x on both sides of proportion: (4x + 8)/6 = (7x - 3)/4

steveopolis

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(4x + 8) (7x-3)
________ = _________
6 4


What I've tried is working up by one number at a time, like trying x as 1, x as 2, and onward to 9, eg.

4x1 = 4
4 + 8 12
12/6 = 2

7x1 = 7
7-3 = 4
4/4 = 1

Anyone have a better hint for me?
 

tkhunny

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(4x + 8) (7x-3)
________ = _________
6 4


What I've tried is working up by one number at a time, like trying x as 1, x as 2, and onward to 9, eg.

4x1 = 4
4 + 8 12
12/6 = 2

7x1 = 7
7-3 = 4
4/4 = 1

Anyone have a better hint for me?
I blame your teacher. Are you doing this online?

How would you solve this?

\(\displaystyle \dfrac{x}{2} = \dfrac{7}{3}\)
 

steveopolis

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How would you solve this?

\(\displaystyle \dfrac{x}{2} = \dfrac{7}{3}\)
I would cross-multiply:

3x = [7x2=]14
14/3 = 4.7 (rounded up)
x = 4.7

I'm upgrading through an official college upgrading program, but we're left largely to our own devices. Sometimes there's an answer key that tells me if I got something right (98%) of the time, but frequently the answer key is incorrect or there is no answer key for a section.
 
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steveopolis

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Can't tell whadda heck you're doing...
My goal is to identify x.
I had no idea where to begin.
The course hadn't introduced this kind of thing before.
So I tried each equation assuming x = 1 (didn't work), x = 2 (didn't work), and so on.

What do you get if you multiply each side by 12?
I'm not sure what you mean by multiplying each side by 12.
If I assume x = 12, they still don't match up.
I don't know how to multiply the numerators by 12, because the equation inside the bracket isn't resolved.
 

steveopolis

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Maybe I'm not supposed to find x (and multiply that by the number it's attached to)?
Maybe I'm supposed to collect like terms?? and then??

Sigh...
 

JeffM

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(4x + 8) (7x-3)
________ = _________
6 4


What I've tried is working up by one number at a time, like trying x as 1, x as 2, and onward to 9, eg.

4x1 = 4
4 + 8 12
12/6 = 2

7x1 = 7
7-3 = 4
4/4 = 1

Anyone have a better hint for me?
Do you mean

\(\displaystyle \dfrac{4x + 8}{6} = \dfrac{7x - 3}{4}.\)

If so, write \(\displaystyle (4x + 8) / 6 = (7x - 3) / 4.\)

And use algebra rather than trial and error.

The least common multiple of the denominators is 12 so multiply both sides of the equation by 12 to clear fractions.

\(\displaystyle \dfrac{4x + 8}{6} = \dfrac{7x - 3}{4} \implies 12 * \dfrac{4x + 8}{6} = 12 * \dfrac{7x - 3}{4} \implies\)

\(\displaystyle 2(4x + 8) = 3(7x - 3) \implies 8x + 16 = 21x - 9 \implies WHAT?\)

Make sure to check your work.
 

Jomo

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I would cross-multiply:

3x = [7x2=]14
14/3 = 4.7 (rounded up)
x = 4.7

I'm upgrading through an official college upgrading program, but we're left largely to our own devices. Sometimes there's an answer key that tells me if I got something right (98%) of the time, but frequently the answer key is incorrect or there is no answer key for a section.
Take my advise and never cross multiply. Just multiply both sides by 3*2.
 

steveopolis

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Hi, JeffM, yes, that's the correct layout. (How do you make the format stay put?)

I don't know why my course would put something in we haven't learned yet. I hadn't learned clearing fractions, but by googling that term, reading here: https://www.varsitytutors.com/hotmath/hotmath_help/topics/solving-multi-step-linear-equations-with-fractions, and then studying what that page did and you did, it seems to be dividing the LCD by the denominator. Okay.

But I still have no idea how to find out what x would be. And that's because these steps haven't appeared in the course yet. What a dumb course I'm in, argh! They offered bonus marks for being able to magically do it, but I think I need to move on, answer it *after* they teach it.

Jomo, I have to cross-multiply (for the query tkhunny gave). The course requires that we show our work in they way they teach it.
 

Jomo

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Hi, JeffM, yes, that's the correct layout. (How do you make the format stay put?)

I don't know why my course would put something in we haven't learned yet. I hadn't learned clearing fractions, but by googling that term, reading here: https://www.varsitytutors.com/hotmath/hotmath_help/topics/solving-multi-step-linear-equations-with-fractions, and then studying what that page did and you did, it seems to be dividing the LCD by the denominator. Okay.

But I still have no idea how to find out what x would be. And that's because these steps haven't appeared in the course yet. What a dumb course I'm in, argh! They offered bonus marks for being able to magically do it, but I think I need to move on, answer it *after* they teach it.

Jomo, I have to cross-multiply (for the query tkhunny gave). The course requires that we show our work in they way they teach it.
Again, please take my advise and never cross multiply. Many students, possibly not you, make horrendous mistakes by cross multiplying. Just multiply both sides by the product of the two denominators (exactly the same as cross multiply) or even better multiply both sides by the lcd. Trust me on this one. I would not lead you wrong.
 

tkhunny

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Hi, JeffM, yes, that's the correct layout. (How do you make the format stay put?)

I don't know why my course would put something in we haven't learned yet. I hadn't learned clearing fractions, but by googling that term, reading here: https://www.varsitytutors.com/hotmath/hotmath_help/topics/solving-multi-step-linear-equations-with-fractions, and then studying what that page did and you did, it seems to be dividing the LCD by the denominator. Okay.

But I still have no idea how to find out what x would be. And that's because these steps haven't appeared in the course yet. What a dumb course I'm in, argh! They offered bonus marks for being able to magically do it, but I think I need to move on, answer it *after* they teach it.

Jomo, I have to cross-multiply (for the query tkhunny gave). The course requires that we show our work in they way they teach it.

"Cross Multiply" isn't a thing. NEVER do anything like it. Just find a common denominator. It doesn't matter AT ALL if there is an 'x' on both sides or on just one side. This is where I blame someone who is not the student. It is often taught that it is somehow harder with the appearance of a second 'x'. Hogwash. It is EXACTLY the same. Find a common denominator and proceed with regular algebra operations. That's all it is. Add "cross multiply" to the list of vulgar or profane words that never should be spoken or written. Apologies for writing it twice to make this statement. :)
 

Jomo

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Add "cross multiply" to the list of vulgar or profane words that never should be spoken or written. :)
I love it! You made my evening!
 

JeffM

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Hi, JeffM, yes, that's the correct layout. (How do you make the format stay put?)

I don't know why my course would put something in we haven't learned yet. I hadn't learned clearing fractions, but by googling that term, reading here: https://www.varsitytutors.com/hotmath/hotmath_help/topics/solving-multi-step-linear-equations-with-fractions, and then studying what that page did and you did, it seems to be dividing the LCD by the denominator. Okay.

But I still have no idea how to find out what x would be. And that's because these steps haven't appeared in the course yet. What a dumb course I'm in, argh! They offered bonus marks for being able to magically do it, but I think I need to move on, answer it *after* they teach it.

Jomo, I have to cross-multiply (for the query tkhunny gave). The course requires that we show our work in they way they teach it.
Dont worry about trying to format math. There is a way to do it on this site, but it is tricky, and you need to focus on math rather than formatting.

When you multiply a fraction by a multiple of the denominator, you can cancel the denominator and get rid of the fraction. But of course you must multiply both sides of the equation, and if you want to get rid of all the fractions, you must multiply by a common multiple of all the denominators.

Clearing fractions is a way to simplify equations. It is the technical justification for cross-multiplication.

If you are required to use cross multiplication without having had it explained to you, that is a terrible way to teach math.

\(\displaystyle \dfrac{4x + 8}{6} = \dfrac{7x - 3}{4} \implies 4(4x + 8) = 6(7x - 3) \text { by cross multiplication.}\)

\(\displaystyle \therefore 16x + 32 = 42x - 18.\)

Are you really unable to solve the last equation above? (By the way that equation has the same solution as the equation that I ended with in my previous post.)
 
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Jomo

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I would cross-multiply:

3x = [7x2=]14
14/3 = 4.7 (rounded up)
x = 4.7

I'm upgrading through an official college upgrading program, but we're left largely to our own devices. Sometimes there's an answer key that tells me if I got something right (98%) of the time, but frequently the answer key is incorrect or there is no answer key for a section.
You have \(\displaystyle \frac{x}{2} = \frac{7}{3}\) You just need to get rid of the 2, so multiply both sides by 2.

\(\displaystyle 2*\frac{x}{2} = 2*\frac{7}{3}\)

\(\displaystyle x= \frac{14}{3}\)

I understand that students are taught to cross multiply. In my opinion it is the MOST inefficient way to solve this problem! Think about it, 1st you move the 3 to the top only to then bring it back to the bottom. Totally a waste of time. I think of this when I think of cross multiplying--- You work for me. Every time I ask you to do anything for me, you always first go and bring the garbage can to your desk, then put it back where it was and then you do what I ask you to do. Then I fire you (for being inefficient).

Two posters now have suggested that you never ever cross multiply. Take our advise as we seen how inefficient it is AND the silly mistakes students make when they use it. As already mentioned--remove the phrase from your vocabulary.
 
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steveopolis

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Two posters now have suggested that you never ever cross multiply. Take our advise as we seen how inefficient it is AND the silly mistakes students make when they use it. As already mentioned--remove the phrase from your vocabulary.
Jomo, I don't have control over the language my teachers are using nor over the content they do (or don't!) provide. I also don't have control over which elements they give or remove a mark for. I have two goals: to learn math really well; to pass these courses with close to 100%.

To gain marks -and not lose marks -I'm required in certain equations to write out the format they taught. When I'm playing at home and after I graduate, I can do whatever I like, yes. But for now I don't have the luxury of ignoring the teachers' requests if I want to meet my second goal as well as my first.

In this problem's case, I don't believe cross-multiplication is the issue. With JeffM expanding on Denis' tip to multiply by 12, I saw how it got to the last stage of what JeffM has presented. However, sadly, no, I don't know how to work it out. The problem is not that I'm attempting to cross-multiply, because I'm not. I have no idea what a person would do next. I believe there's an additional piece I haven't yet been taught.
 

Dr.Peterson

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Jomo, I don't have control over the language my teachers are using nor over the content they do (or don't!) provide. I also don't have control over which elements they give or remove a mark for. I have two goals: to learn math really well; to pass these courses with close to 100%.

To gain marks -and not lose marks -I'm required in certain equations to write out the format they taught. When I'm playing at home and after I graduate, I can do whatever I like, yes. But for now I don't have the luxury of ignoring the teachers' requests if I want to meet my second goal as well as my first.

In this problem's case, I don't believe cross-multiplication is the issue. With JeffM expanding on Denis' tip to multiply by 12, I saw how it got to the last stage of what JeffM has presented. However, sadly, no, I don't know how to work it out. The problem is not that I'm attempting to cross-multiply, because I'm not. I have no idea what a person would do next. I believe there's an additional piece I haven't yet been taught.
I agree with you that you need to be helped with what you have been taught, for now, even though it's not the best. Sometimes the "best" is the enemy of the "good-enough".

I assume that you see that you can just cross-multiply your proportion, (4x + 8)/6 = (7x-3)/4, to get 4(4x + 8) = 6(7x-3), and then expand this to get 16x + 32 = 42x - 18. There's a lot more that could be said, but this is good enough to get by.

I think you are now saying that you don't know how to solve a linear equation with the variable on both sides. The first step here is to subtract 16x from both sides, so that the variable will be on only one side, and then the rest of the work should be familiar. (You could instead subtract 42x from both sides, if you prefer x to stay on the left, and don't mind negatives.)

If this is hard for you, perhaps it will help if you tell us what you have learned so far, so that we can help you use what you know. If I were tutoring you face to face, I would have started by finding out about this, to know where we have to start to get you where you need to be. I'd also be looking through your learning materials to find out how they want you to do things, and how we can help you get past their inadequacies. Perhaps I would find where you were taught what you need to know here, and we could go back to that point and find what went wrong in your learning; or maybe we'd find that they omitted this important step, thinking it was obvious.
 

steveopolis

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I think you are now saying that you don't know how to solve a linear equation with the variable on both sides.
That's right. I know this is the case only because I'd never heard of such a thing ;)
We didn't learn the steps yet.
Instructor threw it onto the assignment with a note, "...two bonus points if you try this."
I imagined that meant if I thought hard enough, or applied what we'd learned in a slightly new way, I could figure it out. Nope!

Things are regularly omitted in this course, but I can never be sure where that's the case and where I'm just not grasping something. There used to be teachers I could check the content with -and they'd acknowledge stuff was just missing or an answer in the key was wrong. Now there aren't teachers available.

The first step here is to subtract 16x from both sides, so that the variable will be on only one side, and then the rest of the work should be familiar. (You could instead subtract 42x from both sides, if you prefer x to stay on the left, and don't mind negatives.)
Okay, this matches what it was saying in that link I posted and tried to figure it out from -unsuccessfully. Yours sounded clearer to me.

Is the answer 1.923? I'll check against the original equation...
 

steveopolis

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So, my best go so far is that the answer is 1.923 (that's rounded off to three decimal places). That seems correct on the original equation IF it's okay to round that, too, to the third decimal place.

Did I get it?
Or no?
 

JeffM

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So, my best go so far is that the answer is 1.923 (that's rounded off to three decimal places). That seems correct on the original equation IF it's okay to round that, too, to the third decimal place.

Did I get it?
Or no?
The exact answer is \(\displaystyle x = \dfrac{25}{13} \approx 1.923.\)

So you worked hard for that. Well done.

One comment. It is usually best in algebra to express an answer exactly unless you are told to give the answer to a certain number of decimal places.
 

steveopolis

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:eek: !!!

How exciting!!!!

Thank you all very much for your patient and enduring help! You guys are real troopers. And now I know a new-to-me math thing! And maybe will get two bonus points :)

THANK YOU!!
 
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