-x=-3#?
1 Answer
Jun 14, 2017
Explanation:
- To solve this specific system, the easiest method is substitution.
#y-2x=-5#
#y-x=-3# - Choose one of the equations .. I will pick
#y-x=-3# . - The objective is to get one of the variables (
#x# or#y# ) by itself on one side of the equation .. Pick a variable to isolate .. I will pick#y# . - To isolate
#y# , ADD#x# to both sides of the equation:
#y-x=-3#
#y=-3+x# - Look at the OTHER equation (the one you did not choose) .. That would be
#y-2x=-5# for me.
See how#y# is EQUAL to#(-3+x)# ?
You can substitute the#y# in#y-2x=-5# with#(-3+x)#
It should look like this:#(-3+x)-2x=-5# - SIMPLIFY! Now we only have ONE variable in the equation.
#(-3+x)-2x=-5#
#-3-x=-5#
#-x=-2#
#x=2# [THIS IS THE ANSWER FOR X] - Take one of the equations and substitute
#x# for#2# to solve for#y# .
#y-2x=-5#
#y-2(2)=-5#
#y-4=-5#
#y=-1# [THIS IS THE ANSWER FOR Y]
If your teacher wants it as
