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

Question

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}x + y = 1 \x^2 + xy - y^2 = -5\end{cases}$

Accepted Answer

Hey everyone in this problem. We are asked to solve the following system of equations Using the substitution method and the equations were given as X plus two, Y is equal to two and X squared plus X Y minus three, Y squared is equal to negative three. Okay, so this is a system of two equations and their nonlinear. So let's go ahead and label these equations. Equation one is X plus two, Y is equal to two. An equation two is X squared plus x Y minus three. Y squared equals negative three. Okay. Alright. We're asked to use the substitution method. And what we want to do in the substitution method is choose one of the equations and one of the variables. Okay, we're gonna isolate that variable in one of the equations and then substitute it into the other equation. So when we look at the equations, when we're in when we have nonlinear equations, we have to be careful. Okay. An equation to you see that we have this this cross term of X. Y. In the middle. So isolating one of these variables gets a little bit messier um than in the linear case. So let's go ahead and choose equation one. That one will be more simple. So equation one, we have X plus two. Y equals two. So we can write X is equal to two minus two Y. Okay, let's call this equation one star so that we can refer to it. So it's the exact same equation as equation one. It's just been rearranged. All right, so now we're gonna substitute one star K. That X equals to minus two Y into equation two. Okay, this is our substitution method. We're substituting into equation two so that we're finding the solution that solves both equations. That satisfies both equations. And what we get when we substitute is we have X squared which is going to be two minus two. Y squared plus x Y two minus two. Y times y minus three. Y squared equals negative three. Alright, so expanding. We're gonna have four minus eight. Y my a plus four Y squared. Sorry? Plus two Y minus two. Y squared minus three. Y squared equals negative three. Okay so on the left hand side we get four minus six, Y minus y squared equals negative three. And moving everything to one side we end up with. Why squared last six? Y minus seven equals zero. Okay, moving down just a bit. So we can factor this. Okay factoring is going to give us y plus seven times y minus one is equal to zero and this gives us a Y values negative seven and positive one. Okay, Alright, so we found these y values but in order to have a solution we need to find the corresponding X values. Okay, so to find the corresponding x values, we need to substitute this back into the equation. Now when we're substituting it back in we can choose either equation equation one or equation too because it has to satisfy both. Okay let's go ahead and choose equation one because that one just looks easier to work with. Okay so we'll choose our first X value or first y value. So y equals one. We're gonna substitute it into equation one. And equation one again is given right here. So we're substituting here, X plus two times Y which is two times one is equal to two. So X plus two is equal to two. This gives us X. Is equal to zero. Okay so have an X value of zero corresponding with the Y value of one. So this gives us the 10.0 comma one. Okay. Doing the same thing with our second Y value substituting Y equals negative seven into equation one. We have X plus two times negative seven is equal to two. X minus 14 is equal to two. So X is equal to 16. Okay so we have another X value and that corresponds with the y. Value of negative seven. So we get the 70.16 negative seven. Okay so if we want to write our solution set we'll write it up here at the top. Okay? We have to include the two points that we found in the two points that we found our zero comma one and 16 comma negative seven. Ok this is going to correspond with answer D and that's it for this one. Thanks everyone for watching. See you in the next video

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}\)x + y = 1 \\(\x\)^2 + xy - y^2 = -5\(\end{cases}$

Key Concepts

Substitution Method

Linear Equations

Quadratic Equations

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