Thread: 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

Yes you can use the same substitution method you used. Just solve one of the two equations for x or y first then substitute.

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

i.e. let 2nd equation be y = (1/3) - (5/3)x and sub that in for y in first or solve first for y and sub into 2nd