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Solve. Remember that graphs can be used to confirm all real solutions.

$$

\begin{aligned}

&x^{2}-y^{2}=16\\

&x+y^{2}=4

\end{aligned}

$$

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Campbell University

Oregon State University

Baylor University

Idaho State University

Okay, so here were provided this system of equations were asked to solve. And the easiest way to solve this would be through the elimination method. Because we can easily eliminate our wife's webs. We have a negative westward him positive, I swear. So those cancel each other out. Now we just treat this way like a a regular edition problem of X word plus X equals 16 plus fours. 20. And let's go ahead and get everything on the same side of our equation so you can have X squared plus X minus 20. He goes to zero. Go ahead and factor this out here, x x something. 11 multiple to get a equals negative. 21 added together equals one. Well, five times four equals negative. 20 and five minus four is in the equal one. So we're gonna have a positive five negative for Let's go ahead and set. Our X is equal to zero, so X plus five equals zero X minus four equals zero. Let's sell for X on both sides. X system equal a negative five and positive for Go ahead and plug this exact into another equations. I'll plug it into my second equation. So when X is negative five Last y squared this for go ahead and add five on both sides so I can get my westward by itself. Y squared equals nine and take the square root. Why is unequal plus or minus three? And let's do another one So we'll have four les y squared equals for it's got a minus form Both sides Get away by itself My scored equal zero Take the square root Well, why was gonna equal zero? So when X equals negative five are wise and be positive three. But when X equals and go five are us are wise also gonna be negative three And when our exes for our why is zero So you have three answers there And here's our solution ups to make it better Here's our solution