1. Solve Each Equation

Solve each equation. The letters a, b and c are constants.

(1) 1 - ax = b, where a DOES NOT = 0.

(2) (a/x) + (b/x) = c, where c DOES NOT = 0.

2. Originally Posted by symmetry
Solve each equation. The letters a, b and c are constants.

(1) 1 - ax = b, where a DOES NOT = 0.
Solve the equation for x, I presume?

It can bother students that there is more than one unknown in an equation. The thing to remember is that all these variables are simply numbers. If it bothers you, plug in a random number for the variable and solve the equation. The steps you have to use to solve it in the form with the variable in it will be almost identical.

$1 - ax = b$

$-1 + 1 - ax = -1 + b$

$-ax = -1 + b$

$\frac{-ax}{-a} = \frac{-1 + b}{-a}$

$x = \frac{-1 + b}{-a} = -\frac{-1 + b}{a} = \frac{b - 1}{a}$

where any one of the three expressions in the last line do the trick. (Most would probably pick the last form, but it's entirely up to you.)

-Dan

3. Originally Posted by symmetry
Solve each equation. The letters a, b and c are constants.

(2) (a/x) + (b/x) = c, where c DOES NOT = 0.
$\frac{a}{x} + \frac{b}{x} = c$

We want to get rid of the fractions, so I am going to multiply both sides by x:

$x \left ( \frac{a}{x} + \frac{b}{x} \right ) = cx$

$x \cdot \frac{a}{x} + x \cdot \frac{b}{x} = cx$

$a + b = cx$

$cx = a + b$

$\frac{cx}{c} = \frac{a + b}{c}$

$x = \frac{a + b}{c}$

I would just like to note that, looking at the original equation that x cannot be allowed to be 0. Thus we not only have the restriction that $c \neq 0$ as given in the problem statement, but also that $b \neq -a$.

-Dan

4. ok

What an interesting set of math notes.

Thanks