Question Video: Solving Systems of Quadratic Equations Algebraically

Find all the real solutions to the system of equations 𝑦 = π‘₯Β² βˆ’ 7, 𝑦 = (π‘₯ + 1) (π‘₯ βˆ’ 5).


Video Transcript

Find all the real solutions to the system of equations 𝑦 equals π‘₯ squared minus seven and 𝑦 equals π‘₯ plus one times π‘₯ minus five.

When we talk about real solutions to systems of equations, we’re talking about the points where they intersect with each other. When we have a system of quadratic equations like this one where the highest exponent is π‘₯ squared, they can have no solutions, meaning they do not intersect. They can have one solution, meaning they intersect once, or two solutions, meaning they intersect twice. The places where they intersect will be the places where their π‘₯- and 𝑦-values are equal to each other.

Of course, we sometimes solve this by graphing both of the equations. But today, we’re gonna look at how we would solve them algebraically. And to do that, we set them equal to each other. We want to know for what π‘₯-value would this equation produce the same 𝑦-value? When would it be true that π‘₯ squared minus seven is equal to π‘₯ plus one times π‘₯ minus five? And now, we’ll try to solve for π‘₯. We first need to expand these two factors. When we do that, we get π‘₯ squared minus five π‘₯ plus π‘₯ minus five. We can combine these π‘₯-terms to have π‘₯ squared minus four π‘₯ minus five.

At this point, we’ll try to get all the π‘₯’s on the same side of the equation. If we subtract π‘₯ squared from both sides, that term is canceled on both sides of the equation. And we’re left with negative seven is equal to negative four π‘₯ minus five. Again, we’re trying to isolate π‘₯. So we add five to both sides, which gives us negative two equals negative four π‘₯. We divide both sides of the equation by negative four, which gives us two over four. And we can simplify that to be one-half.

Based on this, we expect there to be one real solution to occur when π‘₯ equals one-half or when π‘₯ equals 0.5. Either way is fine. What we want to do now is check that it’s true. If we substitute one-half in for π‘₯ in both of these equations, the outcome of 𝑦 should be equal. One-half squared is one-fourth. We have one-fourth minus seven. We can rewrite seven as 28 over four. And then when we subtract, we end up with negative twenty-seven fourths. So for our first equation, when π‘₯ equals one-half, 𝑦 equals negative twenty-seven fourths.

Now we want to solve our second equation. We can rewrite one as two over two and rewrite five as 10 over two. One-half plus two-halves equals three-halves and one-half minus ten-halves is negative nine-halves. To do this multiplication, we multiply the numerators then multiply the denominators, which gives us negative twenty-seven fourths. In both of these cases, when we plug in one-half for π‘₯, 𝑦 equals negative twenty-seven fourths or, in its decimal format, negative 6.75. We’re saying if we graphed both of these equations at the point one-half, negative twenty-seven fourths, both of their graphs would cross this point.

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