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. $\begin{cases}2x^2 + y^2 = 18 \xy = 4\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given the system of equations. The first equation is X squared plus two, Y squared equals 38. And the second equation is X times Y equals negative six. We are asked to solve this system of equations. Well, I want to solve it using substitution. This is kind of a pick your poison. We can either deal with fractions or we can deal with square roots. And since we'll probably be working back and forth, we'll have to deal with both. So I'm going to start off dealing with fractions. I am going to divide both sides by why in the second equation. So I'll have X by itself and X will equal negative six divided by Y. And I am going to substitute negative six divided by Y, wherever there's an X. In the first equation which is right here at the beginning. So now the first equation will read parentheses negative six divided by Y instead of X. This will be squared plus two, Y squared equals 38. And now I need to square that fraction on the left hand side. So negative six squared is 36. And why squared is y squared plus two? Y squared equals 38. Now, I want to get rid of that fraction. So I'm going to multiply everything by the lowest common denominator, which is why squared. So the white squares will cancel out on that first one and the fraction will be gone. Which ye but now we've got 36 plus two Y to the fourth, distributing that Y squared equals 38 Y squared. Okay, well now let's bring over this 38 Y squared to the left hand side. And then let's rewrite everything with their degrees in descending order. So starting with two Y to the fourth minus 38 Y squared plus equals zero. Okay, now we've got this looking mostly in quadratic form And everything here is divisible by two. So I'm going to divide everything by two. So this will now read why to the fourth minus 19, Y squared plus 18 equals still zero. And this is factory ble So I've got what times what gives me Y to the fourth. That's why squared times Y squared and what two numbers when multiplied. Give me a positive 18 but add up to a negative 19. That is going to be negative 18. And negative one. And looking at this factor on the right. Why squared -1? I recognize that that is a difference of squares. So I can factor that out recalling previous lessons on differences of squares. We take the both terms and take the square root of them. So the square root of y squared is why? So the y goes in the first position in each set of parentheses, The square root of one is one. So a one goes in the second position in each set of parentheses and by definition one is plus one is minus. So now I have three factors and I can set each one of them equal to zero. Why squared minus 18 equals zero. Why plus one equals zero and y minus one equals zero. And I can solve for y for each of these. And so for this first one, I'm going to add 18 to both sides and I get why equals a positive 18. Excuse me? Why squared equals a positive 18? Why equals 18 would have been nicer. But now we've got to take the square root of both sides and it's y equals plus or minus and 18. Will that in a factor tree? That's two times +99 is three times three. So this would be three square root two. So one of our solutions for why is plus or minus three square root two. Now solving for the second equation here, subtracting one from both sides and we get y equals negative one. And now for the third equation adding one to both sides and we get a solution of why equals a positive one. Okay, we've got four solutions for why. I'm going to scroll back up to where we can see the whole thing and I'm going to start a solution set and we need four different solutions here for why. And now I can scroll down to where we can see those solutions and plug them back in. So we've got y equals plus or minus three squares of two. So we'll start with negative three square root of two. For one of our y values. And then we have a positive three square root of two. And then we also have Y equals negative one and Y equals a positive one. So now we can plug our Y values into one of our original equations to solve for X. And I'm going to use the X times Y equals negative six equation. I'll write it four times because we've got oops X times Y equals negative six because we have four different Y values to solve for their corresponding X values. And so for this first one we've got X times no longer Y but it's negative three square root of two equals a negative six. And we'll need to divide both sides by negative three squared of two to get X by itself. And that will give us X equals negative six divided by negative three squared of two. Well the negative six and the negative three reduce that negative three is a one and the negative six is a two. So now this equals two divided by the square root of two. And we want to get the radical out of the denominator. So we're going to multiply both. I'll change colors, multiply both the numerator and the denominator by the squared of two. So squared of two over the square root of to that's really just one. And now this will read X equals two times the square root of two in the numerator. And in the denominator, the squirt of two times the square root of two is the squirt of four, which is two. So we've got two squared of two divided by two. Well those two's cancel out. They reduced to ones. And so we have just the square root of two. So when Y equals negative three squared of two, X equals the square root of two. Okay, we got one of them. Now this next one will be very similar. So this is X times no longer Y but now it's three squared of two Equals A - six. And to get the X by itself will divide both sides by three squared of two. So we'll have X equals negative 6/3 squared of two. These reduce the three becomes a one and the negative six becomes a negative two. And we'll need to multiply. Well here I'll say this equals -2 divided by the squared of two. And again we need to multiply both sides, both the top and the bottom. Excuse me by the square root of two. In order to get that radical out of the denominator. So we have X equals negative two times the square root of two in the numerator. And squared of two times squared of two is the square to four which is two in the denominator. And then again these reduce the twos reduced to ones. So we have a negative square root of two. So when Y equals positive three squared of two, X equals negative square root of two. Okay, we've got two of them. These next two are not quite as complex, so for X times Y we've got X times not why? But we're substituting in a negative one. So we have X times negative one equals negative six. We can divide both sides by negative one and we get X equals a positive six. Just draw some lines here in case this is too confusing. Okay so we solved that when Y equals negative one, X equals six. And now we can substitute into this last one, X times just a positive one for the Y value. While X times one is just X. So we've got X equals negative 61 Y equals one. Alright. We did it Well done. We look at our answer choices and this matches with answer choice B. We'll catch you on the next one.

In Exercises 29–42, solve each system by the method of your choice. $\begin{cases}\)2x^2 + y^2 = 18 \\(\xy\) = 4\(\end{cases}$

Key Concepts

Systems of Equations

Nonlinear Equations

Substitution Method

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