In Exercises 47–48, solve each system by the method of your choic...

Question

In Exercises 47–48, solve each system by the method of your choice. (x - y)/3 = (x + y)/2 - 1/2(x + 2)/2 - 4 = (y + 4)/3

Exercise 48: Two variable system of linear equations to solve.

Accepted Answer

Welcome back. I am so glad you're here. We are given these two equations. In a system of equations. The first one is X plus Y divided by five equals x minus y, divided by three minus four. And the second equation is x minus one divided by three plus Y plus one divided by four plus seven, divided by six equals zero. And we get to use any method, we want to solve this. But oh my goodness, look at all of these fractions. Okay, what I want to do first is do some work on each of these equations, get them to look a little bit less fraction E and then we can then we can figure out what method we want to solve to use to solve this system of equations. So for equation one, let's get rid of these fractions by multiplying everything by the lowest common denominator. Our denominators are five and three. So the lowest common denominator is 15, whole multiply all three of these by 15 and then the five and the 15 cancel out to one and three. The three in the 15 cancel out to one and five. So now rewriting this, we have a three times X plus y equals a five times x minus y -4 times 15 is 60. Now we can distribute the three and the five, so three times x plus y is three X plus three, Y equals five times X minus y is five, x minus five Y. And then bring down that minus 60. And let's bring these three X. The three X over to the right hand side and the three y over to the right hand side and Actually let's bring the 60 over to the left hand side. So we'll have 60 equals five x minus three X. Which is two X. And negative five, Y minus three, Y. Is minus eight Y. And that looks like a much nicer problem to deal with. Now let's do the same thing for equation two, let's get rid of these fractions lowest common denominator 434 and six is 12. So we're going to multiply all four of these times 12 and then do our reductions 12 and three reduces to one and four. Four and 12 reduces to one and three and six and 12 reduces to one and 2 12 times zero is just going to be zero. And so now we've got four times x minus one plus three times Y plus one and plus two times 14 which are two times seven which is 14. Now we can distribute the four and the three, so four times x minus one is four x minus 43 times Y plus one is plus three, Y plus three plus 14 equals zero. And we can bring that four X. And the plus three Y. And now just to combine all of these constant terms, so we have a negative four plus three plus 14. That is plus 13 equals zero. And then let's bring that 13 over to the right hand side. So we've got our X term for X plus or Y term. Three Y equals our constant term negative 13. Okay now we can stack these up as a much nicer system of equations. So one putting the X and Y terms on the left hand side. We've got two X minus eight. Y equals 60. And our equation two is four, X plus three Y equals negative 13. This is so much nicer. I am going to use the addition method. I am going to multiply the first equation times negative two and then add that to the second equation to get the X terms to cancel out. So negative two times two X is negative for X negative two times negative eight Y is plus 16, Y Equals -2 times 60 is negative. 120. Now adding these together. The X terms cancel out. Oops, I kind of crossed off the wise. Okay, the X terms cancel out three, Y plus 16, Y is 19, Y equals negative 13 plus negative 120 is negative. 133 Can divide both sides by 19 and we get one of our solutions, Y equals negative 133 divided by 19 is negative seven. Yeah, we got one. We got ry now to solve for X. The easier one to use would probably be this second equation and we can substitute that back and up here for our Y term because there's only one Y term and then we're not dealing with two wise or two exes. So this would be four times x minus one. I'll distribute that four again. So four x minus four plus three times not why in the parentheses there, but negative seven plus one plus 14 equals zero. And doing what's inside the parentheses there is negative seven plus one will give us a negative six. And now distributing that three that three times negative six will give us a minus 18. And now it's just combining all of these constant terms. So negative four minus 18 is negative plus 14, that's minus eight. So we have four X minus eight equals zero. We can add 8 to both sides, write this off to the side a little bit more. We get four X equals eight, dividing both sides by four and we get X equals two. We have solved for X and Y. All right. We look at our answer choices and this matches with answer choice. A Well done. We'll catch you on the next one

In Exercises 47–48, solve each system by the method of your choice. (x - y)/3 = (x + y)/2 - 1/2(x + 2)/2 - 4 = (y + 4)/3

Key Concepts

Systems of Linear Equations

Substitution Method

Elimination Method

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