How to Solve an Equation with Fractions and Variables?

[ai93]

I have an equation that I am having trouble on
x+3 over 4 = x-3 over 5 + 2
any suggestions?

 

[MarkFL]

You equation is:

\(\displaystyle \frac{x+3}{4}=\frac{x-3}{5}+2\)

Correct?

 

[ai93]

You equation is:

\(\displaystyle \frac{x+3}{4}=\frac{x-3}{5}+2\)

Correct?

Yes it is

 

[MarkFL]

Yes it is

Okay good. :D

Now, we want to get rid of those pesky denominators...what number can we multiply both sides of the equation by, such that we will rid ourselves of those denominators?

 

[ai93]

Okay good. :D

Now, we want to get rid of those pesky denominators...what number can we multiply both sides of the equation by, such that we will rid ourselves of those denominators?

Do we find a LCM, so times the first equation by 5 and the second by 4 so the denominator are both 20?

 

[MarkFL]

Do we find a LCM, so times the first equation by 5 and the second by 4 so the denominator are both 20?

You are on the right track...20 is the LCM of the two denominators, so multiplying both sides by 20 will get rid of the denominators. So what do you have after multiplying both sides by 20?

 

[ai93]

You are on the right track...20 is the LCM of the two denominators, so multiplying both sides by 20 will get rid of the denominators. So what do you have after multiplying both sides by 20?

So I wouldn't times each side by 4 and 5 for the LCM 20 and cancel out?
\(\displaystyle \frac{x+3}{4(x5)} = \frac{x-3}{5(x4)}+2 \) to equal
\(\displaystyle \frac{x+3}{20}=\frac{x-3}{20}+2\) then cancel out the denominators. I would be left with x+3=x-3+2. I am guessing I would move either x+3 or x-3?

 

[MarkFL]

This is what I am suggesting:

\(\displaystyle 20\left(\frac{x+3}{4}=\frac{x-3}{5}+2\right)\)

After multiplying each term on both sides of the equation by 20, we obtain:

\(\displaystyle 5(x+3)=4(x-3)+40\)

Now, it's a matter of distributing, and collecting like terms, and solving for $x$. :D

 

[ai93]

This is what I am suggesting:

\(\displaystyle 20\left(\frac{x+3}{4}=\frac{x-3}{5}+2\right)\)

After multiplying each term on both sides of the equation by 20, we obtain:

\(\displaystyle 5(x+3)=4(x-3)+40\)

Now, it's a matter of distributing, and collecting like terms, and solving for $x$. :D

When multiplying out should you always times top first then divide by the bottom. So in this example, 20 * x+3 then / by 4 which = 5? Just a general question, you wouldn't times 20 * (x+3) which = 20x+60. I hope you understood that bit.

So to solve for x. We get (5x+15) = (4x-12) + 160.
(5x+15) - (4x-12) = 160
?

 

[MarkFL]

This is how I think of it...

\(\displaystyle 20\left(\frac{x+3}{4}=\frac{x-3}{5}+2\right)\)

\(\displaystyle 20\cdot\frac{x+3}{4}=20\cdot\frac{x-3}{5}+20\cdot2\)

\(\displaystyle 4\cdot5\cdot\frac{x+3}{4}=4\cdot5\cdot\frac{x-3}{5}+20\cdot2\)

\(\displaystyle \cancel{4}\cdot5\cdot\frac{x+3}{\cancel{4}}=4\cdot\cancel{5}\cdot\frac{x-3}{\cancel{5}}+20\cdot2\)

\(\displaystyle 5(x+3)=4(x-3)+40\)

 

[ineedhelpnow]

You could multiply 20 by the whole thing and then divide by 5 so like 20x+60 and then divide by 4 and you get 5x+15 but it's easier to simply first. just do 20/4 which is 5 and then you have 5(x+3)

You have 5x+15=4x-12+40.
$5(x+3)=4(x-3)+40$
$5x+15=4x-12+40$
$5x-4x=-12-15+40$
$x=?$

 

[ai93]

This is how I think of it...

\(\displaystyle 20\left(\frac{x+3}{4}=\frac{x-3}{5}+2\right)\)
\(\displaystyle 20\cdot\frac{x+3}{4}=20\cdot\frac{x-3}{5}+20\cdot2\)

\(\displaystyle 4\cdot5\cdot\frac{x+3}{4}=4\cdot5\cdot\frac{x-3}{5}+20\cdot2\)

\(\displaystyle \cancel{4}\cdot5\cdot\frac{x+3}{\cancel{4}}=4\cdot\cancel{5}\cdot\frac{x-3}{\cancel{5}}+20\cdot2\)

\(\displaystyle 5(x+3)=4(x-3)+40\)

In the third step why did you start to times 4*5, when before it was just 20?
I see how you got to 5(x+3)=4(x-3)=40. Now like you said earlier solve for x?
Times out the bracket maybe?:
\(\displaystyle (5x+15)=(4x-12)+40\) now I am lost..

 

[ineedhelpnow]

In the third step why did you start to times 4*5, when before it was just 20?
I see how you got to 5(x+3)=4(x-3)=40. Now like you said earlier solve for x?
Times out the bracket maybe?:
\(\displaystyle (5x+15)=(4x-12)+40\) now I am lost..

because 20 is the same as 4*5 and he wanted to show you how to simplify it so the the two 4's could be canceled out which leaves you with 5.

 

[ai93]

u could multiply 20 by the whole thing and then divide by 5 so like 20x+60 and then divide by 4 and u get 5x+15 but its easier to simply first. just do 20/4 which is 5 and then u have 5(x+3)

u have 5x+15=4x-12+40.
$5(x+3)=4(x-3)+40$
$5x+15=4x-12+40$
$5x-4x=-12-15+40$
$x=?$

I understand now! From 5(x+3)=4(x−3)+40 times it out for 5x+15=4x−12+40. Then just move like terms! I sometimes forget it can be that easy..
So \(\displaystyle x=13!\)
Thanks!

 

[ineedhelpnow]

No problem. Just be careful when multiplying out. Mistakes can easily be made. I still don't know where that 160 came from :p (There are like 8 people viewing this thread. I bet they're all in your class (Giggle) )

 

[ai93]

no problem. just be careful when multiplying out. mistakes can easily be made. i still don't know where that 160 came from :p (there are like 8 people viewing this thread. i bet theyre all in ur class (Giggle) )

That 160 came from 4(x−3)+40. 4*40 is 160 haha. They could be, I could be helping them. It is some tricky maths! (Giggle) :D