Sign In
Does either of the equations have an isolated variable in it?
(6,- 2)
When solving a system of equations using the Substitution Method, there are three steps.
For this exercise, y is already isolated in both equations, so we can skip straight to solving! Since the expression equal to y in the second equation is simpler, let's use that for our initial substitution.
(I):y= 2x-14
(I): LHS-x=RHS-x
(I): Commutative Property of Addition
(I): Subtract terms
(I): LHS+14=RHS+14
(I): Add terms
Great! Now, to find the value of y, we need to substitute x=6 into equation (II) in the given system.
(II): x= 6
(II): Multiply
(II): Subtract term
The solution, or point of intersection, to this system of equations is the point (6,- 2).
To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.
(I), (II): x= 6, y= - 2
(II): Multiply
(I), (II): Subtract term
Because both equations are true statements, we know that our solution is correct.