Re: unsure on the next step?

You made some mistakes

$\displaystyle 3b(x+3b)=a(a-x)$

$\displaystyle 3bx+9b^2=a^2-ax$

Add "ax" to both sides

$\displaystyle 3bx+ax+9b^2=a^2+ax-ax=a^2$

Subtract "9b^2" from both sides

$\displaystyle x(3b+a)+9b^2-9b^2=a^2-9b^2$

$\displaystyle (3b+a)x=(a-3b)(a+3b)$

Now divide both sides by (3b+a) to get the expression for "x"

$\displaystyle \frac{(3b+a)x}{(3b+a)}=\frac{(a-3b)(a+3b)}{(3b+a)}$

and simplify the last line.

As the denominator cannot be zero,

the given condition is

$\displaystyle a\ne\ -3b$

Re: unsure on the next step?

so would i be correct in saying x =(a-3b) so x =-a+3b

Re: unsure on the next step?

Quote:

Originally Posted by

**Orlando** so would i be correct in saying x =(a-3b) so x =-a+3b

$\displaystyle x =(a-3b)$ does not imply $\displaystyle x =-a+3b$

Re: unsure on the next step?

Quote:

Originally Posted by

**Orlando** so would i be correct in saying x =(a-3b) so x =-a+3b

If you had

-x=(a-3b)

then -(-x)=-(a-3b)

x=-a-(-3b)

x=3b-a

But x=(a-3b) so x=a-3b