1. ## algebra problems with fractions and letters

1.) solve
b-4ac
x= ————
2a
for "a"

2.)simplify:
(5y)/(4xy2)•(3)/(5x)•(-3x2y)/(3xy)

* the / means its a fraction.

2. ## Re: algebra problems

1)$\displaystyle x=\frac{b-4ac}{2a}\,\,$ multiply both sides by 2a

so you get $\displaystyle (2a)x=(\frac{b-4ac}{2a})*2a\,\,$ so the 2a's cancel each other

$\displaystyle 2ax=b-4ac$ add 4ac to both sides

$\displaystyle 2ax+4ac=b-4ac+4ac$ in LHS take (a) as common factor

$\displaystyle a(2x+4c)=b$ divid both sides by the term (2x+4c)

$\displaystyle \frac{a(2x+4c)}{2x+4c}=\frac{b}{2x+4c}$

$\displaystyle a=\frac{b}{2x+4c}$

3. ## Re: algebra problems

1st one: easier if you use criss-cross multiplication (example: if a/b = c/d, then ad = bc):
2ax = b - 4ac : 2ax + 4ac = b : a(2x + 4c) = b : a = b / (2x + 4c)

2nd one: start by cancelling where possible; like 5y / (4xy^2) = 5 / (4xy)

4. ## Re: algebra problems with fractions and letters

wow! thank you so much! i understand!!（＾∇＾）

5. ## Re: algebra problems

thank you!!! i understand! thank you for helping with both problems ><