# Can someone explain how this linear equation was solved [proof provided]

Can someone please explain to me the steps taken in the proof provided to solve the linear equation? : http://i.imgur.com/2N52occ.jpg "Proof"

What I don't understand is how he removed the denominator of both fractions (3,5 respectively). I know that 3 * 5 is 15, so he just multiplied everything by 15?

Now I understand that he found the least common multiples of 3 and 5 (which I kinda already knew he was trying to do), however, did he use the 5 from the 2nd part of the equation (2m / 5) or from the first part (5/1). I think that is where I lost track.

• yes. Just multiply everything by 15. You don't have to... but that's how whoever did it, did it – user44197 Jan 9 '14 at 3:42
• Yes, essentially. – 2012ssohn Jan 9 '14 at 3:42

## 1 Answer

Yes. The proof just multiplied everything by $15$ which is the least common multiple of $3,5$.

Note that in this case $3\times 5=15$ is the least common multiple of $3,5$, but if you have two denominators $3,6$, then the least common multiple is $6$. ($3\times 6=18$ is not the least common multiple.)

In general, if you have several fractions in an equation, you can multiply everything by the least common multiple of all of the denominators to get an equation without any fraction.

So, you'll be able to solve it more easily.

EDIT : To answer another quation.

The answer is the second. You can write as $$\frac 51-\frac{2m+7}{3}=\frac{2m}{5}.$$

Hence, the denominators are only $1,3,5$.

$$15\times \frac 51-15\times \frac{2m+7}{3}=15\times \frac{2m}{5}$$ $$15\times 5-5(2m+7)=3(2m).$$

• @Wish : I added a bit. Does this help? – mathlove Jan 9 '14 at 4:05
• Yes thank you I see it more clearly now :). – Wish Jan 9 '14 at 4:06
• You are welcome! my pleasure. – mathlove Jan 9 '14 at 4:07
• Hey again, can I ask one more thing please? 1 goes into 15, 15 times - so the first denominator should be multiplied by 15, however shouldn't the second denominator be multiplied by 5 and the last denominator be multiplied by 3 since that is how many times they go into 15, and not all just multiplied by 15? I found it more comprehensible this way, but I am not sure if it is valid. Thank you so much for your time :) – Wish Jan 9 '14 at 4:49
• We have to multiply everything by the same number, so first we multiply everything $15$. Then, since $15/3=5, 15/5$, you can see the added edit of mine. – mathlove Jan 9 '14 at 4:55