1. ## Simple fractional equation

So here's the equation.

Solve:

$7 \frac {1}{2}x -\frac {1}{2}x = 3 \frac {3}{4} + 39$

I know x = 12 because of the answer key, but I keep getting the wrong answer. I really need a step-by-step explanation.

thanks

2. Originally Posted by dreamer09
So here's the equation.

Solve:

$7 \frac {1}{2}x -\frac {1}{2}x = 3 \frac {3}{4} + 39$

I know x = 12 because of the answer key, but I keep getting the wrong answer. I really need a step-by-step explanation.

thanks
Hi dreamer09,

You have terms on the left side of your equation with the variable x in them. Combine these. On the right side, you just have numbers. Combine these.

$7\frac{1}{2}x-\frac{1}{2}x=3 \frac{3}{4}+39$

$7x=42 \frac{3}{4}$

Now, if you divide $42 \frac{3}{4}$ by 7, you will not get 12. You will, in fact, get $\frac{171}{28}$. Maybe you miscopied the problem. Check it again.

3. Originally Posted by masters
Hi dreamer09,

You have terms on the left side of your equation with the variable x in them. Combine these. On the right side, you just have numbers. Combine these.

$7\frac{1}{2}x-\frac{1}{2}x=3 \frac{3}{4}+39$

$7x=42 \frac{3}{4}$

Now, if you divide $42 \frac{3}{4}$ by 7, you will not get 12. You will, in fact, get $\frac{171}{28}$. Maybe you miscopied the problem. Check it again.
Oh sorry, yes I left something out.

$7\frac{1}{2}x-\frac{1}{2}x=3 \frac{3}{4}x+39$

4. Originally Posted by dreamer09
Oh sorry, yes I left something out.

$7\frac{1}{2}x-\frac{1}{2}x=3 \frac{3}{4}x+39$
Ok, that's better. Combine the tems on the left side first:

$7x=3 \frac{3}{4}x+39$

Next, subtract $3 \frac{3}{4}x$ from both sides of the equation.

$3 \frac{1}{4}x=39$

Finally divide both sides by $3 \frac{3}{4}$ to arrive at your desired conclusion.

Now, if it's the arithmetic that's giving you problems, you might want to convert those mixed fractions to improper fractions. Might make it easier to solve.

$7x=\frac{15}{4}x+39$

Multiply each term by 4.

$28x=15x+156$

Subtract 15x from both sides

$13x=156$

Divide by 13.

$x=12$