Intermediate Algebra problem with fractions

[Kristyg921]

I am having trouble with this problem because of the different fractions.

[(t-2)/5]+5t= (7/5) - (t-2)/2

The first time I tried this problem I got t=2/3, but when I plugged it back in the answer did not work.

 

[Mrspi]

I am having trouble with this problem because of the different fractions.

[(t-2)/5]+5t= (7/5) - (t-2)/2

The first time I tried this problem I got t=2/3, but when I plugged it back in the answer did not work.

If there are "different fractions," you use the common denominator for all of the fractions as a multiplier to eliminate the fractions. The common denominator for halves and fifths is 10....so, multiply each term of the equation by 10:

10*[(t-2)/5] + 10*5t = 10*(7/5) - 10*(t - 2)/2

Each denominator divides into the multiplier, leaving you with this:

2(t - 2) + 10*5t = 2*7 - 5(t - 2)

Can you take it from here? If you're still having trouble, please repost and show all of the work you've done to try to finish the problem.

 

[JeffWest]

Hey Kristyg921,

The way that I like to do this problem is to break up each fraction into its own. As shown:

\(\displaystyle \frac{t-2}{5}=\frac{t}{5}-\frac{2}{5}\)

and

\(\displaystyle \frac{t-2}{2}=\frac{t}{2}-\frac{2}{2}\)

With these put into the big equation it reads:

\(\displaystyle \frac{t}{5}-\frac{2}{5}+5t= \frac{7}{5} - \frac{t}{2}-\frac{2}{2}\)

With it set up like this, there is one thing that must be done. Combine like terms using common denominators.

Show you work if you have any more questions.

 

[masters]

I am having trouble with this problem because of the different fractions.

\(\displaystyle \frac{t-2}{5}+5t=\frac{7}{5}-\frac{t-2}{2}\)

The first time I tried this problem I got \(\displaystyle t=\frac{2}{3}\), but when I plugged it back in the answer did not work.

If there are "different fractions," you use the common denominator for all of the fractions as a multiplier to eliminate the fractions. The common denominator for halves and fifths is 10....so, multiply each term of the equation by 10:

\(\displaystyle 10\left(\frac{t-2}{5}\right)+10(5t)=10\left(\frac{7}{5}\right)-10\left(\frac{t-2}{2}\right)\)

Each denominator divides into the multiplier, leaving you with this:

\(\displaystyle 2(t - 2) + 10(5t) = 2(7) - 5(t - 2)\)

Can you take it from here? If you're still having trouble, please repost and show all of the work you've done to try to finish the problem.

Just tweaked your post a bit, Mrspi. Don't hate me. I'll bake you a pie.

 

[masters]

I am having trouble with this problem because of the different fractions.

\(\displaystyle \frac{t-2}{5}+5t=\frac{7}{5}-\frac{t-2}{2}\)

The first time I tried this problem I got \(\displaystyle t=\frac{2}{3}\), but when I plugged it back in the answer did not work.

If there are "different fractions," you use the common denominator for all of the fractions as a multiplier to eliminate the fractions. The common denominator for halves and fifths is 10....so, multiply each term of the equation by 10:

\(\displaystyle 10\left(\frac{t-2}{5}\right)+10(5t)=10\left(\frac{7}{5}\right)-10\left(\frac{t-2}{2}\right)\)

Each denominator divides into the multiplier, leaving you with this:

\(\displaystyle 2(t - 2) + 10(5t) = 2(7) - 5(t - 2)\)

Can you take it from here? If you're still having trouble, please repost and show all of the work you've done to try to finish the problem.

Just tweaked your post a bit, Mrspi. Don't hate me. I'll bake you a \(\displaystyle \Pi\).

 

[Kristyg921]

Thank you for your help.

 

[JeffWest]

Your Welcome