Thread: algebra problems

please help me with these algebra problems! thank you!

1.) solve

b-4ac

x= ————

2a

for "a"

---

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

* the / means its a fraction.

1) multiply both sides by 2a

so you get so the 2a's cancel each other

add 4ac to both sides

in LHS take (a) as common factor

divid both sides by the term (2x+4c)

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)

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

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