Solve for j a=(j+b)/(j-c)

$a = \frac{j + b}{j - c}$

Rewrite the equation as $\frac{j + b}{j - c} = a$ .

$\frac{j + b}{j - c} = a$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$j - c, 1$

Remove parentheses.

$j - c, 1$

The LCM of one and any expression is the expression.

$j - c$

Multiply each term in $\frac{j + b}{j - c} = a$ by $j - c$ to eliminate the fractions.

Multiply each term in $\frac{j + b}{j - c} = a$ by $j - c$ .

$\frac{j + b}{j - c} (j - c) = a (j - c)$

Simplify the left side.

Cancel the common factor of $j - c$ .

Cancel the common factor.

$\frac{j + b}{j - c} (j - c) = a (j - c)$

Rewrite the expression.

$j + b = a (j - c)$

Simplify the right side.

Apply the distributive property.

$j + b = a j + a (- c)$

Rewrite using the commutative property of multiplication.

$j + b = a j - a c$

Solve the equation.

Subtract $a j$ from both sides of the equation.

$j + b - a j = - a c$

Subtract $b$ from both sides of the equation.

$j - a j = - a c - b$

Factor $j$ out of $j - a j$ .

Factor $j$ out of $j^{1}$ .

$j \cdot 1 - a j = - a c - b$

Factor $j$ out of $- a j$ .

$j \cdot 1 + j (- a) = - a c - b$

Factor $j$ out of $j \cdot 1 + j (- a)$ .

$j (1 - a) = - a c - b$

Divide each term in $j (1 - a) = - a c - b$ by $1 - a$ and simplify.

Divide each term in $j (1 - a) = - a c - b$ by $1 - a$ .

$\frac{j (1 - a)}{1 - a} = \frac{- a c}{1 - a} + \frac{- b}{1 - a}$

Simplify the left side.

Cancel the common factor of $1 - a$ .

Cancel the common factor.

$\frac{j (1 - a)}{1 - a} = \frac{- a c}{1 - a} + \frac{- b}{1 - a}$

Divide $j$ by $1$ .

$j = \frac{- a c}{1 - a} + \frac{- b}{1 - a}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$j = - \frac{a c}{1 - a} + \frac{- b}{1 - a}$

Move the negative in front of the fraction.

$j = - \frac{a c}{1 - a} - \frac{b}{1 - a}$

Simplify terms.

Combine the numerators over the common denominator.

$j = \frac{- a c - b}{1 - a}$

Factor $- 1$ out of $- a c$ .

$j = \frac{- (a c) - b}{1 - a}$

Factor $- 1$ out of $- b$ .

$j = \frac{- (a c) - (b)}{1 - a}$

Factor $- 1$ out of $- (a c) - (b)$ .

$j = \frac{- (a c + b)}{1 - a}$

Simplify the expression.

Rewrite $- (a c + b)$ as $- 1 (a c + b)$ .

$j = \frac{- 1 (a c + b)}{1 - a}$

Move the negative in front of the fraction.

$j = - \frac{a c + b}{1 - a}$

Solve for j a=(j+b)/(j-c)

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