1. ## combining equations... help please??? :(

Okay I thought I was getting better at but I keep seeming to get some questions wrong!! My math teacher never seems to have time to explain either!!! Could someone pretty please tell me where I'm going wrong?? Is there something I'm not understanding please?

Write an equation for x that does not involve p if:

1.) a = ym/p and 2.) m = pz^2

So I thought rearrange equation 2 to make p the subject and then substitute into equation 1 but I got the question wrong???

rearranging 2 gives p = m/z^2

substituting into equation 1

a = ym/mz^2

simplifying

a = y/z^2 but I was told this was wrong help??

Combine a = bcd and e = ct to give an equation for d that does not involve c

I thought

rearrange a = bcd to make d the subject

d = a/bc

rearrange e = ct to make c the subject

c = e/t

so

d = a / b(e/t) right??

Help please cause I'm confused and a bit frustrated that I keep getting questions like this wrong lol

2. Hi RainbowRevolution,

For the first one we have,

$a=\frac{my}{p}~~~(*)$

$m=pz^2~~~(**)$

If we sub in $(**)$ straight into $(*)$ we get

$a=\frac{pyz^2}{p}$

Clearly the $p$ cancels to give $a=yz^2$ . One problem is the question says to write for x however there is no x term present....

You where right upto the point where you said

$a=\frac{my}{mz^2}~~~(***)$

Actually you had $p=\frac{m}{z^2}$ not $p=mz^2$ and thus equation $(***)$ should actually be

$a=\frac{my}{\frac{m}{z^2}}$ which cancels to

$a=yz^2$

The second looks right to me but we can do a bit of tidying up. We have

$d=\frac{a}{b\left(\frac{e}{t}\right)}$ and thus,

$d=\frac{at}{be}$

Hope this helps.

3. Ah yeah sorry I typed the first question out wrong when I was copying it from the question sheet, it was find an "equation for a" not x, oops! and that does help thank you sorry I wrote the first question wrong lol

Thank you !