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. $\begin{cases}\frac{3}{x^2} + \frac{1}{y^2} = 7 \\frac{5}{x^2} - \frac{2}{y^2} = -3\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations. The first equation is nine divided by x squared plus three divided by y squared equals six. And the second equation is three divided by x squared minus six divided by y squared equals negative five. And to solve for the system of equations, I am going to take the first equation and multiply it by two. And then use the addition method to get my y squared terms to cancel out and then I can solve for X. So taking the first equation times to nine over X squared times two is 18 over X squared plus three over Y squared times two is six over Y squared Equals six times 2 is adding the second equation in the blue equation together. Three over X squared plus 18 over X squared is over X squared the y squared terms cancel out. And then this equals negative five plus 12, which is seven. And then using cross multiplication here to quickly get rid of the fraction. This will be seven, X squared equals 21 Dividing both sides by seven gives us x squared equals three. Taking the square root of both sides. And we get X equals plus or minus the square root of three. So X can equal a positive squared of three or x can equal a negative squared of three. And now we need to solve for the corresponding y values. So we can choose either equation from our original system of equations. I'm going to choose the 1st 19 over X squared plus three over y squared equals six. And I'm going to plug in those X values that we know to see what the corresponding Y values are. So let's color code this solving for the positive square root of three. First for our X value. So this is now nine over the squared of three squared plus three over y squared equals six. And the square root of three squared is just three. So this is 9/3 plus three over y squared equals six. And well 9/3. That's three. So we've got three plus three over y squared equals six. And now we can subtract three from both sides and this will give us three over y squared equals six minus three is three. Again we can use cross multiplication to get rid of the fraction real quick. And so we'll have three, Y squared equals three, dividing both sides by three. We get y squared equals three divided by three is one. Taking the square root of both sides. And that's why equals plus or minus the square root of one is one. So when X equals the square root of three, why can equal a positive one? Or why can equal a negative one? Alright, now let's plug in for the negative square root of three for x. So we'll have nine over negative squared of three squared instead of x squared plus three over Y squared equals six. And now if we've got negative squared of three times negative squared of three, well the negative times the negative that's like a negative one times a negative one in front of that square root in front of the radical. That is going to be a positive one. So this is going to equal a positive square root of nine which is positive three. So this is going to be much the same. This is 9/3. Again, plus three over y squared equals six. And actually from here on out it's going to be identical problem solving steps and we're going to come up with why equals plus or minus one when X equals the negative square root of three. So again, why can equal one or why can equal negative one? Now we look at our answer choices and this matches with answer choice B. Well done. We'll catch you on the next one.

In Exercises 47–52, solve each system by the method of your choice. $\begin{cases}\]\frac{3}{x^2}\) + \(\frac{1}{y^2}\) = 7 \\\(\frac{5}{x^2}\) - \(\frac{2}{y^2}\) = -3\(\end{cases}$

Key Concepts

Systems of Equations

Substitution and Elimination Methods

Handling Rational Expressions and Variables in Denominators

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