Be careful how you write this! You are dividing y- c by m so that would be written (y- c)/m= x. What you wrote would normally be interpreted as y- (c/m)= x which is not correct.

No, mx+ cx= (m+c)x. To solve (m+ c)x= y- c, you get x= (y- c)/(m+ c)ii) y=mx+cx+c

y-c = mx+cx

I'm not sure where to go after this, I would come out with a final answer of y-c/mc = 2x?

I don't know what you mean by this. Is that supposed go be y= ((x+ c)/a)/b or y= x+ (c/a)/b?iii) y=x+c/a/b - once again i'm not even sure where to go with this to start with

The equation x^2= y- c hasiv) y=x^2+c

y-c=x^2

sqroot(y-c)=xtwosolutions, x= sqrt(y-c), which you giveandx= -sqrt(y- c).

[quote]v) y=e^Lx -1

y+1 = e^Lx

not sure where to go with this, I imagine you use logs somehowAssuming you mean e^(Lx), yes. Thedefinitionof the (natural) logarithm is that it is the inverse function to e^x: log(e^x)= x. So if you take the logarithm of both sides, log(y+ 1)= log(e^(Lx))= Lx. Now divide both sides by L.

So am I! I have no idea what "cos" and "sin", could possibly mean without an argument! Is it possible you meant cos^2(x)+ sin^2(x) or xsin^2(x)+ xcos^2(x)? If it is either of those, use the fact that sin^2(x)+ cos^2(x)= 1.vi) y=xcos^2 + xsin^2

Completely stumped

vii) y=log(e^x)

Not sure where to begin

All help is apprecaited! many thanks