I missed my math class on tuesday, and this is one of the problems we have for homework. I have no idea how to figure this out. Any help, much appriciated.

$\displaystyle

\frac{3x-5}{4}+\frac{x}{2}=\frac{1}{3}

$

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- May 16th 2009, 06:09 PMelitescouterHelp solve this problem
I missed my math class on tuesday, and this is one of the problems we have for homework. I have no idea how to figure this out. Any help, much appriciated.

$\displaystyle

\frac{3x-5}{4}+\frac{x}{2}=\frac{1}{3}

$ - May 16th 2009, 06:32 PMmathaddict
- May 16th 2009, 07:04 PMVonNemo19
Multiply both sides by four

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

solve for x

$\displaystyle 5x-5=\frac{4}{3}$

$\displaystyle 5x=\frac{4}{3}+\frac{15}{3}$

$\displaystyle 5x=\frac{19}{3}$

$\displaystyle x=(\frac{19}{3})(\frac{1}{5})$

Does that help? - May 17th 2009, 12:12 AMAlderDragon
$\displaystyle

\frac{3x-5}{4}+\frac{x}{2}=\frac{1}{3}$

The first thing you'll need to do is make both fractions on the left side of the equation have the same denominator. Notice how the one on the far left has a denominator of 4 while the other has 2. Simply double the fraction with the denominator of 2. Since this is a fraction, it will not change the value, as long as you do the same thing to the top as you do to the bottom. So we end up with:

$\displaystyle \frac{3x-5}{4}+\frac{2x}{4}=\frac{1}{3}$ because $\displaystyle (2)(2) = 4$ and $\displaystyle (x)(2) = 2x$

It's important to know that you can combine fractions that have the same denominator. For example, if you had:

$\displaystyle \frac {1} {5} + \frac {3} {5}$ you can combine them as: $\displaystyle \frac {4} {5}$

The denominator stays the same, while the top numbers are added.

So, let's focus on the left side of the problem quickly. We have:

$\displaystyle \frac{3x-5}{4}+\frac{2x}{4}$

Since both of the fractions have the same denominator (4), we can add the top values together:

$\displaystyle \frac{3x-5+2x}{4}$

The value in the numerator can be simplified by combining 2x and 3x:

$\displaystyle 3x-5+2x = 5x-5$

So now, our equation looks like this:

$\displaystyle \frac {5x-5}{4} = \frac {1}{3}$

The next step is to multiply both sides by 4. Why? Because we are dividing by 4 on the left side. It is the same concept as addition and subtraction. Multiplying the left side by 4 will cancel out the denominator and leave you with just the value of the numerator. It's simply going one step closer to getting the variable alone.

$\displaystyle \frac {5x-5}{4}(4) = \frac {1}{3}(4)$

$\displaystyle 5x-5 = \frac {1}{3}(4)$

Now, let's focus on the right side of the equation. We have one-third multiplied by four. Instead of looking at it like that, think of it like this:

$\displaystyle (\frac{1}{3})(\frac {4}{1}) = \frac{4}{3}$

Because,

$\displaystyle 4 = \frac {4}{1}$

We're almost done. Now the equation looks like this:

$\displaystyle 5x-5=\frac {4}{3}$

The next thing we will do is add 5 to both sides to get rid of the -5 on the left side of the equation. But, we will need to change it before subtracting it from the right side of the equation:

$\displaystyle 5x-5+5=\frac {4}{3}+5$

$\displaystyle 5x=\frac {4}{3}+\frac{5}{1}$

Before you add two fractions, their denominators have to be the same. So we change $\displaystyle \frac{5}{1}$ to $\displaystyle \frac{15}{3}$ by multiplying both top and bottom by 3, so that both fractions on the right side of the equation have the same denominator.

$\displaystyle 5x=\frac{4}{3}+\frac {15}{3}$

$\displaystyle 5x=\frac{19}{3}$

Next, divide both sides of the equation by 5 to isolate the x. It can get confusing on the right side of the equation, since we have a fraction already. It would really be " 15 divided by 3 divided by 5" which ends up getting messy. There is an easy way of doing this.

Dividing by a fraction is the same thing as multiplying by its reciprocal. The reciprocal is a fraction that, when multiplied with the original number, equals 1.

The reciprocal of 5 is $\displaystyle \frac{1}{5}$ because $\displaystyle (5)(\frac{1}{5}) = 1$

So this means instead of dividing a fraction, we can multiply it by its reciprocal instead.

What I'm saying is this:

$\displaystyle \frac{\frac{19}{3}}{5}$

Is the SAME thing as:

$\displaystyle (\frac{19}{3})\frac{1}{5}$

Reciprocals are easy to point out. It's generally just a fraction with the number 1. Examples of reciprocals:

$\displaystyle 7 ... \frac{1}{7}$

$\displaystyle 25 ... \frac{1}{25}$

$\displaystyle 7282 ... \frac{1}{7282}$

The same is true for the other way. Multiplying by 1/5 is the same thing as dividing by 5.

So, look back to the original problem:

$\displaystyle 5x=\frac{19}{3}$

$\displaystyle \frac{5x}{5}=\frac{\frac{19}{3}}{5}$

$\displaystyle x = (\frac{19}{3})(\frac{1}{5})$

$\displaystyle x = \frac{19}{15}$

So your final answer is $\displaystyle x=\frac{19}{15}$.

Hope I helped :). If you have any questions go ahead and message me.

- May 17th 2009, 09:19 AMelitescouter
WOW AMAZING.... Thank you guys soooo much! This has helped me a ton!!!! I would of never got it by my self!

Thanks again!!!!! - May 17th 2009, 10:47 AMSodapop
This is how I would do it:

$\displaystyle \frac{3x-5}{4}+\frac{x}{2}=\frac{1}{3}$

Put 12 as the common denominator:

$\displaystyle \frac{3(3x-5)}{12}+\frac{6x}{12}=\frac{4}{12}$

$\displaystyle \frac{3(3x-5)+6x}{12}=\frac{4}{12}$

Then, simplify everything to remove the fractions:

$\displaystyle 3(3x-5)+6x=4$

Then, remove the parenthesis:

$\displaystyle 9x-15+6x=4$

Then, simply solve for $\displaystyle x$:

$\displaystyle 15x-15=4$

$\displaystyle 15x=19$

$\displaystyle x=\frac{19}{15}$

I know this has already been answered, but I just wanted to show you that many ways are possible to solve using algebra. Everybody uses his own way, and remember, there is no good or bad way, just use what you find the easiest for you! ;)