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. $\begin{cases}3x^2 + 4y^2 = 16 \2x^2 - 3y^2 = 5\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations. The first equation is two, X squared plus three, Y squared equals 19. And the second equation is three, X squared minus two, Y squared equals nine. We are asked to solve for the system of equations. Well, since we have variables with the same degrees, I am actually going to use the addition method to solve the system of equations. I'm going to multiply everything in the first equation, times two and everything in the second equation times three which will get our y squared terms to cancel out. So multiplying everything in the first equation, times two that will be a two times two X squared which is four, X squared plus six, Y squared equals 19 times two, which is 38. Multiplying the second equation times 33 X squared times three is nine, X squared minus six, Y squared equals three times nine is 27. Now adding these two equations together, four x squared plus nine, X squared is 13 X squared. The y squared is canceled out and 38 plus 27 is 65. Now, to solve for X squared, dividing both sides by 13 gives us x squared equals 65 divided by 13 is five. And so for x squared equals five. We need to take the square root of both sides and that gives us X equals plus or minus the square root of five. Alright, we've got two X solutions. So I'm going to start a solution set up here and so our two solutions for x are a negative square root of five and a positive square root of five. And now we can plug this back into either equation. I'm going to choose the second equation. And so we'll have to do um three X squared minus two, Y squared equals nine. Will have to do this two times because we have two X solutions. So for the first one this will no longer be three times X squared but it will be three times Negative Square root of five squared minus two, Y squared equals nine. Now this negative square root of five square that's taking the negative square to five times the negative squared of five and the negative times the negative. That's a positive a positive one. Now it's like a negative one times negative one. So it's a positive one. And that's the square root of five times five, which is 25. So this equals five. So now this reads three times five minus two, Y squared equals nine. Well three times five is 15, so 15 minus two, Y squared equals nine. And we can subtract 15 from both sides trying to get that y by itself. So we have negative two, Y squared equals nine minus 15 which is negative six, Dividing both sides by -2 gives us a y squared on the left equals negative six divided by negative two is a positive three. Now again we need to take the square root of both sides and we get y equals plus or minus the square root of three. So when X equals negative square to five, we actually have two different solutions for why? Why can equal a negative square root of three? Or why can equal a positive square root of three. So now let's substitute in that positive square root of five for X. And it's going to be a very similar process to solve for y. Here. Again, the square to five squared is going to be just a positive five. And now look this looks identical to what we're doing on the left. So this will be 15 minus two. Y squared equals nine. We are subtracting 15 from both sides and we've got negative two, Y squared equals negative six. We are dividing both sides by negative two. And we've got why squared equals a positive three. We're taking the square root of both sides. And we have got why equals again plus or minus the square root of three. So that tells us that when X equals a positive square root of five, why could equal either a negative squared of three or a positive square root of three. And this is our solution set. We look at our answer choices in this matches with answer choice. D 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}\)3x^2 + 4y^2 = 16 \\2x^2 - 3y^2 = 5\(\end{cases}$

Key Concepts

Systems of Equations

Solving Systems by Substitution or Elimination

Handling Quadratic Equations in Systems

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