In Exercises 29–42, solve each system by the method of your ch...

Question

In Exercises 29–42, solve each system by the method of your choice. $\begin{cases}x^2 + y^2 + 3y = 22 \2x + y = -1\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given the system of equations and asked to solve it. Our first equation in our system is X squared minus eight X plus Y squared equals nine. Our second equation is three X plus Y equals 17. Alright, I am going to solve this system of equations using substitution. I'm going to subtract three X from both sides of equation two. So now equation two will read Y equals 17 minus three X. And I can use substitution by anywhere. There's a Y in equation one, I can put in 17 minus three X. So now equation one will read X squared minus eight X plus and now not Y squared but it will be 17 minus three, X squared equals nine. So bringing down the X squared minus eight X. I need to foil out this 17 minus three X times 17 minus three X. And so our firsts will be 17 times 17 is 289. Outside 17 times negative three X is minus 51 X. Inside same thing minus 51 X. And our lasts negative three X times negative three X. Is plus nine X squared equals nine. Now we've got a quadratic here. So I want to bring this nine over to the left hand side. So it will be equal to zero on the right. And I need to combine all of my like terms. So my X squared we have X squared plus nine X squared, gives me 10 X squared. Now all of my X is negative eight X minus 51 X minus another 51 X. That will give me a minus X. And the 289 minus nine will be plus 280 all equal zero. Well everything here is divisible by 10. So let's do that. 10 X squared divided by 10 is x squared minus X is minus 11, X plus 280 is plus equals zero, divided by 10 is still zero. Now I can factor this what times what gives me X squared That's X times X and what? Two numbers when multiplied give me a positive 28 but when added together give me a negative 11. That would be a minus four and a minus seven. So now I have two factors, I can set them both equal to zero, X minus four equals zero and x minus seven equals zero. Getting the X by itself will add four to both sides for the first one and add seven to both sides for the second one. So I'll get to solutions, I'll get X equals a positive four and I'll get X equals a positive seven. And now I can set up a solution set and plug in my X values. So I've got an X can be four and an X can be a seven and now I'm going to use equation 23, X plus Y equals 17. To plug in my ex values and solve for Y. So now instead of three times X it'll be three times four plus why not another four plus Y equals 17. So this is 12 plus Y equals 17. Subtracting 12 from both sides, get Y equals 17 minus 12 is five. So when X equals four, Y equals five. Now plugging in the seven for X three times seven plus Y equals 17, 3 times seven is 21 plus Y equals 17. Subtracting 21 from both sides, Y equals 17 minus is negative four. So when X equals seven, Y equals negative four, we look at our answer choices and this matches with answer choice. A Well done. We'll catch you on the next one.

In Exercises 29–42, solve each system by the method of your choice. $\begin{cases}x^2 + y^2 + 3y = 22 \\2x + y = -1\end{cases}$

Key Concepts

Solving Systems of Equations

Substitution Method

Quadratic Equations and Circles

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