Solve equation...Assistance please

[abbycat]

I see that the quadratic in the last one factors out to (x-5)(x-8)...

 

[pka]

I see that the quadratic in the last one factors out to (x-5)(x-8)...

What you posted is false.
\(\displaystyle \dfrac{3}{x-5}+\dfrac{7}{x-8}\ne\dfrac{5}{x^2-13x+40}\)

BUT \(\displaystyle \dfrac{3}{x-5}+\dfrac{7}{x-8}=\dfrac{10x-59}{x^-13x+40}\)

 

[abbycat]

What you posted is false.
\(\displaystyle \dfrac{3}{x-5}+\dfrac{7}{x-8}\ne\dfrac{5}{x^2-13x+40}\)

BUT \(\displaystyle \dfrac{3}{x-5}+\dfrac{7}{x-8}=\dfrac{10x-59}{x^-13x+40}\)

I'm sorry, but I don't quite follow...The attachment is the problem that I was given

 

[pka]

I'm sorry, but I don't quite follow...The attachment is the problem that I was given

It is not our fault that you do not follow a reply if you did not post a correct question.
If you mean to solve for \(\displaystyle x\) then by all means say so.
That is trivial: \(\displaystyle 10x-59=5\).

 

[soroban]

Hello, abbycat!

I see that the quadratic in the last one factors out to (x - 5)(x - 8) . . . . Good!

. . \(\displaystyle \text{Solve for }x\!:\;\;\dfrac{3}{x-5} + \dfrac{7}{x-8} \:=\:\dfrac{5}{x^2-13x + 40}\)

\(\displaystyle \displaystyle \text{We have: }\:\frac{3}{x-5} + \frac{7}{x-8} \:=\:\frac{5}{(x-5)(x-8)}\)


\(\displaystyle \text{Multiply by }(x-5)(x-8)\!:\)

. . \(\displaystyle (x-5)(x-8)\!\cdot\!\dfrac{3}{x-5} + (x-5)(x-8)\!\cdot\!\dfrac{7}{x-8} \:=\) .\(\displaystyle (x-5)(x-8)\!\cdot\!\dfrac{5}{(x-5)(x-8)}\)

. . \(\displaystyle 3(x-8) + 7(x-5) \:=\:5 \quad\Rightarrow\quad 3x - 24 + 7x - 35 \:=\:5 \quad\Rightarrow\quad 10x \:=\:64\)

. . \(\displaystyle x \:=\:\dfrac{64}{10} \quad\Rightarrow\quad \boxed{x \:=\:\dfrac{32}{5}}\)

 

[abbycat]

Why thank you! After factoring, I was at a standstill and could not figure out where to go from there. This process is very clear to me now!