Solve Equation: \frac{x}{2}+3=\frac{x+2}{3}-x

[headbang]

Ok have other solve Equations..
\(\displaystyle \frac{x}{2}+3=\frac{x+2}{3}-x\)

 

[MarkFL]

We should get rid of those denominators...what is the smallest number we can multiply both sides by so that the denominators will be gone?

 

[headbang]

Ok hav to be 6.. 2*3

\(\displaystyle \frac{2*3*x}{2}+6*3=\frac{x*2*2*3}{3}-x*6\)

\(\displaystyle 3x+18=4x-6x\)

\(\displaystyle 3x-4x-6x+18=\)

\(\displaystyle -7x=-18\)

\(\displaystyle 7x=18\)

OK ?

 

[topsquark]

Ok hav to be 6.. 2*3

\(\displaystyle \frac{2*3*x}{2}+6*3=\frac{x*2*2*3}{3}-x*6\)

\(\displaystyle 3x+18=4x-6x\)

\(\displaystyle 3x-4x-6x+18=\)

\(\displaystyle -7x=-18\)

\(\displaystyle 7x=18\)

OK ?

There's a problem in line 1. The RHS shold be
\(\displaystyle \frac{(x + 2) \cdot 2 \cdot 3}{3} - 6x\)

 

[MarkFL]

Not quite...you did pick the correct value, the LCM of 2 and 3...but you multiplication wasn't quite right...

\(\displaystyle 6\left(\frac{x}{2}+3\right)=6\left(\frac{x+2}{3}-x\right)\)

\(\displaystyle 3x+18=2(x+2)-6x\)

\(\displaystyle 3x+18=2x+4-6x\)

Try it from there...

 

[Prove It]

Ok have other solve Equations..
\(\displaystyle \frac{x}{2}+3=\frac{x+2}{3}-x\)

My attitude is always SIMPLIFY before trying to do anything.

Each side is the sum of a fractional amount (even whole numbers are fractions with 1 as the denominator). How do you add fractions?

Then I would make use of the fact that if two fractions that have the same denominator are equal, then the numerators must be equal to (as the denominator tells you the TYPE of parts you have, if both sides have the same type of parts, then the only way that the fractions can be equal is if they have the same NUMBER of these type of parts, in other words, if the numerators are equal).

 

[headbang]

\(\displaystyle 3x+18=2x+4-6x\)

\(\displaystyle 3x-2x+6x=4-18\)

\(\displaystyle \frac{7x}{7}=\frac{-14}{7}\)

\(\displaystyle x=-2\)An other try... Is it ok?

And thank you all so much for helping me, trying to get my head around this math.. You are all worth your weight in gold...

 

[MarkFL]

\(\displaystyle 3x+18=2x+4-6x\)

\(\displaystyle 3x-2x+6x=4-18\)

\(\displaystyle \frac{7x}{7}=\frac{-14}{7}\)

\(\displaystyle x=-2\)An other try... Is it ok?

And thank you all so much for helping me, trying to get my head around this math.. You are all worth your weight in gold...

Your algebra looks good, but let's check your answer in the original equation:

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

\(\displaystyle -1+3=0+2\)

\(\displaystyle 2=2\)

Yes, your solution works! (Yes)