In Exercises 47–52, solve each system by the method of your ch...

Question

In Exercises 47–52, solve each system by the method of your choice. $\begin{cases}2x^2 + xy = 6 \x^2 + 2xy = 0\end{cases}$

Accepted Answer

Hey, everyone. Here we are asked to work out the following system of nonlinear equations where the first given equation is X squared minus nine X Y is equal to 12. And the second equation is X squared plus nine have X Y is equal to zero. Here we have four answer choice options. Answer choice A X is equal to negative two, Y is equal to four nines and X is equal to two, Y is equal to negative four nines. Answer choice B X is equal to two, Y is equal to four nines and X is equal to negative two, Y is equal to negative four nights. Answer choice C X is equal to four, N Y is equal to two and X is equal to next negative four N Y is equal to negative two. And finally answer choice D X is equal to negative four, N Y is equal to two and X is equal to four N Y is equal to negative two. So to begin solving this problem, we first want to solve both of our given equations for X Y. So beginning with our first equation which is we recall is X squared minus nine X Y is equal to 12. So again, here we want to solve for X Y. So first, we just want to subtract X squared from both sides and we get negative nine X Y is equal to 12 minus X squared. Now we want to get rid of the coefficient for X Y by dividing both sides by negative nine. And we get that X Y is equal to the quantity of X squared minus 12 divided by nine. Now, we want to repeat this process for the second equation which was X squared plus nine halves, X Y is equal to zero. So again, solving for X Y, we first subtract X squared on both sides. And we have nine halves, X Y is equal to negative X squared. And now dividing both sides by nine halves to get X Y alone, we get that from our second equation. We have X Y is equal to negative two NS X squared. So now that we have solved both equations for X Y, we can simply set them equal to each other and solve for X. So we have from the first equation, the quantity of X squared minus 12 divided by nine is equal to. And our second equation gives us negative two ns X squared. So now to begin solving for X, we first need to notice that both sides of the equation have nine in the denominator. So both nines cancel. And now rewriting, we have X squared minus 12 is equal to negative two X squared. Now we can move our negative two X squared to the left side of the equation by adding two X squared to both sides of our equation. So we have three X squared minus 12 is equal to zero. So now we just want to factor out X so that we can find our X values. So we have three times the quantity of X minus two times the quantity of X plus two is equal to zero. Now, we can solve four values for X by setting each set of parentheses equal to zero individually. So first, we have X minus two is equal to zero. And then we will have X plus two is equal to zero. So solving our first equation, we can just add two to both sides and we get one value for X as positive two. And then from our second equation, we subtract two from both sides. And we see that our other value for X is negative two. So now with these values for X that we obtained, we can just plug them into one of our given equations and solve for our Y values. So we can begin by taking our answer of positive two for X and plugging it into our first equation for example. So again, using our first equation and plugging in two for X, we have two squared minus nine times two times Y is equal to 12. Now we just need to solve for Y so two squared is just four, which means we can just subtract four from both sides and we get negative nine times two is 18. So we have negative 18, Y is equal to 12 minus four, which is just eight. Now we can divide both sides by negative 18 to get Y alone. And simplifying, we see that one of our values for Y is negative four nines. And now we can repeat this process using the negative two value that we have for X. The other solution we have for X. And so again, plugging this value into our first equation, we now have the quantity of negative two squared minus nine times negative two times Y is equal to 12. And now just simplifying our values, we have four minus nine, negative nine times negative two actually gives us plus. So we have four plus 18. Y is equal to 12. Now just solving four Y, we first subtract four from both sides and we have 18, Y is equal to eight. Now just dividing both sides by 18 to get Y alone, we see that our other value for Y is positive for ns. So now we see that our solutions are first for when we plug in positive two X, we get two and negative four nines. So we have two and negative four nines as one solution. And then we have negative two and positive four nines as our second solution. So just writing out these two solutions here, we see that the first solution again is when X is equal to two and Y is equal to negative four nights. And then our second solution is when X is equal to negative two and Y is equal to positive four nights. So with our two solutions, we are left here with answer choice. A thanks for watching. I hope you found this video helpful and I'll see you again next time.

In Exercises 47–52, solve each system by the method of your choice. $\begin{cases}\)2x^2 + xy = 6 \\(\x\)^2 + 2xy = 0\(\end{cases}$

Key Concepts

Systems of Nonlinear Equations

Substitution Method

Factoring and Quadratic Equations

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