View Full Version : Solving equations with fractions

naevino

01-20-2010, 11:13 PM

Hello, I need help with a question that I am working on. I tried to multiply both sides of the equation by the LCD. But I am not sure how to do that in this type of problem.

Here is the problem: Solve.

3(x-5)/2 = 2(x+5)/3

Here is what i got

9(3x-15) = 4(2x+10)

Thanks for you time and help!

mmm4444bot

01-21-2010, 12:23 AM

Hi naevino:

You're correct; multiply both sides of the given equation by the LCD.

The LCD is 6, right?

\frac{6}{1} \cdot \frac{3(x - 5)}{2} \;=\; \frac{6}{1} \cdot \frac{2(x + 5)}{3}

Is this what you tried?

(Your result looks like you accomplished multiplying the lefthand side by 81 and the righthand side by 12.)

Do you see the cancellations that take place above, to get rid of the fractions in this exercise?

Okay! I think I see what you might have done. AFTER you canceled the denominators, you multiplied by 3 twice, and you multiplied by 2 twice.

In other words, when you multiplied 3*[3(x - 5)], you multiplied both the 3 and the (x - 5) by 3. That's not correct.

We multiply what's inside the square brackets by 3:

3*3*(x - 5) = 9(x - 5)

That's all there is to it. (Remember the Order of Operations?)

Likewise: 2*[2*(x + 5)] is 2*2*(x + 5) = 4(x + 5)

Now you've got the following.

9(x - 5) = 4(x + 5)

Let us know, if you still need help.

If you get a value for x, you can check it by substituting it for x in the original equation, followed by doing the arithmetic on each side to see whether or not they're equal.

Cheers ~ Mark

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