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 - 1)^2 + (y + 2)^2 = 10\end{cases}$

Accepted Answer

Hey everyone in this problem we are asked to solve the system of equations given below. Using the substitution method. And the equations were given our x minus Y equals negative seven and X plus three L squared plus y minus four L squared is equal to eight. Okay. The first thing we're gonna do is label these equations. So we have equation one is x minus Y equals minus seven. An equation to X plus three all squared plus y minus four. All squared is equal to eight. Okay. Labeling them is just gonna help us refer to them later. Alright. We're told to use the substitution method and in the substitution method, what we wanna do is we want to take one of the equations, one of the variables. Okay we want to isolate a variable and then substitute it into the other equation. Okay so we can choose either equation one or equation too and we can choose either extra Y. To isolate. Now what we wanna do, we have to be careful, this is these are nonlinear equations. Okay so if you look at say equation to we're going to have X squared terms and we're gonna have X terms. So it can be hard to isolate the variable. So let's go ahead and start with equation one. Okay that one looks a little bit simpler to isolate. So we'll use equation one. So if we take equation one, X minus Y equals seven equals negative seven. Story and we can rearrange this. Okay and get X is equal to negative seven plus Y. Okay and let's call this equation one star because we want to refer to this equation. So we're gonna call this equation one star. And now we can do our substitution method. We've isolated one of the variables from equation one. Now we can substitute it into equation two so that we find solutions that solve both of these equations that satisfy both equations and solve the system. Okay, so let's substitute one star into equation two. This is going to give us so we have X plus three. In this case X is negative seven plus Y plus three squared plus y minus four squared is equal to eight. So this is going to give us oops, y minus four squared plus y minus four squared is equal to four. Oops is equal to eight, sorry two y minus four squared is equal to eight. Okay And now we can divide by two. Let me move down so we have a bit more room to work, we can divide by two and we're gonna be left with y minus four squared is equal to four. Okay? Alright so a couple of things we can do, we can take the square root here or if we didn't see that um we can expand the left hand side and solve, so let's do it that way this time. So we have y squared minus eight X plus 16 on the left hand side. We have four on the right hand side. Okay? And I wrote eight X. And that should be eight. Y. We don't have any excess in this equation. Right now we're just working with why we're trying to solve for y. Okay so we have y squared minus eight. Y. Moving the four to the left minus 12 is equal to zero. Okay? And we need a bit more room so if we factor this were confined factors y minus two times y minus six is equal to zero. Which gives y values of two or six. Okay so we find two different y values. Now this isn't the solution. Remember for a solution we need X values and y values. So we're gonna take those Y values and find the corresponding X values. Now in order to find the corresponding X values. We're gonna substitute into our equations. Now we can substitute into either equation one or equation too and let me move up here and I'm just gonna write what? Whoops. I'm just gonna write Y is equal to two or six. Okay those values that we found up here so that we can refer to them. Alright Let's substitute into equation one So that one's a little bit simpler. So we'll substitute into equation one. So we're gonna substitute our first y value of two into equation one and we're gonna get X minus Y. Which is two is equal to negative seven. Okay so X is equal to negative five. Well this gives us the point we've seven in the y Value to we got an X. Value of minus five. So this gives us the point negative 52. Alright. Moving to the next equation or the next y value substitute Y equals six and we can substitute it into equation one. Again we're gonna have X minus Y. Which is x minus six is equal to negative seven, so X is equal to negative one. Okay? So now we have the point negative 16. Alright so we have two points that we found. So we can write our solution set and our solution set is going to be the set of the points we found negative five and two. A negative one and six. Okay. And this is going to correspond with answer C. Thanks everyone for watching. I hope this video helped see you in the next one.

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

Key Concepts

System of Equations

Substitution Method

Solving Quadratic Equations

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