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. $\begin{cases}x^2 + 4y^2 = 20 \x + 2y = 6\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations. The first equation is X squared plus nine, Y squared equals 225. And the second equation is X - y equals 75. And we are asked to solve this system of equations. Well, I'm going to solve it using substitution. I'm going to add 21 Y to both sides of the second equation so that I'll have X by itself, X equals 75 plus 21 Y. And now wherever I have an X in the first equation here, I can substitute 75 plus Y. So the first equation will now read parentheses plus 21 Y and parentheses squared plus nine Y squared equals 225. And now I do need to foil this whole thing out, 75 plus 21 Y times plus 21 Y plus nine Y squared equals 225. And now 1st 75 times 75 is 5625. Yeah, I did these ahead of time. I did not memorize these and for outside times 21 Y is plus 1575 Y Insides likewise plus 575 Y and lasts 21 Y times 21 Y is plus 441 Y squared plus nine, Y squared equals 225. Now we've got a quadratic. So I want to have the right hand side of the equation equal to zero so that hopefully we can factor, I'm going to subtract 225 from both sides. And we need to combine like terms. So let's start with the highest degree. Let's start with the Y squared 441 Y squared plus nine, Y squared is 450 Y squared next with the wise 1575 Y plus 1575 Y is plus 3150 Y And -225 is plus 5400. That all equals zero. Now by just a wonderful miracle of a coincidence of a probably how it was designed. All of these terms are divisible by 450. So this will leave us with dividing everything by 450. It's actually just Y squared plus seven, Y plus 12 equals still zero. And look at that. We can factor that. What times what gives us why square? That's why times why what two numbers when multiplied, Give us a positive 12 and add up to a positive seven. That's going to be plus four plus three. Now we have two factors, we can set each of them equal to zero, Y plus four equals zero, Y plus three equal zero. And solve for Y. Subtracting four from both sides on the first one and subtracting three from both sides. On the second one I get y equals negative four for one solution and Y equals negative three. For another solution. Now I can set up a solution set and fill in the Y values. So the first one I've got Y equals negative four. And the second one I've got y equals negative three. And I can plug these Y values into either equation but I'm going to use the second one so X minus 21 Y equals 75. I'm going to plug the Y values in so that I can solve for X. So for the first one plugging negative four in for y, that's X minus 21 times negative four instead of y equals 75. And that's x negative 21 times negative four is going to be a positive 84. So X plus 84 equals 75. Subtracting 84 from both sides and that gives me X equals negative nine. So when y equals negative four, X equals negative nine. Now solving for the second one, solving for Y equals negative three. So substituting a negative three and for y x negative times negative three is going to be a positive 63. So X plus 63 equals 75. Subtracting 63 from both sides of the equation gives us X equals a positive 12. So when y equals negative three, X equals 12. We look at our answer choices and this matches with answer choice C. 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}\)x^2 + 4y^2 = 20 \\(\x\) + 2y = 6\(\end{cases}$

Key Concepts

Systems of Equations

Substitution Method

Solving Nonlinear Equations

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