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Solve the system by either the substitution or the elimination method.$$\left\{\begin{array}{l}{\frac{m}{4}+\frac{n}{3}=-\frac{1}{12}} \\{\frac{m}{2}-\frac{5}{4} n=\frac{7}{4}}\end{array}\right.$$
(1,-1)
Algebra
Chapter 4
Systems of Linear Equations and Inequalities
Section 3
Solving Systems of Equations by Elimination (Addition)
Equations and Inequalities
Systems of Equations and Inequalities
Campbell University
Oregon State University
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okay. I think it would be great to start this problem by eliminating the fractions in both equations. And so I'm going to multiply the first equation by 12 and the second equation by four. So that gives us three M plus four n equals negative one and to M minus five in equal seven. Now, if we're going to continue with elimination, we want opposite terms on em or opposite terms on end. And we could go either way. I think what I'll do is go ahead and multiply the first equation by five and the second equation by four. And there are other options that would work. And this way I can eliminate the N terms. So now my equations are 15 m plus 20 n equals negative five and eight. M minus 20 n equals 28 and we can add these together and the end terms cancel and we have 23 m equals 23 that works out nicely. M equals one now to find the value of N. We can take either equation one here or equation to and substitute our M value in there. And so I'm going to go with the equation. one doesn't matter which and I have three times EMS, though three times one plus four n equals negative one. So three plus four n equals negative one. Subtract three from both sides and four n equals Negative four divided by four and we get an equals negative one. Now, before I finish, I should check my answer by putting these numbers into equation to the other equation. So that would be two times m. So two times one minus five times in five times negative one equals seven. So this would be to plus five equals seven. And that checks. So now we write her answer as an ordered pair m comma in negative one. Sorry. Positive one first for M common negative one m in one negative one.
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