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

Question

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}x^2 + y^2 = 25 \x - y = 1\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given a system of equations with two equations here, X squared plus Y squared equals 10. And our second equation X plus Y equals two. And we are asked to solve this using substitution. Alright, well we've got options but let's solve for Y here will subtract X from both sides of the equation. So our new equation two will read Y equals two minus X. And now we can substitute this in for Y. In the first equation will put in two minus X. Where that Y is in the first equation. So our new equation one will read X squared plus no longer Y squared but now two minus X squared equals 10. And so we can bring down that first X squared plus And now we do have to foil this out to minus X times two minus X equals 10. So we have X squared plus and our first two times two is four, our outsides two times negative X is minus two. X insides negative X times two is minus two X. And our last negative X times negative X is plus X squared still equals 10. Okay, now we can combine like terms we have two X squared, so that is two X squared and we've got minus two X minus two X. That's going to be minus four X plus four equals 10. Well, everything here is divisible by two. So I'm just going to divide all of our terms by two. So now we'll have X squared minus two, X plus two equals five. And because we've got a quadratic here, let's set this equal to zero. We'll subtract five from both sides of the equation. So we'll have x squared minus two. X and two minus five is minus three equals zero now. And we can factor this. So what times what gives us X squared, that's X times X and what two numbers when multiplied, give us a negative three. And when added together give us a negative two, that would be a minus three plus one. And we can set both of those two factors equal to zero. So we'll have x minus three equals zero and X plus one equals zero and solving for x minus three equals zero will add three to both size and we'll get an answer of X equals three. There's one part and now for X plus one equals zero, will subtract one from both sides and we'll get X equals negative one. Okay, we have two X solutions. If we're writing this as a solution set, it's the beautiful squiggly and then we know our X solutions. The x's go in the first spot. So one X solution will be three and one X solution will be negative one. And now we have to figure out what the y solutions are for each of those X solutions. So we have X, then y. And so now let's go back to equation too and let's plug in our X solutions and find out what are applicable why solutions are. So four X equals three. Will plug a three in for X. And then plus Y equals two. And to solve for y will subtract three from both sides and we'll get y equals two minus three is negative one. So when X equals three, Y equals negative one, that is a crooked one. Alright Y equals negative one. Now we will go with equation to again and we'll plug in negative one for X to see what are y solution will be so still negative one plus Y equals two and we can add one to both sides and we'll get why equals two plus one is three. So when X equals negative one, why equals three. And that is our solution set. We look at our answer choices and they just put them in a different order. The ordered pairs in a different order. But that matches with answer choice C. Well done. We'll catch you on the next one.

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}\)x^2 + y^2 = 25 \\(\x\) - y = 1\(\end{cases}$

Key Concepts

Substitution Method

Solving Quadratic Equations

Systems of Nonlinear Equations

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