You can put this solution on YOUR website! and
Finding point(s) of intersection is what solving systems of equations is about. With your system we can use the Substitution Method. Both equations are already solved for y. So we can substitute the first equation's expression for y into the second equation giving:
This is a quadratic equation (because if the squared terms). But before we start solving, let's get rid of the fraction by multiplying both sides by 2:
which simplifies to:
Now we can proceed. With quadratic equations we want one side to be zero. So I'll subtract the entire right side from both sides. (If this confuses you, just subtract the three terms one by one.)
Next we factor (or use the Quadratic Formula). This factors into
From the Zero Product Property we know that this product is zero only if one of the factors must be zero. So: or
Solving these we get: or
At this point it is natural to feel success. But all we've done so far is find the x coordinates of the points of intersection. We need the y coordinates for each x. We use one of the original equations to find the y's. I'll use the first one so that I can avoid the fraction in the second equation:
Find the y for x = -8/3:
So one point of intersection is (-8/3, 140/9)
Find the y for x = 4:
So the second point of intersection is (4, 0)
(