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

Question

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

Accepted Answer

Welcome back. I am so glad you're here. We are given these two equations. Were given x minus Y equals one and y equals x squared minus three X plus two. And we are asked to solve using substitution. Well we already have one of our equations set equal to Y. This Y equals x squared minus three X plus two. So where there's a Y. In the first equation we can substitute in this X squared minus three X plus two from the second equation. So we can rewrite equation one as x minus. Then X squared minus three, X plus two instead of x minus Y equals one. Alright now we just need to solve for X first we're going to distribute that negative sign to all three of the terms inside of parentheses. So X minus X squared plus three X. Because negative times negative and minus two equals one. Combine like terms are exes here. So we'll start off with that negative X squared and X plus three X is plus four. X minus two equals one. We've got a quadratic. So let's set this equal to zero. So that perhaps we can factor. So we've got negative X squared plus four, X minus three equals zero. And now let's divide everything by negative one. So that we don't have a negative X squared at the beginning. So this will now be a positive X squared minus four X plus three equals still zero. And now let's factor what times what gives us x square. That's an X times an X. What two numbers when multiplied? Give us a positive three. But when added together give us a negative four. That will be a minus three minus one. And now we can set both of these factors equal to zero to solve for X. So for x minus three equals zero we can add three to both sides and we get a solution of X equals three. There's one of our X is for x minus one equals zero. We can add three to both sides, Not three. We just did that on the previous one. Sorry, we can add one to both sides of the equation and then we get X equals positive one. So we have two X solutions. We draw fancy schmancy squiggly lines for our solution set and we know that our X comes first followed by ry so R two X solutions are three and one. Now we just need to find out what our why solutions are when X equals three and X equals one. And let's use equation one for that, X minus Y equals one. So substituting our three in for X here. So three -Y equals one. Well we can subtract three from both sides and then we have negative Y equals negative two and then divide both sides by negative one and we get y equals two. So when X equals three, then y equals two. Now again x minus Y equals one but we're going to solve for when X equals one. So one minus Y equals one. Okay, well we can subtract one from both sides. We get a negative Y equals zero. I mean to get y by itself, technically we divide both sides by negative one and we get Y equals zero divided by negative one is still zero. We get Y equals zero. So when X equals one, Y equals zero, we look at our answer choices and even though they've ordered their ordered pairs in a different order, this 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 + y = 2 \\(\y\) = x^2 - 4x + 4\(\end{cases}$

Key Concepts

System of Equations

Substitution Method

Quadratic Equations

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