My Math Forum (http://mymathforum.com/math-forums.php)
-   Algebra (http://mymathforum.com/algebra/)
-   -   solve algebraically (http://mymathforum.com/algebra/20025-solve-algebraically.html)

 drewm July 4th, 2011 01:58 PM

solve algebraically

I have not taken a math course in over a year and was wondering if someone could help me to recall a few things.

I need to solve for y: 2m= (3y-4)/(5-x)

For m: 2m= (3y-4)/(5-x)

for x: 2m= (3y-4)/(5-x)

Can someone help me understand how to do this? Thank you!

 filipd July 4th, 2011 02:12 PM

Re: solve algebraically

If it asks you to solve let's say for x, that means that on left side of equality you need to have x: (x=...). So out of that equation you have to express x. Same is for m or y. Does this help?

 MarkFL July 4th, 2011 03:06 PM

Re: solve algebraically

I'll do the first part, and see if you can apply the technique to the other two parts.

Solve $2m=\frac{3y-4}{5-x}$ for y.

We wish to isolate y one one side of the equation and everything else on the other side. So the first thing we need to do is get rid of the denominator on the right side by multiplying both sides by that denominator so that it will calcel out on the right side:

$2m(5-x)=\frac{3y-4}{5-x}(5-x)$

Now divide out or cancel the factor common to the denominator and numerator on the right:

$2m(5-x)=3y-4$

Now we need to remove the 4 being subtracted on the right by adding 4 to both sides:

$2m(5-x)+4=3y-4+4$

Since -4 + 4 = 0, and a + 0 = a, we may now write:

$2m(5-x)+4=3y$

Now we need to remove the coefficient of y which is 3 by dividing both sides by 3:

$\frac{2m(5-x)+4}{3}=\frac{3y}{3}=\frac{3}{3}y=y$

Thus, we find:

$y=\frac{2m(5-x)+4}{3}$

 All times are GMT -8. The time now is 09:35 AM.