Solving Multivariable Equations: 1/n=2/m-3/q for q (my steps diff. from calc.)

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spaceshowfeature1

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Hello! I just solved this equation for q: 1/n=2/m-3/q.
1. Multiply the lcd (nmq): mq=-3nm+2nq
2. Add 3nm to the left side: mq+3nm=2nq
3. Subtract mq from the left side: 3nm=2nq-mq
4. Factor out q: 3nm=q(2n-m)
5. Divide each side by 2n-m: q=3nm/2n-m

The calcutor I put this problem through gave me a different set of steps, and an answer of -3nm/m-2n. After looking at the answer, I figured out that if I multiply my answer by -1, I will get the calculator’s answer. Is my answer correct? Thanks and good day!
 
spaceshowfeature1 said:
Hello! I just solved this equation for q: 1/n=2/m-3/q.
1. Multiply the lcd (nmq): mq=-3nm+2nq
2. Add 3nm to the left side: mq+3nm=2nq
3. Subtract mq from the left side: 3nm=2nq-mq
4. Factor out q: 3nm=q(2n-m)
5. Divide each side by 2n-m: q=3nm/2n-m

The calcutor I put this problem through gave me a different set of steps, and an answer of -3nm/m-2n. After looking at the answer, I figured out that if I multiply my answer by -1, I will get the calculator’s answer. Is my answer correct? Thanks and good day!
Click to expand...
\(\displaystyle mq = -\ 3nm + 2nq \implies\)

\(\displaystyle 3nm + mq = 3nm - 3nm + 2nq \implies\)

\(\displaystyle 3nm + mq = 2nq \implies\)

\(\displaystyle 3nm + mq - mq = 2nq - mq \implies\)

\(\displaystyle 3nm = 2nq - mq \implies\)

\(\displaystyle 3nm = q(2n - m)\implies\)

\(\displaystyle \dfrac{q(2n - m)}{2n - m} = \dfrac{3nm}{2n - m} \text { if } 2n \ne m \implies\)

\(\displaystyle q = \dfrac{3nm}{2n - m} \text { if } 2n \ne m.\)

So you did a very nice job, but do you see why you must impose the condition that

\(\displaystyle 2n \ne m\)?

Notice that in your last line, you should write 3mn / (2n - m). The parentheses are important when writing in line.

And it is not a multiplication by minus 1, but a multiplication by a special form of 1.

\(\displaystyle q = \dfrac{3mn}{2n - m} = \dfrac{3mn}{2n - m} * 1 = \dfrac{3mn}{2n - m} * \dfrac{-\ 1}{-\ 1} = \dfrac{-\ 3mn}{m - 2n}.\)

GOOD WORK.
 
Last edited:
JeffM said:
\(\displaystyle mq = -\ 3nm + 2nq \implies\)

\(\displaystyle 3nm + mq = 3nm - 3nm + 2nq \implies\)

\(\displaystyle 3nm + mq = 2nq \implies\)

\(\displaystyle 3nm + mq - mq = 2nq - mq \implies\)

\(\displaystyle 3nm = 2nq - mq \implies\)

\(\displaystyle 3nm = q(2n - m) \implies \dfrac{q(2n - m)} \implies\)

\(\displaystyle \dfrac{3nm}{2n - m} \text { if } 2n \ne m \implies\(\displaystyle

\(\displaystyle q = \dfrac{3nm}{2n - m} \text { if } 2n \ne m.\)

So you did a very nice job, but do you see why you must impose the condition that

\(\displaystyle 2n \ne m\)?

Notice that in your last line, you should write 3mn / (2n - m). The parentheses are important when writing in line.

And it is not a multiplication by minus 1, but a multiplication by a special form of 1.

\(\displaystyle q = \dfrac{3mn}{2n - m} = \dfrac{3mn}{2n - m} * 1 = \dfrac{3mn}{2n - m} * \dfrac{-\ 1}{-\ 1} = \dfrac{-\ 3mn}{m - 2n}.\)

GOOD WORK.\)\)
Click to expand...
\(\displaystyle \(\displaystyle
Thanks! I love proofs :D\)\)
 
spaceshowfeature1 said:
I just solved this equation for q: 1/n=2/m-3/q.

1. Multiply the lcd (nmq): mq=-3nm+2nq
2. Add 3nm to the left side: mq+3nm=2nq
3. Subtract mq from the left side: 3nm=2nq-mq
4. Factor out q: 3nm=q(2n-m)
5. Divide each side by 2n-m: q=3nm/2n-m
Click to expand...
Slightly simpler/easier if you start this way:
1/n = 2/m - 3/q
Step 1:
3/q = 2/m - 1/n
Get my drift?
 
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