How to solve 9=5+2(ab+bc+ca)

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.

Solve 9=5+2(ab+bc+ca) | Equatix

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Question

$$9=5+2(ab+bc+ca)$$

Solve for a

$a=-\frac{bc-2}{b+c}$
$b\neq -c$

Steps for Solving Linear Equation

Use the distributive property to multiply $2$ by $ab+bc+ca$.

$$9=5+2ab+2bc+2ca$$

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

$$5+2ab+2bc+2ca=9$$

Subtract $5$ from both sides.

$$2ab+2bc+2ca=9-5$$

Subtract $5$ from $9$ to get $4$.

$$2ab+2bc+2ca=4$$

Subtract $2bc$ from both sides.

$$2ab+2ca=4-2bc$$

Combine all terms containing $a$.

$$\left(2b+2c\right)a=4-2bc$$

Divide both sides by $2b+2c$.

$$\frac{\left(2b+2c\right)a}{2b+2c}=\frac{4-2bc}{2b+2c}$$

Dividing by $2b+2c$ undoes the multiplication by $2b+2c$.

$$a=\frac{4-2bc}{2b+2c}$$

Divide $4-2bc$ by $2b+2c$.

$$a=\frac{2-bc}{b+c}$$

Show Solution

Solve for b

$b=-\frac{ac-2}{a+c}$
$a\neq -c$

Steps for Solving Linear Equation

Use the distributive property to multiply $2$ by $ab+bc+ca$.

$$9=5+2ab+2bc+2ca$$

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

$$5+2ab+2bc+2ca=9$$

Subtract $5$ from both sides.

$$2ab+2bc+2ca=9-5$$

Subtract $5$ from $9$ to get $4$.

$$2ab+2bc+2ca=4$$

Subtract $2ca$ from both sides.

$$2ab+2bc=4-2ca$$

Combine all terms containing $b$.

$$\left(2a+2c\right)b=4-2ca$$

The equation is in standard form.

$$\left(2a+2c\right)b=4-2ac$$

Divide both sides by $2a+2c$.

$$\frac{\left(2a+2c\right)b}{2a+2c}=\frac{4-2ac}{2a+2c}$$

Dividing by $2a+2c$ undoes the multiplication by $2a+2c$.

$$b=\frac{4-2ac}{2a+2c}$$

Divide $4-2ca$ by $2a+2c$.

$$b=\frac{2-ac}{a+c}$$

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