Exercises 57–59 will help you prepare for the material covered...

Question

Exercises 57–59 will help you prepare for the material covered in the next section. Solve: $\begin{cases}A + B = 3 \2A - 2B + C = 17 \4A - 2C = 14\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given three equations and we are to work out the values of A. B and C. I'm going to start off by writing those equations. We have A plus B equals two and I left a bunch of space there because this next one has three terms five A plus eight B minus C equals one. And then we have six A. And then space because there's no B term minus three C equals zero. I'll label these as equations 12 and three. And what we're going to do is we're going to work on these equations a little bit to get some of our terms to cancel out and then we'll write new equations. So the first term I want to cancel out is the C. Term. And I'm going to do that by taking the second equation here and multiplying it by a negative three and then adding that to our third equation. So first I'm going to write what the second equation times negative three would be so negative three times five A is negative 15 A negative three times or positive eight B is negative 24 B negative three times negative C. Is plus three. C equals negative three times one is negative three. And now I'm going to add that red row two, equation three to give us new equation 46 a minus A. Is minus nine A. Bring down that -24 b. And negative three C plus three. C cancels out that +00 plus negative three is negative three. Right now I've got a new equation four that only has A. And B terms just like equation one. So I'm going to do some work on equation one and then add that to equation four to get a new equation five, I'm going to take equation one and multiply it by 24 to get RB terms to cancel out. 24 times A. Is 24 A. 24 times B is plus 24 B. and 24 times two is 48, adding equation four plus the blue is negative nine A plus 24 A. Is 15 A negative 24 B plus 24 B. That's nothing they cancel out negative three plus 48 is 45. Now I've got 15 a equals 45. And I can solve for a. I'll divide both sides by 15 And we've got a equals three. Now I can solve for B. And for C by substituting that A back in. So if A. Is three, I'm going to take the first equation and substitute three in for A. So that'll be three plus B equals two and subtracting three from both sides. We now have B Equals 2 -3 is -1. Alright. We solved for A. And B. And now we can solve for C. I am going to substitute that A equals three into our third equation here. So that will give us a six times three minus three C equals 06 times three is 18. So we have 18 minus three C equals zero. I will subtract 18 from both sides and that reads negative three C equals negative 18 And divide both sides by -3 and C equals negative 18 divided by negative three C equal six. We've solved for A B and C. You can always go back in and check your solutions by plugging A. B and C into the original three equations. So for the first equation A plus B three plus negative one equals two. Yes that's true. For the second equation five times three plus eight times negative one minus six as C equals 1. 15 minus eight minus six equals one. Yes that's true. And the third equation six times three is a minus three times six is C equals 0, 18 minus 18 equals zero. Yes that's true. So we know that our solutions work. We look at our answer choices in this matches with answer choice. A. Well done. We'll catch you on the next one.

Exercises 57–59 will help you prepare for the material covered in the next section. Solve: $\begin{cases}\)A + B = 3 \\2A - 2B + C = 17 \\4A - 2C = 14\(\end{cases}$

Key Concepts

Systems of Linear Equations

Substitution and Elimination Methods

Manipulating and Combining Equations

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