# Thread:

1. ## Simultaneous Equation [C1]

I've been going through a textbook and I've came across Simultaneous equations by this point. I can do many, such as:

c = 7d - 2 [1]
3c - 4d = 11 [2]

To do this I subbed c in [2] and did this method:

3(7d-2) - 4d = 11
21d - 4d - 6 = 11
17d = 17
d = 1

And then went back to 3c - 4d = 11
I found that 3c = 15
and c = 5.

However the last question is:

6x - 5y = 27
3y = 1 - 5x

Now I'm pretty sure you don't use the same method as last time so what can I do here

2. For example: from the second equation: $y=\frac{1-5x}3$ and then substitute this into the first one: $6x-\frac 53(1-5x)=27\leftrightarrow 18x-5(1-5x)=81\leftrightarrow 18x-5+25x=81\leftrightarrow 43x=86\leftrightarrow x=2$ substitute this into $y=\frac{1-5x}3$: $y=\frac {1-5\cdot 2}3=-3$.

3. rearrange so that the x and y on same side

6x - 5y = 27
3y + 5x = 1

then multiply top equation by 3(for y) and bottom equation by 5 so you get

18x - 15y = 81
15y + 15x = 5

now i think you will be able to solve it

4. i made a slight mistake, the bottom equation should say 15y + 25x = 5

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