Help with "Solving a system of linear equations using elimination with multiplication"

[150mph]

Hi all. First post. I’m working in ALEKS learning system Prep for College Algebra through my online university and their explanations are often lacking. Hope this is correct place to post.

Pls. see red arrow on attached image where I’m confused:

“We’ll multiply the first equation by 7 and the second by -6.”

I think I get why I’d multiply by ( –6 ) in the second equation - because it’s the opposite of the positive 6 next to the variable in the first equation. Maybe this isn’t why it’s done, because why wouldn’t I multiply the parts of the 1st equation by ( – 7 ) rather than (7 ) ???

What do I need to know to do this kind of problem correctly?

TIA

 

[tkhunny]

Some texts will tell you to SUBTRACT the two equations. If this is the case, then you should multiply by +7 and +6 -- and then subtract.

Some texts will tell you to ADD the two equations. If this is the case, then you should multiply by +7 and -6 (or -7 and +6) -- and then add.

Personally, I think there is a lot more to go wrong with subtraction. Thus, picking the sign of the multiplication that will allow addition to work is the better way to go.

 

[150mph]

Thanks. The question in my addition version is : How do I look at the 2 equations in the Problem, and decide which positive or negative numbers go in which place to come up with "We’ll multiply the first equation by 7 and the second by -6.”

 

[Steven G]

I agree with tkhunny that there is a lot more to go wrong with subtraction. for some students. But in my opinion if this is true for you then you should practice subtraction. It is not always about getting the right answer (by avoiding your weak points) but rather what you learn while arriving at the right answer. tkhunny is a tutor and I respect that greatly ( I have been a tutor for over three decades) but there is a difference between being a tutor and a teacher. A tutors goal is to get the best grade out of the student on their test, while (imo) a teachers job is to teach the material well. Yes, it is slight difference but it is there.

On this forum I made the decision early on that I would behave more like a teacher than a tutor (although on a site like this it is not always easy to do).

To do your problem you just want to get the same number in front of x (or the y) while ignoring signs. If the signs end up the same than you subtract, if they are different then you add. IMO you should put no energy at all in determining whether to multiply, for example, by 12 or -12. Just multiply by 12

 

[tkhunny]

This is why you practiced the Least Common Multiple (LCM) in your previous studies! Make sure that idea is stuck very firmly in your brain. It is your moment to let those skills shine!

Slight Digression:

For one who has ALWAYS advocated making sure you can deal with WHATEVER you encounter, it is difficult to see my recommendations construed as avoidance of things that have more to go wrong. Let that not ever be said of me. If there are 10 ways to proceed, then get good at all 10. Might one of those 10 be the easiest and the least likely to go wrong? Yes. Shall we avoid the other nine if this is the case? No. One cannot control what one is asked to do in an academic setting, and even less so (or maybe less voluntarily) in an employment setting.

There is a third option, other than tutor or teacher, as has be described above. The other option is programmer. When writing computer programs, one really does have to pick how to to code whatever it is. You can code to make choices, at times, but certainly not always. There are a few things relevant to this discussion to consider when coding:

1) Does one way take more or fewer calculations? If you're doing 10,000,000 calculations, even the choice between addition and subtraction can make a difference. It's not a matter of avoiding one way or another, it's about getting the job done.

2) Does one way take more or less CPU? or RAM? Most things are easy and ridiculously fast, these days, and no one cares anymore about such considerations. Once in a while, though, something pops up where old-school conservative coding is the way to go.

3) What sort of precision are you looking for? We hold to the dream of infinite decimal places when we write numbers, but when a computer sees it, it only goes so far. This is nicely annoying for things as common as sevenths or ninths. When subtraction prevails, you may find things that you think are zero, but the internal comparison has a different opinion.

4) We just had in this forum an interesting question about numerical solutions of difficult problems. The prevailing methodology requires a fraction to be calculated. The challenge is to make sure the denominator of that fraction stays away from zero. Subtraction is substantially more likely to lead to a zero condition than addition. As a practical matter, subtraction in the denominator should be rewritten to an equivalent expression that doesn't have this problem.

End of Digression

 

[tkhunny]

... there is a difference between being a tutor and a teacher. A tutors goal is to get the best grade out of the student on their test, while (imo) a teachers job is to teach the material well. Yes, it is slight difference but it is there.

Good language. I have always wondered why I seem to annoy more students than most. Frankly, this explains it almost entirely. Using your language, those annoyed by me want a tutor and I'm just not on that page at all - seeking instead the greater breadth of understanding that a teacher would advocate. I expect the student to learn and to care. A tutor gets paid the same either way and may not hold those ideals. Unfortunately, in some respects, this elicits from me discourses where simple answers would have been sufficient (or closer to ideal).

You made my life better, today, having laid an old philosophical / social question to rest.

 

[150mph]

I’m missing something and still unsure what I’m looking for in The Problem by knowing “…you just want to get the same number in front of x (or the y) while ignoring signs.”

Here’s the next one:

Solve the following system of equations: 7x+8y=21 and -9x-7y=-4

So, to eliminate the variable x in preparation to add the equations as the lesson creators direct in this case, I’ll take the -9 from the 2nd equation —because its the number beside the variable x (right?)—and multiply both sides of the first equation by 9 (because it doesn’t matter if its 9 or -9 right?), then take 7 from the 1st equation and multiply both sides of the 2nd equation by 7 like so:

9(7x+8y)=9(21) and 7(-9x-7y)=7(-4)

63x+72y=189

-63x-49y=-28

Adding these 2 equations gets

0+23y=161 then 23y divided by 23 = 161 divided by 23 --—> y= 7

I put the 7 in the original :

7x+8(7)=21 --—> 7x+56=21 --—> 7x+56(-56)=21(-56) --—> 7x=-35 --—>

x=-5, y = 7

I know this is the correct answer (cheated and looked) but IDK if my thinking is correct.

 

[Dr.Peterson]

Yes, your work is correct.

It isn't always necessary to multiply each equation by the coefficient of x in the other; if they have a common factor, you can use smaller numbers to make the coefficients "equal and opposite" so they will cancel. But what you did will always work.

Here is an explanation of the overall process.

 

[Steven G]

7x+56(-56)=21(-56) --—> 7x=-35

When you write 56(-56) that means 56*(-56)=-3136. You meant to write 7x+56-56=21-56

 

[tkhunny]

The fundamental premise is that you can REPLACE any equation with a linear combination of the existing equations and have an equivalent system. (Well, don't pick zero multiplied by both!) Keep always in mind that there are TWO equations. It does not matter so much exactly how you combine them. Find a combination that helps you on your way toward a solution. Does it need to jump right to the solution? No. You can take a step or two. No need to fret about style.

For this one:

7x+8y=21
-9x-7y=-4

The second has an ungainly appearance, so let's replace it by (the sum of both).

7x+8y=21
-2x+y=17

That's a little better. Did we get anywhere?

Let's replace the first by the sum of (the first) and (3 times the second)

x+11y=72
-2x+y=17

Ah, the x's now look interesting. Let's replace the second by (the second) plus (twice the first).

x+11y=72
23y=161

Finally, we got rid of the x in one of them. Let's replace the second with (the second divided by 23)

x+11y=72
y=7

Really, if you're hung up on the precise way to proceed, then you are not seeing the idea very clearly. You can create equivalent systems all day long by making these kinds of alterations.

Let's replace the first by (the first) minus (11 times the second).

x=-5
y=7

Did I take ANY of the steps recommended in the book's example or in your work? Sort of toward the end, I suppose. But notice how it just didn't matter what path I took. Unique answers don't care how you find them! (I've heard that, somewhere.)

Practice doing these sorts of manipulations. Don't worry about what numbers to use. If all you want or need is the answer or the quickest way to the answer, then you should consider those LCM ideas and pursue that path. Develop a neat and clean style so it's easier to keep track of what you are doing. I recommend keeping track of both equations throughout your process. This keeps you from having to go back to see where you left something.

Good work so far. Practice more and gain greater confidence.

 

[150mph]

I appreciate all the insight. I'm still learning the syntax and keyboard…

Even though I got the answer, when I tried the next problem, I got stuck again using my new logic, which is:

“To eliminate x, grab the number next to x in one statement and use it to multiply both sides of the other statement. Do this for top and bottom statement” and “Forum guys say go ahead and just use positive numbers —for now—because that works, because it doesn’t matter —for now— if these new multipliers are positive or negative if they give the right answer”)

Solve the following system of equations:
-7x+9y=10
-4x+3y=10

4(-7x+9y)=4(10) —> 28x+36y=40
7(-4x+3y)=7(10) —> -28x+21y=70

-28x+36y=40
-28x+21y=70
_________________
-56x+57y=110

So I’m lost after here, and checked the answer they (ALEKS) gives (attached), which throws another curve. Why are they choosing -3 ? It seems that either the logic I'm using is off or they're describing a 2nd method now...

 

[Steven G]

To add to tkhunny's excellent post I want to add that you can get rid of fractions this way. You can multiply any equation by any non-zero number you feel like. It would be best if that number clears all the fractions

 

[Otis]

4(-7x+9y)=4(10) —> 28x+36y=40
7(-4x+3y)=7(10) —> -28x+21y=70

Hi 150mph. You have a sign error above; that 28 needs to be -28.

     -28x + 36y  =  40
—    -28x + 21y  =  70
------------------------
             15y = -30

Because the coefficients on x are now both -28, you eliminate the x-term from the system by subtracting (as shown above).

Continue …

PS: There are many ways to solve this system (by first adjusting coefficients and then adding or subtracting equations). Your approach is fine, and you don't need to match their method. Try to double-check your arithmetic and copying onto paper, after each step, to confirm your work as you go.

?

 

[150mph]

...You have a sign error above; that 28 needs to be -28. ...

I saw that missing minus sign after I posted but it was correct in the next step down where I did the actual calculating.

Because the coefficients on x are now both -28, you eliminate the x-term from the system by subtracting...

Thanks - this is probably what Jomo meant with "If the signs end up the same than you subtract, if they are different then you add
." This type of problem seems unusually confusing to me because I don't think this solution is shown in the lesson authors' solution, which is why I'm here asking.

Just to be clear, I'm trying my best to use and follow the authors' exact methods and thinking because there isn't time to post each problem I have and ask for help, so I'm looking for the simplest way to understand why they chose the method given.

Thanks for taking the time to help me.

 

[Steven G]

-7 and -7 have the same sign. What do you do to get 0 (ie add or subtract)? Remember x-x = 0 for ANY x-value.
-7 and 7 have different signs. Do you add them or subtract them to get 0.

Here is what I think about your comment above. Next semester, you may have a teacher or book that does the problem differently. I think that you should decide on the method that you like the best. I never think that a teacher should insist on how a student should think. Unfortunately teachers have to show students how to do new problems and if they do it in more than one way many students get confused. Just the other day Romsek and I solved a probability problem in completely different ways yet we got the same answer. We have different styles and different ways of thinking but in the end we got the same answer.

 

[tkhunny]

the authors' exact methods and thinking

There's the difficulty. The author's thinking dictates a flexible methodology related to each specific problem. Examples are not given for memorization. That would be utterly foolish. The examples are only demonstrations of how one might proceed, given the specific example. No amount of examples can prepare you for ALL the problems you will encounter in your career.

It's always a least common multiple. Sometimes, achieving that requires adjustment of only one equation.
Sometimes, we work first on x, other times on y. (or w or q or whatever the variables are.)
Sometimes we add, other times subtract.
Sometimes multiplication is required, other times division.
Sometimes we manipulate just one equation, sometimes two (or more if there are that many).
Sometimes, we see something entirely invalid and we just plain stop working on it because the solution is obvious.

There are so many ways to proceed. You need the concept, not the memorization.

 

[Otis]

… "If the signs end up the same than you subtract, if they are different then you add." This type of problem seems unusually confusing to me …

Hi. It won't be confusing, once you've practiced enough. We call this the 'elimination method' of solving a system of two equation because we eliminate either the x-terms or the y-terms. How do we eliminate the terms? By getting zero.

Think about it. If we have 4x and -4x, how do we get zero? Do we add them, or do a subtraction? We must add them, to get zero. That's the only way. Opposite numbers (like 4 and -4) always add to make zero.

If we have 4x and 4x, how do we get zero? We subtract because those numbers are the same. Identical numbers (like 4 and 4) always subtract to make zero. (It would be no different with -4x and -4x; those numbers are the same -- we must subtract.)

In the elimination method, we need to get both coefficients on the x-term (or on the y-term) to be the same number (to eliminate by subtraction) OR to be opposite numbers (to eliminate by addition). That's always the process, when solving a system of two equations using the elimination method: You eliminate either the x-terms or the y-terms via addition or subtraction. The choice is entirely yours to make.

Here's the system, again:

-7x+9y=10
-4x+3y=10

You may proceed in one of four ways.

1) Eliminate the x-terms. Multiply the equations, to get -28x and -28x (the coefficients are the same number, so subtract one equation from the other).

2) Eliminate the x-terms. Multiply the equations, to get -28x and 28x (the coefficients are opposites; so add the equations).

3) Eliminate the y-terms. Multiply one equation, to get 9y and 9y (the coefficients are the same; subtract equations).

4) Eliminate the y-terms. Multiply one equation, to get -9y and 9y (the coefficients are opposites; add equations).

From exercise to exercise, solving a system via elimination always begins in one of the four ways above. The only difference from exercise to exercise is the actual numbers used. The strategy does not change.

… I'm trying my best to use and follow the authors' exact methods …

There's no need for that. The author had the same choices that you have, and they could have chosen to proceed by using any one of the four basic approaches.

… I'm looking for the simplest way to understand why they chose [to find the solution as shown].

You don't need to understand their choice because it's not important. Their choice was arbitrary. That is, they could have chosen one of four different ways to begin. Just pick one; it doesn't matter which (at this point in your studies) because each of the four ways lead to the same solution.

After you have more practice, and you recognize the same patterns used over and over, THEN you'll be in a position to care about one choice perhaps being a little less work than another choice. The main thing for you right now is to get as much practice as you need, to reach the point where you see the pattern (opposites add to make zero; identical numbers subtract to make zero), so that you can simply go at it.

PS: Don't forget to check your solution candidates, to confirm that they work. Also, the more organized your work is, the easier it will be for you to go back and double-check your arithmetic and steps, in case a mistake was made

?

 

[150mph]

Thanks all for taking the time to help.

I think I've been missing that answering this type of problem requires making a choice of addition or subtraction in order to get zero

We call this the 'elimination method' of solving a system of two equation because we eliminate either the x-terms or the y-terms. How do we eliminate the terms? By getting zero.

I appreciate the encouragement and advice. It must be frustrating to hear my novice thinking! This is #136 of 212 daunting new and obscure topics I have to learn to pass a placement test and get into Precalculus (there's zero advanced math in my career as an Art Director, despite this requirement). So right now, memorizing how to do each (e.g., y=mx+b) — and what a correct answer even looks like - is critical for me. The dozens of hours with pen and paper help me get it right so I can move on to the next problem, but memorizing what I did to solve each problem come quiz time is everything.

 

[Otis]

… 212 daunting new and obscure topics I have to learn to pass a placement test and get into Precalculus …

Hello 150mph. Have you completed any algebra courses? Beginning and Intermediate Algebra are generally prerequisites for Precalculus. What type of school do you attend?

?

 

[150mph]

I'm familiar with that progression. I'm finishing up ASU Online, BS in Web Design. ASU is like many schools that require passing only a College Algebra placement test with a C as a prereq to Precalc. (It's hard!) This is what most students in my program do.