Can't do this basic problem!

Jun 2009
186
5
Mars
I cannot do this problem.
Question:
0.7 = s/335 - s/5300
Find s.

do I do it as a quadratic or how else.
Thanks for help.
 
Nov 2009
927
260
Wellington
Hello,
just substract the two fractions by taking the greatest common divisor of 335 and 5300. Then you are left with a linear equation.
 
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Dec 2009
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I cannot do this problem.
Question:
0.7 = s/335 - s/5300
Find s.

do I do it as a quadratic or how else.
Thanks for help.
Or....

just multiply both sides by 335 and then multiply both sides by 5300
as the sides will remain equal.

\(\displaystyle 0.7=\frac{s}{335}-\frac{s}{5300}\)

\(\displaystyle 0.7(335)=s-\frac{335s}{5300}\)

\(\displaystyle 0.7(335)(5300)=5300s-335s=4965s\)

\(\displaystyle 1242850=4965s\)

\(\displaystyle s=\frac{1242850}{4965}\)
 
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Jun 2009
186
5
Mars
Wait.
The second step where you moved just multiplied both sides doesn't make any sense. Shouldn't you make fraction. Doing it that way I got the answer 444447165.16
Is that right or is Archie right.
 
Nov 2009
927
260
Wellington
\(\displaystyle 0.7=\frac{s}{335}-\frac{s}{5300}\)

\(\displaystyle 0.7=\frac{5300 \times s}{335 \times 5300}-\frac{335 \times s}{5300 \times 335}\)

\(\displaystyle 0.7=\frac{5300 \times s - 335 \times s}{335 \times 5300}\)

\(\displaystyle 0.7=\frac{s(5300 - 335)}{335 \times 5300}\)

\(\displaystyle 0.7=\frac{4965s}{1775500}\)

\(\displaystyle 0.7 \times 1775500=4965s\)

\(\displaystyle \frac{0.7 \times 1775500}{4965}=s\)

\(\displaystyle s \approx 250.322 \)

Does it make sense now ?
 
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Dec 2009
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Wait.
The second step where you moved just multiplied both sides doesn't make any sense. Shouldn't you make fraction. Doing it that way I got the answer 444447165.16
Is that right or is Archie right.
If you have

\(\displaystyle \frac{3}{5}-\frac{2}{5}=x\)

then you can just multiply both sides by 5 to get 3-2=5x

It's the same as combining the fractions to get \(\displaystyle \frac{3-2}{5}=x\)

You can either combine the fractions or just multiply both sides by the denominators.

\(\displaystyle x=\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\)

\(\displaystyle 2x=1+\frac{2}{3}\)

\(\displaystyle 6x=3+2=5\)

is another way.

If x=5, then 2x=10 by multiplying both sides by 2, they are still equal and saying the same thing.

5=5, 5(2)=5(2), (3+2)=5, 2(3+2)=2(5)

If two sides are the same, then if you multiply all terms of both sides
by the same number, they will still be equal.

You should get used to adding fractions, so that it becomes easy,
however, multiplying by the denominators in this case is faster.
 
Last edited:
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Jun 2009
186
5
Mars
WOW great explanation, sorry I have trouble with the basics.
 
Jun 2009
186
5
Mars
So Bacterius if I multiply one side by 335 I don't have to multiply other side by 335. Correct.
 
Nov 2009
927
260
Wellington
So Bacterius if I multiply one side by 335 I don't have to multiply other side by 335. Correct.
Consider \(\displaystyle \frac{3}{5}\). This is the same as \(\displaystyle \frac{3 \times 7}{5 \times 7}\) (since the seven cancels out). This trick allows us to set two different fractions to the same denominator, in order to add them easily.

Note : you are not actually doing anything to the equation by multiplying by 335 inside the fraction : you are only rewriting a fraction in a different form (but it is equivalent) to make it simpler to add. Therefore, since the equation is left unchanged, doing anything to the other side would be superfluous. Simple example :

\(\displaystyle \frac{3 \times 7}{5 \times 7} = \frac{3}{5} \times \frac{7}{7} = \frac{3}{5} \times 1 = \frac{3}{5}\)

See ? It is basically equivalent to multiplying the fraction by 1, but adding multiplicative information to allow fraction addition. Since multiplying by 1 does not change anything, there is nothing to change on the other side of the equation ;)
 
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