### Video Transcript

Find all the real solutions to the system of equations four π¦ plus π₯ squared minus four π₯ minus 40 equals zero and two π¦ minus seven-halves π₯ squared minus six π₯ plus four equals zero.

The solution of these systems of equations is the place, if we graphed both of these equations, where they intersect. And when we say that algebraically, weβre looking for the places where they are equal to one another. Since we have been given two equations set equal to zero, one way to solve them algebraically would be to take these two equations and set them equal to each other.

There is one reason I wouldnβt want to use this strategy. And thatβs because I have negative seven-halves π₯ squared as one of the terms. And so I go back and I look at that equation. And I see that if I multiply the entire equation by two, we eliminate having to deal with a fractional coefficient of our π₯ squared term. We can rewrite our second equation as four π¦ minus seven π₯ squared minus 12π₯ plus eight equals zero. And once we do that, we see that we have four π¦ and four π¦ in both of these equations. This makes me think that I would want to solve with the elimination method.

If we want to eliminate the π¦-term, we can multiply our new equation through by negative one, which gives us the equation negative four π¦ plus seven π₯ squared plus 12π₯ minus eight equals zero. And then we have two equations where the π¦-terms are additive inverses of one another. If we line these two equations up vertically so that the terms are aligned with the like terms, we can add these two equations together. Four π¦ plus negative four π¦ equals zero. π₯ squared plus seven π₯ squared is eight π₯ squared. Negative four π₯ plus 12π₯ equals eight π₯. And negative 40 plus negative eight equals negative 48.

Using this new equation, we can find the π₯-values that make this equation true. These will be the places where the real solutions will occur. First, we recognize that all three of these terms are divisible by eight. And so we can divide through by eight. We now get a simplified form of π₯ squared plus π₯ minus six equals zero. Since we have a quadratic here, we know that there could be two solutions. And we want to try and factor to find the solutions.

We have our two π₯-terms, which when multiplied together equal π₯ squared. And from there, we need two values that when multiplied together equal negative six but when added together equal positive one. I know that three times two equals six. If one of these values is negative and one is positive, then weβll get negative six. And if we add negative two and three together, we get positive one. So our two factors are π₯ minus two and π₯ plus three.

To solve for π₯, we set both of these factors equal to zero and find that is true when π₯ equals two or when π₯ equals negative three. This is saying if we plug in π₯ equals two into both of our original equations, the resulting π¦-value should be the same for each of them. We wanna check if thatβs true and find out what that π¦-value would be. If we plug in π₯ equals two to our first equation, we end up with four π¦ minus 44 equals zero, which is four π¦ equals 44. And that shows that π¦ equals 11.

This means that when π₯ equals two in our first equation, π¦ equals 11. And we expect that to be the case for our second equation as well. But which of these three should we use? Theyβre all three equivalent expressions for the second equation we started with. I would choose to use the one without the fractional coefficient of π₯ squared. But any of these three equations will work.

We plug in two, which gives us four π¦ minus 28 minus 24 plus eight equals zero, which reduces to four π¦ minus 44. This is the same thing we had in our first equation, which simplifies to four π¦ equals 44, and that π¦ equals 11. This confirms one of the solutions, but we need to repeat this process for π₯ equals negative three. When we plug in π₯ equals negative three to the first equation, we get four π¦ plus nine plus 12 minus 40 equals zero, which is four π¦ minus 19 equals zero. We add 19 to both sides. And if four π¦ equals 19, then π¦ will equal 19 divided by four, which is 4.75.

Now, we check our second equation, expecting π¦ to be equal to 4.75 again. When we plug in negative three for π₯, we get four π¦ minus 63 plus 36 plus eight equals zero, which is four π¦ minus 19 equals zero. And we see that weβre working with the same equation four π¦ equals 19, which makes π¦ equal to 4.75. This confirms our second solution. And if we wanted to write them in the more common coordinate form, we would say that there are solutions at two, 11 and negative three, 4.75.