How to solve 3e + 5 = 29 - oc

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 3e + 5 = 29 - oc | Equatix

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Question

$$3e+5=29-oc$$

Solve for c

$c=\frac{3\left(8-e\right)}{o}$
$o\neq 0$

Steps for Solving Linear Equation

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

$$29-oc=3e+5$$

Subtract $29$ from both sides.

$$-oc=3e+5-29$$

Subtract $29$ from $5$ to get $-24$.

$$-oc=3e-24$$

The equation is in standard form.

$$\left(-o\right)c=3e-24$$

Divide both sides by $-o$.

$$\frac{\left(-o\right)c}{-o}=\frac{3e-24}{-o}$$

Dividing by $-o$ undoes the multiplication by $-o$.

$$c=\frac{3e-24}{-o}$$

Divide $3e-24$ by $-o$.

$$c=-\frac{3\left(e-8\right)}{o}$$

Show Solution

Solve for o

$o=\frac{3\left(8-e\right)}{c}$
$c\neq 0$

Steps for Solving Linear Equation

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

$$29-oc=3e+5$$

Subtract $29$ from both sides.

$$-oc=3e+5-29$$

Subtract $29$ from $5$ to get $-24$.

$$-oc=3e-24$$

The equation is in standard form.

$$\left(-c\right)o=3e-24$$

Divide both sides by $-c$.

$$\frac{\left(-c\right)o}{-c}=\frac{3e-24}{-c}$$

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

$$o=\frac{3e-24}{-c}$$

Divide $3e-24$ by $-c$.

$$o=-\frac{3\left(e-8\right)}{c}$$

Show Solution

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Solve for c

Steps for Solving Linear Equation

Solve for o

Steps for Solving Linear Equation