help! i cant figure out this problem for the life of me!

• Aug 24th 2008, 07:50 AM
math shark
help! i cant figure out this problem for the life of me!
4/9(x+3)=g

solve for x. state any restrictions on the variables.
• Aug 24th 2008, 08:16 AM
skeeter
1. multiply both sides by 9/4

2. subtract 3 from both sides

you're done.
• Aug 24th 2008, 08:17 AM
Chop Suey
Is your question $\displaystyle g = \frac{4}{9(x+3)}$

Given:$\displaystyle g = \frac{4}{9(x+3)}$

Multiply by (x+3) on both sides: $\displaystyle g(x+3) = \frac{4}{9}$

Divide by g on both sides: $\displaystyle (x+3) = \frac{4}{9g}$

Subtract 3 from both sides: $\displaystyle x = \frac{4}{9g} - 3$

As for the domain restriction, you should know that division by zero is not allowed, so you have to restrict any values of x that will make the denominator zero.

$\displaystyle x+3 \neq 0$

$\displaystyle \therefore x \neq -3$
• Aug 24th 2008, 08:19 AM
Jhevon
Quote:

Originally Posted by math shark
4/9(x+3)=g

solve for x. state any restrictions on the variables.

the idea is to get x on one side of the equation by itself. so we need to get rid of the other things on the same side with x. we do this by performing the opposite operations on these terms.

first step, there is a 4/9 multiplying the expression with x in it, so to get rid of it, we divide by 4/9, which is the same as multiplying by 9/4. but since this is an equation, whatever we do to one side, we must do to the other, to maintain equality. so the first step is:

$\displaystyle \frac 94 \cdot \frac 49 (x + 3) = \frac 94g$

$\displaystyle \Rightarrow x + 3 = \frac 49g$

(the 9/4 cancels the 4/9, do you see that?)

now we have on one side, x + 3, how do we get the x by itself? what's the next step?

EDIT: geez, two times too slow.... i'm not a morning person, that's my excuse
• Aug 24th 2008, 08:40 AM
Chop Suey
Quote:

Originally Posted by Chop Suey
Is your question $\displaystyle g = \frac{4}{9(x+3)}$

Given:$\displaystyle g = \frac{4}{9(x+3)}$

Multiply by (x+3) on both sides: $\displaystyle g(x+3) = \frac{4}{9}$

Divide by g on both sides: $\displaystyle (x+3) = \frac{4}{9g}$

Subtract 3 from both sides: $\displaystyle x = \frac{4}{9g} - 3$

As for the domain restriction, you should know that division by zero is not allowed, so you have to restrict any values of x that will make the denominator zero.

$\displaystyle x+3 \neq 0$

$\displaystyle \therefore x \neq -3$

I also forgot to say that g shouldn't be zero either.
• Aug 24th 2008, 08:46 AM
math shark
no the question is only:

4/9*(x+3)=g