hey just need to know if im on the right track with this: 7/(4x+2) + 3/(2 - 3x) = 0
- Thread starter MrBubbles
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- Feb 4, 2004
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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 [imath]x = \frac{20}{9}[/imath] makes the original equation true. So check, plugging into the left-hand side:
[imath]\qquad \dfrac{7}{4\left(\frac{20}{9}\right) + 2} + \dfrac{3}{2 - 3\left(\frac{20}{9}\right)}[/imath]
[imath]\qquad \dfrac{7}{\left(\frac{98}{9}\right)} + \dfrac{3}{\left(\frac{-42}{9}\right)}[/imath]
[imath]\qquad \left(\frac{7}{1}\right)\left(\frac{9}{98}\right) + \left(\frac{3}{1}\right)\left(-\frac{9}{42}\right)[/imath]
[imath]\qquad \frac{9}{14} - \frac{9}{14}[/imath]
[imath]\qquad 0[/imath]
So the solution checks.
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.