1. ## Solving equations?

I have an equation;

7 = 2
3+a a - 2

can I solve it this way?

6a + 2a = 8a
7a - 14 7a -14

= a - 14

The method I have used is to cross multiply, but I am unsure?

2. ## Re: Solving equations?

Originally Posted by David Green
I have an equation;

7 = 2
3+a a - 2
You can say $\frac{7}{3+a}=\frac{2}{a-2}$
$\\ 7a-14 =6+2a$.

Now solve for $a$ .

3. ## Re: Solving equations?

Originally Posted by Plato
You can say $\frac{7}{3+a}=\frac{2}{a-2}$
$\\ 7a-14 =6+2a$.

Now solve for $a$ .
Does that mean that now there is two (a) I have to factor one out?

4. ## Re: Solving equations?

Originally Posted by David Green
Does that mean that now there is two (a) I have to factor one out?
No, this is a linear equation with variables on both sides. You need the two terms with "a" on one side, and everything else on the other side.

5. ## Re: Solving equations?

Originally Posted by David Green
Does that mean that now there is two (a) I have to factor one out?
I don't even know what that question means, much less how to answer it.
What level are you on?
What are you studying?
Can you solve $7a-14=6+2a$ for $a~?$

6. ## Re: Solving equations?

Originally Posted by Plato
I don't even know what that question means, much less how to answer it.
What level are you on?
What are you studying?
Can you solve $7a-14=6+2a$ for $a~?$
I am only a beginner

Introduction to maths

a(7 - 14)(6 + 2) ???

7a - 14 = 6 + 2a
7a - 2a = 6 + 14
5a = 20
a = 4

how am I doing!

7. ## Re: Solving equations?

Originally Posted by David Green
I am only a beginner

Introduction to maths
\begin{align*} \frac{7}{3+a}&=\frac{2}{a-2} \\ 7a-14 &=6+2a\\ 5a&=20\text{ subtract 2a and add 14} \\ a&=4\end{align*}

8. ## Re: Solving equations?

Originally Posted by Plato
\begin{align*} \frac{7}{3+a}&=\frac{2}{a-2} \\ 7a-14 &=6+2a\\ 5a&=20\text{ subtract 2a and add 14} \\ a&=4\end{align*}
Thank you, so I did get it right with a little help, after all "every little helps"