please help me with these algebra problems! thank you!
1.) solveb-4acx= ————2afor "a"
2.)simplify:
(5y)/(4xy^{2})•(3)/(5x)•(-3x^{2}y)/(3xy)
* the / means its a fraction.
please help me with these algebra problems! thank you!
1.) solveb-4acx= ————2afor "a"
2.)simplify:
(5y)/(4xy^{2})•(3)/(5x)•(-3x^{2}y)/(3xy)
* the / means its a fraction.
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)
$\displaystyle \frac{a(2x+4c)}{2x+4c}=\frac{b}{2x+4c}$
$\displaystyle a=\frac{b}{2x+4c}$
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)