hey just need to know if im on the right track with this: 7/(4x+2) + 3/(2 - 3x) = 0

[MrBubbles]

so im to solve, and state assumptions on x.
i am i on the correct track?

 

[stapel]

so im to solve, and state assumptions on x.View attachment 37039
i am i on the correct track?

You can check the answer to any "solving" exercise by plugging your value back into the original problem. If your case, you want to see if $x = \frac{20}{9}$ makes the original equation true. So check, plugging into the left-hand side:

$\qquad \dfrac{7}{4\left(\frac{20}{9}\right) + 2} + \dfrac{3}{2 - 3\left(\frac{20}{9}\right)}$

$\qquad \dfrac{7}{\left(\frac{98}{9}\right)} + \dfrac{3}{\left(\frac{-42}{9}\right)}$

$\qquad \left(\frac{7}{1}\right)\left(\frac{9}{98}\right) + \left(\frac{3}{1}\right)\left(-\frac{9}{42}\right)$

$\qquad \frac{9}{14} - \frac{9}{14}$

$\qquad 0$

So the solution checks.

 

[MrBubbles]

thank u so much.

 

[lookagain]

so im to solve, and state assumptions on x.
i am i on the correct track?

suki, you are certainly on the right track, with a correct final solution,
and you were shown how your solution checks. Let me write a slight
redo with certain equals signs that do not belong. Also, let me
address the "assumptions on x" you mentioned.

$\displaystyle \dfrac{7}{4x + 2} \ + \ \dfrac{3}{2 - 3x} \ = \ 0$

$\displaystyle 7(2 - 3x) \ + \ 3(4x + 2) \ = \ 0$

$\displaystyle 14 - 21x + 12x + 6 \ = \ 0$

$\displaystyle -9x + 20 \ = \ 0$

$\displaystyle 9x \ = \ 20$

$\displaystyle x \ = \ \dfrac{20}{9}$

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For the fractions in the first line, neither denominator may equal zero.
We can find out restrictions on x by stating each denominator is not
equal to zero and solve that respective nonequation:

$\displaystyle 4x + 2 \ne 0$

$\displaystyle 4x \ne -2$

$\displaystyle x \ne \dfrac{-1}{2}$

Also,

$\displaystyle 2 - 3x \ne 0$

$\displaystyle -3x \ne -2$

$\displaystyle x \ne \dfrac{2}{3}$

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Notice how the solution of $\displaystyle \ x \ = \ \dfrac{20}{9} \$ is different from either of
these two restricted values.

 

[MrBubbles]

im loving the help. thank u.