Variables on both sides

GizmoBlueBell

New member
Joined
Jul 11, 2016
Messages
1
Hi,

Struggling with the following question, just don't seem to get to the right answer. Unfortunately the PC won't carry on with new problems until this one is solved.
Not sure where it is going wrong.

1/6 (a-4) = 1/3(2a + 4)
Thought to first get rid of fractions first by finding common denominator.

I've tried answers of 4 , -4, -2, -2.66 and -2.67 but the computer doesn't accept them.....

Any assistance/guidance please. Thanks
 
Hi,

Struggling with the following question, just don't seem to get to the right answer. Unfortunately the PC won't carry on with new problems until this one is solved.
Not sure where it is going wrong.

1/6 (a-4) = 1/3(2a + 4)
Thought to first get rid of fractions first by finding common denominator.

I've tried answers of 4 , -4, -2, -2.66 and -2.67 but the computer doesn't accept them.....

Any assistance/guidance please. Thanks

Good thought - but

how did you do it?

What did you get after getting rid of the fractions?

Can you please share your work?
 
Hi,

Struggling with the following question, just don't seem to get to the right answer. Unfortunately the PC won't carry on with new problems until this one is solved.
Not sure where it is going wrong.

1/6 (a-4) = 1/3(2a + 4)
Thought to first get rid of fractions first by finding common denominator.

I've tried answers of 4 , -4, -2, -2.66 and -2.67 but the computer doesn't accept them.....

Any assistance/guidance please. Thanks

First of all, is it:

\(\displaystyle \frac{1}{6}(a-4) = \frac{1}{3}(2a+4)\)

or

\(\displaystyle \frac{1}{6(a-4)}=\frac{1}{3(2a+4)}\) ??
 
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