# How to distribute a fraction?

• B
I'm given the problem 14x+5y=31, 2x-3y=-29

This is a set of linear equations that I am asked to solve. That being said I am getting hung up on how to distribute a fraction. The answer is x=9, however, I am getting x=36 and the step I do not understand is distributing the fraction.

We then get setup with 2*((31-5y)/14)-3y = -29 by solving for x.

We then distribute the 2 this gives us ((62-10y)/14)-3y = -29.

This can also be written as 1/14 *(62-10y)-3y = -29.

Now why can't we distribute this is the inverse? e.g. 1/14*14/1 = 1

(62-10y)-3y = -29*14 -> (62-10y)-3y = -406

That gets us 62-13y = -406 then do some basic arithmetic ->

-62, -62

-13y = -465, divide by -13 -> y = 36

How is this method incorrect? I'm very frustrated because I have spent hours on this. I can see another method of performing this would be multiply by the inverse ->

(62-10y)/14)-3y = -29

Multiple by the inverse ((62-10y)/14) * ((14)/(62-10y)) ->

62-10y-3y = -29* ((14)/(62-10y)) and then I go nowhere.

I will be incredibly happy if someone can help me to understand this. Thanks kindly.

Svein
(62-10y)-3y = -29*14 -> (62-10y)-3y = -406
You forgot to multiply 3y by 14!

• Voltux
You forgot to multiply 3y by 14!
Svein,

I would like to say thank you so very much for pointing this out to me. I believe you truly made my evening.

I indeed performed the calculations and get -468/-52 = 9.

Can you cite the rule or fundamental misunderstanding that I have missed? I was under the impression that we can distribute once like I demonstrated.

Thank you again, Svein. Thank you.

Svein
Can you cite the rule or fundamental misunderstanding that I have missed? I was under the impression that we can distribute once like I demonstrated.
Think of an equation as an old-fashioned scale (the one you use for weighing things). You start with the scale in balance (that is what the equal sign tells you). Now you can add or subtract the same amount on both sides - the scale will still balance. You can also double the amount on both sides - the scale will still balance. In short - as long as you do the same thing on both sides, the scale will keep on balancing.

In your case - yes, you can multiply both sides of the equal sign by 14. But "both sides" means exactly that. Therefore, encapsulate both the left and right side in parenthesis before you multiply!

• pbuk
Gold Member
... but that is not usually the best way to solve linear simultaneous equations which I suggest is:
1. You can see that ## 2x - 3y = -29 ## can be written ## 14x - 21y = -203 ##
2. Subtract this from the first equation to get ## 14x + 5y -14x - (-21y) = 31 - (-203) ##
3. Simplify to ## 26y = 234 \implies y = 9##
4. Substitute into the second equation to get ## 2x - 3 \times 9 = -29 \implies 2x - 27 = -29 \implies 2x = -29 + 27 = -2 \implies x = -1 ## (I could have chosen the first equation, but this one looked easier).
5. Check by substituting into the first equation to get ## 14 \times (-1) + 5 \times 9 = 31 \implies -14 + 45 = 31 ## OK
Fewer steps means fewer mistakes! (although I have just edited a typo in step 3!)

... but that is not usually the best way to solve linear simultaneous equations which I suggest is:
1. You can see that ## 2x - 3y = -29 ## can be written ## 14x - 21y = -203 ##
2. Subtract this from the first equation to get ## 14x + 5y -14x - (-21y) = 31 - (-203) ##
3. Simplify to ## 26y = 234 \implies y = 9##
4. Substitute into the second equation to get ## 2x - 3 \times 9 = -29 \implies 2x - 27 = -29 \implies 2x = -29 + 27 = -2 \implies x = -1 ## (I could have chosen the first equation, but this one looked easier).
5. Check by substituting into the first equation to get ## 14 \times (-1) + 5 \times 9 = 31 \implies -14 + 45 = 31 ## OK
Fewer steps means fewer mistakes! (although I have just edited a typo in step 3!)
And another way is making the first equation into

y = (2x+29)/3
and substitute y into the second equation,then you get y out of the equation.
(Not really useful in this particular set of equations,but seldom gets more efficient then the add-subtracting method that @pbuk mentioned,for instance:
y = 2x+1
3x+2y = 9

• Janosh89
symbolipoint
Homework Helper
Gold Member
No reason to wonder how or why to "distribute fractions". You present a system of equations.
14x+5y=31, 2x-3y=-29

A neat way to handle this is try to at least start with elimination of a variable. The x looks like a nice choice this way.

Keep first equation, but multiply second equation by 7.
14x+5y=31, 14x-21y=-203

Now as so written, subtract first equation from second equation.
14x-21y-14x-5y=-203-31

-26y=-234

26y=234

y=9

There is one of the variables, now solved. Any use of fractions occurring was no large struggle; only the simplest of arithmetic, even if some electronic tool was used.

How you go about solving for x, is your choice. Use elimination if you want, or try substitution in either of the original equations.

• YoungPhysicist
Svein
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