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

Question

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}y^2 = x^2 - 9 \2y = x - 3\end{cases}$

Accepted Answer

Hey everyone welcome back in this problem. We are asked to solve the system using the substitution method. And the system we're given is y squared equals x squared minus four and three Y is equal to two, X plus four. And the first thing we're gonna do is just go ahead and label those equations. So we have equation one is y squared equals x squared minus four and equation two is three, Y is equal to two X plus four. Well you notice we have two equations and these are nonlinear equations. Okay, we have squared terms in equation one. So this is nonlinear system. Okay. We're asked to use the substitution method with the substitution method. The first thing we want to do is we want to choose one of the equations. We're going to isolate one of the variables in that equation and substitute it into the other equation. We can choose either equation, we can choose either variable. Now when we have nonlinear equations like one it can be a little bit more difficult to isolate one of the variables K. When you're taking square roots and it gets just a little bit messy. So let's go ahead and isolate a variable and equation two, that's going to be a little bit easier and then substitute it into equation one, we're gonna take equation too. We have three, Y equals two, X plus four and let's go ahead and isolate y. So we get y is equal to two thirds X plus four thirds. We're gonna call this equation to star. So it's the exact same as equation to. We've just divided by three all the way across case it gives the same information just been rearranged a little bit. And now we're gonna substitute that information about equation to star into equation one. Okay, this allows us to find solutions that satisfy both of those equations simultaneously. All right, so we're gonna get Y squared and r Y. Is now two thirds X plus 4/3. Okay, so all that squared is equal to X squared minus four. And what you notice is that with our substitution we now have just an equation with X. And so we can solve for X. Okay, so on the left hand side expanding, we get 4/9 X squared Plus 16/9 x Plus 16/9 is equal to X squared minus four. Okay, multiplying everything left and right hand side by nine, we're gonna get four X squared plus 16. X plus 16 is equal to nine, X squared minus 36. Alright, so let's give ourselves a little bit more room here Rearranging we get five x word minus 16, X minus 52 is equal to zero. Okay factoring this five X minus 26 times X plus two is equal to zero. Okay? And if you're on a test or something and you can't factor this, you're not seeing the factors are just not working out. Remember that you can always use the quadratic formula in order to find the roots if you need to. Okay so from this, From the left hand route we're going to get X is equal to 26/ and from this X plus two we're gonna get X is equal to negative two for a route. Ok, alright so we found two values for X. Now in order to have a solution we need to have corresponding values for y. Okay so let's go ahead and substitute these back into our equation. We can choose to substitute it back into equation one or equation two. Okay in this case let's choose to substitute into equation two star because we've already simplified it, it's going to make it easier and it's just gonna be a little bit less work for us. Okay so again you can do either equation in this case we're choosing to so substitute X equals, we'll start with the first 26/ into equation two star we get y is equal to 2/3 times x which is 26/ plus four thirds. This is gonna give us 52 over 15 plus 4/3 can be written as 20/15. Okay so we have a common denominator We get 72/15 Which is 24/5. And so we can write this um point that we found as 26/5 comma 24/5 and we see in the possible answers were given that they've changed everything into decimal so you just want to go ahead and change these into a decimal as well? Okay so this is the x value is 5.2 and the y value is four point. Okay, so we found one point here. Now we need to go ahead and find the other point. So now let's substitute the second X value X equals -2. We're gonna substitute it into equation two star again as we get y is equal to two thirds times X which is minus two plus four thirds. Okay this is gonna give us negative four thirds plus four thirds which is equal to zero. Okay and so we get the point negative two and zero. Alright so writing the solution, the solution is going to be the set of points that we found which is going to be negative. 20 And 5.2. 4.8. Okay, so that is our solution. It's two points and that's going to correspond with answer B. Thanks everyone for watching. See you in the next video

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}\)y^2 = x^2 - 9 \\2y = x - 3\(\end{cases}$

Key Concepts

Substitution Method

Solving Quadratic Equations

Checking Solutions in Systems

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