Probability
Full Member
- Joined
- Jan 26, 2012
- Messages
- 425
In a activity I was asked to multiply out the brackets. It seems I did not go far enough from what I thought was correct.
[MATH]{2}(\frac{1}{2}{A^2}+\frac{3}{2})=[/MATH]
I thought;
[MATH]{2}(\frac{1}{2}{A^2}+\frac{3}{2})={2}\frac{1}{2}{A^2}+{2}\frac{3}{2}[/MATH]
Which I thought was multiplying out the brackets. The answer given is [MATH]({A^2}+{3})[/MATH]
My idea now seem to be based around how am I going to cancel down the fractions to end up with [MATH]{+}{3}?[/MATH]
So the first thought I had was to convert the mixed fractions to proper fractions (top heavy);
[MATH]\frac{5}{2}{A^2}+\frac{7}{2}=[/MATH]
From this point on-wards using trial and error I can end up with [MATH]{A^2}-\frac{3}{5}[/MATH] which is obviously incorrect.
If I were to cancel down [MATH]\frac{5}{2}{A^2}={A^2}[/MATH] but I can't do the same to [MATH]\frac{7}{2}[/MATH] so from this point I'm stuck!!
[MATH]{2}(\frac{1}{2}{A^2}+\frac{3}{2})=[/MATH]
I thought;
[MATH]{2}(\frac{1}{2}{A^2}+\frac{3}{2})={2}\frac{1}{2}{A^2}+{2}\frac{3}{2}[/MATH]
Which I thought was multiplying out the brackets. The answer given is [MATH]({A^2}+{3})[/MATH]
My idea now seem to be based around how am I going to cancel down the fractions to end up with [MATH]{+}{3}?[/MATH]
So the first thought I had was to convert the mixed fractions to proper fractions (top heavy);
[MATH]\frac{5}{2}{A^2}+\frac{7}{2}=[/MATH]
From this point on-wards using trial and error I can end up with [MATH]{A^2}-\frac{3}{5}[/MATH] which is obviously incorrect.
If I were to cancel down [MATH]\frac{5}{2}{A^2}={A^2}[/MATH] but I can't do the same to [MATH]\frac{7}{2}[/MATH] so from this point I'm stuck!!