Question

how do you solve (3m-2)/(m+1)=4-(m+2)/(m-1)? if you could show steps please :)

Answer 1

Multiply through by (m+1) gives
\[3m-2=4(m+1)-(m+2)/(m ^{2}-1)\]
then rearrange so the fraction is on one side and the other terms on the other
\[(m+2)/(m ^{2}-1)=4m+4-(3m-2)=m+6\]
Then multiply up the bottom of the fraction
\[m+2=(m+6)(m ^{2}-1)=m ^{3}+6m ^{2}-m-6\]
Then rearrange so the equation is equal to zero
\[m ^{3}+6m ^{2}-2m-8=0\]
Then solve this polynomial, though I come out with long fractions - possibly I misinterpreted the question :S I assumed the 4 was not in the fraction, is that right?

Answer 2

You have a small mistake
\[\frac{m+2}{m-1} (m+1) \neq \frac{m+2}{m^2-1}\]

Answer 3

Damn, I knew I had done something, sorry! I will leave it to you lovely people to solve then, it seems I am too tired to keep maths in my head for now :) Basically for any equation like that, you get rid of the simple fraction by multiplying up, rearrange the equation so the remaining fraction is on one side, multiply up again, rearrange for zero and solve the resultant polynomial.

Answer 4

\[\frac{3m-2}{m+1}=4-\frac{m+2}{m-1} \iff 3m-2=4(m+1)-\frac{(m+2)(m+1)}{m-1},m\neq \left\{ -1; 1 \right\}\]
\[\iff (3m-2)(m-1)=4(m+1)(m-1)-(m+2)(m+1)(m-1), m\neq \left\{ -1;1 \right\}\]
\[\iff 3m^2-3m-2m+2=4(m^2-1)-(m+2)(m^2-1),m\neq \left\{ -1;1 \right\}\]
\[\iff 3m^2-5m+2=4m^2-4-(m^3+2m^2-m-2),m\neq \left\{ -1;1 \right\}\]
\[\iff 3m^2-5m+2=2m^2-m^3+m-2,m\neq \left\{ -1;1 \right\}\]
\[\iff m^3+m^2-6m+4=0,m\neq \left\{ -1;1 \right\}\]

I THINK... although I'm not sure about it all...

Answer 5

Um, sorry, second line, is
\[\iff (3m-2)(m-1)=4(m+1)(m-1)-(m+2)(m+1),m\neq \left\{ -1;1 \right\}\]
Then keep going, it's much easier to solve...