We want all equations in terms of “y.” The first equation (the quadratic) is already in terms of y, but the second equation (linear equation) is not in terms of y.
Solving second equation in terms of y:
Add x to both sides:
y=-4+x
Use the commutative property if addition to rearrange the Rayo to y=mx+b form:
y=x-4
So, we have a quadratic function and a linear equation.
Pay close attention to the equations below:
y=x²-3x-4
y=x-4
Both equations are equal to y, and because a system of equation solves for interception points on the two functions (the point where the two functions contain the same x and y-coordinates), the y-coordinates must be equal to each other. Therefore, the equations can be set equal to each other:
(x-4)=x²-3x-4
Factor the quadratic:
(x-4)=(x+1)(x-4)
Now, let’s solve for x:
Divide both sides by (x-4):
(x-4)/(x-4)=[(x+1)(x-4)]/(x-4)
Simplify:
1=(x+1)
Subtract 1 to both sides:
1-1=x
0=x
Apply the symmetric property:
x=0
Now that we know the two equations intercept at x=0, let’s plug 0 back into “y” to find its value. Since both equations are set equal, plug x=0 in for either one.
y=(0)-4
y=-4
So, a solution set is (0, -4) and (4, 0)