Solving equations w fractions (not quite getting it, help!)

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Solving equations w fractions (not quite getting it, help!)

Hi,
Can someone please explain to me how to do this problem step-by-step

Solve for b: H=5/3(a+2ab)

I know the answer is b=3H-5a/10
However, I'm just unsure how to get there, more specifically how H became 3H.
More, more specifically I don't know how to multiply the fraction with the equation... for the right side I know I convert the a and 2ab into fractions by giving them the same denominator as 5/3 (i.e. 5 * a/3 + 2ab/3)

..But, how did 3 get multiplied with H.. shouldn't it be 5/3*H = 5/3 * H/3? (why is H getting multiplied with the denominator??)

please explain to me how this works
thanks!
onmyown

Posts: 18
Joined: Fri Dec 17, 2010 10:16 pm

onmyown wrote:Can someone please explain to me how to do this problem step-by-step

Solve for b: H=5/3(a+2ab)

....But, how did 3 get multiplied with H.. shouldn't it be 5/3*H = 5/3 * H/3? (why is H getting multiplied with the denominator??)

You need to get "b" by itself. It is currently in the denominator:

. . . . .$H\, =\, \frac{5}{3(a\, +\, 2ab)}$

To undo this, you need to multiply through by the current denominator. This will get the "b" up on the left-hand side.

To get the "b" by itself, you need to get rid of the "3" which is now on top, so you'll need to divide through after you subtract off the 3aH term.

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm

Re:

stapel_eliz wrote:
onmyown wrote:Can someone please explain to me how to do this problem step-by-step

Solve for b: H=5/3(a+2ab)

....But, how did 3 get multiplied with H.. shouldn't it be 5/3*H = 5/3 * H/3? (why is H getting multiplied with the denominator??)

You need to get "b" by itself. It is currently in the denominator:

. . . . .$H\, =\, \frac{5}{3(a\, +\, 2ab)}$

To undo this, you need to multiply through by the current denominator. This will get the "b" up on the left-hand side.

To get the "b" by itself, you need to get rid of the "3" which is now on top, so you'll need to divide through after you subtract off the 3aH term.

Thank you. It all makes sense now!
onmyown

Posts: 18
Joined: Fri Dec 17, 2010 10:16 pm