Simultaneous Equation

Hi. I have the working out here for a sim equation but I’m not understanding the process. Explain like I’m five please and thank you.

Y/3 (< fraction) + 12 = y + 12 / 2 (< all of it is a fraction)
Then we eliminate the fractions to get a simple sim equation.
Starting on the right we remove the denom of 2 and multiply the y and the 12 on the left hand side by this 2
So 2y/3 + 24 = y+12 (my first question is why we don’t multiply the 3 by 2 as well?)
Then 2y+72=3(y+12) - removing the 3 on the left hand side (now why do we multiply the number on the same side (24 > 72) by 3 and we didn’t do that for the 2? Also why do we multiply the numerator?)
2y+72= 3y+36
Then I know how to work it out from there.

Is this suppose to be
$\dfrac{y}{3} + 12 = \dfrac{y + 12}{2}$ ?

So your first step is to multiply both sides by 2. This will clear the fraction on the RHS because the denominator is 2.

On the LHS you are multiplying by 2 i.e. doubling, so you only multiply the numerator by 2. If you multiplied the denominator of 3 by 2 you would be finding an equivalent fraction to $\dfrac{y}{3}$, which is NOT what you want!

I appreciate the help but I’d like insight on my other questions too.

Are you now happy that after the 1st step you now have:
$\dfrac{2y}{3} + 24 = y + 12$ ?

You now have 2 terms on the LHS and one of them involves a fraction with the denominator 3, In order to clear this fraction you can multiply both sides by 3 but remember that you have to multiply both sides by 3 - you are basically just applying the rule 3(a + b) = 3a + 3b to both sides.

The explanation here is really just the same as I gave for the first case of multiplying by 2. Remember that there is a difference between "multiplying everything by a number to clear a fraction" and "finding a fraction equivalent to a given one"

Why do we multiply the 24 by the denom of 3 when we’re already multiplying just by cancelling the fraction? So technically you’re multiplying two things on the LHS by 3. Now, we multiply 24 by 3 and they’re in the same side of the equation.. so why don’t we multiply y+12 on the rhs by 2?

Break it down into components so you can see what's going on

You have A + B = y + 12 where $A = \dfrac{2y}{3}$ and $B = 24$

Now multiply both sides by 3 to get

$3(A + B) = 3(y + 12)$

Now multiply out the brackets on both sides:

$3A + 3B = 3y + 36$

Now substitute back in for A and B:

$3A = 3 \times \dfrac{2y}{3} = 2y$ and $3B = 3 \times 24 = 72$

Hence $2y + 72 = 3y + 36$

Break it down into components so you can see what's going on

You have A + B = y + 12 where $A = \dfrac{2y}{3}$ and $B = 24$

Now multiply both sides by 3 to get

$3(A + B) = 3(y + 12)$

Now multiply out the brackets on both sides:

$3A + 3B = 3y + 36$

Now substitute back in for A and B:

$3A = 3 \times \dfrac{2y}{3} = 2y$ and $3B = 3 \times 24 = 72$

Hence $2y + 72 = 3y + 36$

I’m still confused about the rhs. Why isn’t that 2(y/3 +12) = 2(y+12)? We haven’t multiplied anything on the rhs by 2, is this because it’s a fraction? Multiplying it would change the fraction?
But then why did we multiply the first one by 2 and not the second? i.e 2y/3? That in itself is changing the fraction

Also you didn’t multiply 2y\3 even though you put the 3 behind its brackets

OK, can I just ask what level you're currently working at? Are you a GCSE student or are you returning to education after a gap?


The operations I'm performing on your equations are fairly basic GCSE manipulations, but you seem to be having a bit of trouble following the logic, so I need to try to understand how much algebra you understand already because maths is a foundational subject and if you haven't grasped the basics then you are going to struggle moving on to more difficult stuff

I’ve returned to education in June after 5 years educationless. I’m 18 now.