SOLUTION: The directions say: Solve the equation. Check your solution. {{{2m/(m+4) = 3/(m-1)}}} 1. so what i did was cross multiply which gave me: {{{2m(m-1)=3(m+4)}

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The directions say: Solve the equation. Check your solution. {{{2m/(m+4) = 3/(m-1)}}} 1. so what i did was cross multiply which gave me: {{{2m(m-1)=3(m+4)}      Log On


   

Question 440606: The directions say: Solve the equation. Check your solution.
2m%2F%28m%2B4%29+=++3%2F%28m-1%29
1. so what i did was cross multiply which gave me: 2m%28m-1%29=3%28m%2B4%29
2. then i distributed the 2m and 3 giving me: 2m%5E2-2m+=+3m%2B12
3. then i made it into a quadratic equation: 2m%5E2-5m-12=0

*** From here on I am lost, I tried factoring but i couldnt find anything that multiplies to -12 and adds
to -5
*** In the back of the book they showed the problem worked out, and from
here they managed to factor it to %282m%2B3%29%28m-4%29+=+0 ; I'm confused
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2m^2 - 5m -12 = 0
----
2m^2 -8m+3m -12 = 0
----
2m(m-4)+3(m-4) = 0
---
(m-4)(2m+3) = 0
m = 4 or m = -3/2
======================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
the sure-fire way is to use the quadratic formula
m+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
when equation is in the form
+ax%5E2+%2B+b%2Ax+%2B+c+=+0+
2m%5E2+-+5m+-+12+=+0+
+a+=+2+
+b+=+-5+
+c+=+-12+
m+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A2%2A%28-12%29+%29%29%2F%282%2A2%29+
m+=+%28+5+%2B-+sqrt%28+25+%2B+96+%29%29+%2F+4+
m+=+%28+5+%2B-+sqrt%28+121+%29%29+%2F+4+
+m+=+%28+5+%2B+11%29%2F4+
(1) +m+=+4+
and the (-) square root,
+m+=+%285+-+11%29%2F4+
(2) +m+=+-3%2F2+
Now just rewrite (1) and (2) to get factors
+m+-+4+=+0+
+m+%2B+3%2F2+=+0+
Multiply both sides by 2
+2m+%2B+3+=+0+
+%28m+-+4%29%2A%282m+%2B+3%29+=+0+
This works no matter how strange the coefficients are