How to solve Le + b, He а b

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

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

Solve for a

$a=\frac{bc}{b+c}$
$c\neq 0\text{ and }b\neq 0\text{ and }b\neq -c$

Steps for Solving Linear Equation

Variable $a$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $abc$, the least common multiple of $a,b,c$.

$$bc=ac+ab$$

Swap sides so that all variable terms are on the left hand side.

$$ac+ab=bc$$

Combine all terms containing $a$.

$$\left(c+b\right)a=bc$$

The equation is in standard form.

$$\left(b+c\right)a=bc$$

Divide both sides by $b+c$.

$$\frac{\left(b+c\right)a}{b+c}=\frac{bc}{b+c}$$

Dividing by $b+c$ undoes the multiplication by $b+c$.

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

Variable $a$ cannot be equal to $0$.

$$a=\frac{bc}{b+c}\text{, }a\neq 0$$

Show Solution

Solve for b

$b=-\frac{ac}{a-c}$
$c\neq 0\text{ and }a\neq 0\text{ and }a\neq c$

Steps for Solving Linear Equation

Variable $b$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $abc$, the least common multiple of $a,b,c$.

$$bc=ac+ab$$

Subtract $ab$ from both sides.

$$bc-ab=ac$$

Combine all terms containing $b$.

$$\left(c-a\right)b=ac$$

Divide both sides by $c-a$.

$$\frac{\left(c-a\right)b}{c-a}=\frac{ac}{c-a}$$

Dividing by $c-a$ undoes the multiplication by $c-a$.

$$b=\frac{ac}{c-a}$$

Variable $b$ cannot be equal to $0$.

$$b=\frac{ac}{c-a}\text{, }b\neq 0$$

Show Solution

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Solve for a

Steps for Solving Linear Equation

Solve for b

Steps for Solving Linear Equation