In Exercises 1–18, solve each system by the substitution metho...

Question

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}y = x^2 - 4x - 10 \y = -x^2 - 2x + 14\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given two equations. First equation Y equals three, X squared minus three X minus eight. And the second equation Y equals negative two X squared plus seven X plus seven. And we are to solve using substitution. Well, both of these already have Why? Equal for them? And we know that. Why is equal to Why? So actually the right hand side for each equation is equal to the right hand side. For the other equation. So we can set the two right hand sides equal to each other. We can say that three X squared minus three X minus eight equals negative two X squared plus seven X plus seven. And now we can solve for X. Well, I am going to start off by bringing all of the things on the right hand side over to the left hand side so that it will be equal to zero on the right hand side. So adding two X squared to both sides, subtracting seven X from both sides and subtracting seven from both sides. So now the left hand side will read five, X squared minus 10, X minus 15 equals And on the right hand side that will be zero. Well, every single term here is divisible by five. So I'm going to divide everything by five which will leave us with X squared minus two X minus three equals zero. And now we can factor the left hand side. What times what gives us X squared. That's X times X. And what? Two numbers when multiplied? Give us a negative three. But add up to a negative two, that would be a minus three plus one. We can set both of these factors equal to zero to solve for X. For the first one, x minus three equals zero will add three to both sides And we will get x equals three. There's one solution and for X plus one equals zero. We'll subtract one from both sides and we'll get X equals negative one. There's another solution. So we've solved for X and we can set that up in a solution set with a fancy squiggly line. They're mine has some extra squiggles and we know that we have the X value and then the Y value and we have our x values X can be three or X can be negative one but we need to figure out what are y values are when X equals three and when X equals negative one. So we're going to go back up to our original equations and I'm going to use equation one where y equals three X squared minus three x minus eight and we can first substitute a three in four X. So that will read why equals three times three squared -3 times three minus eight. So three times three squared, well three squared is nine, so three times nine is 27 minus three times three is nine, so minus nine minus eight, 27 minus nine is 18 minus another eight is 10. So when X equals three then y equals 10. Now we can go through that whole process again and substitute in negative one for X. So that would be why equals three times negative one squared -3 times negative one. I don't know why I did that. The parentheses in black for the first time for each one, not for the second. I don't know. And then -8 and three times negative one squared negative one squared is positive one. So we have three times positive one which is three and then negative three times negative one is a positive three. So three plus three minus eight. Well three plus three is six and six minus eight is negative two. So when X equals negative one then y equals negative two. Well, we look at our answer choices and even though the ordered pairs are in a different order, this matches with answer choice B Well done. We'll catch you on the next one.

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}\)y = x^2 - 4x - 10 \\(\y\) = -x^2 - 2x + 14\(\end{cases}$

Key Concepts

System of Equations

Substitution Method

Quadratic Equations

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