Solve 5 + 3(2oc + i ^ 2) = 6 - h(oc + 1) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$5+3(2oc+i^{2})=6-h(oc+1)$$

Solve for c

$\left\{\begin{matrix}c=-\frac{h-4}{o\left(h+6\right)}\text{, }&h\neq -6\text{ and }o\neq 0\\c\in \mathrm{C}\text{, }&h=4\text{ and }o=0\end{matrix}\right.$

Steps for Solving Linear Equation

Calculate $i$ to the power of $2$ and get $-1$.

$$5+3\left(2oc-1\right)=6-h\left(oc+1\right)$$

Use the distributive property to multiply $3$ by $2oc-1$.

$$5+6oc-3=6-h\left(oc+1\right)$$

Subtract $3$ from $5$ to get $2$.

$$2+6oc=6-h\left(oc+1\right)$$

Use the distributive property to multiply $h$ by $oc+1$.

$$2+6oc=6-\left(hoc+h\right)$$

To find the opposite of $hoc+h$, find the opposite of each term.

$$2+6oc=6-hoc-h$$

Add $hoc$ to both sides.

$$2+6oc+hoc=6-h$$

Subtract $2$ from both sides.

$$6oc+hoc=6-h-2$$

Subtract $2$ from $6$ to get $4$.

$$6oc+hoc=4-h$$

Combine all terms containing $c$.

$$\left(6o+ho\right)c=4-h$$

The equation is in standard form.

$$\left(ho+6o\right)c=4-h$$

Divide both sides by $ho+6o$.

$$\frac{\left(ho+6o\right)c}{ho+6o}=\frac{4-h}{ho+6o}$$

Dividing by $ho+6o$ undoes the multiplication by $ho+6o$.

$$c=\frac{4-h}{ho+6o}$$

Divide $4-h$ by $ho+6o$.

$$c=\frac{4-h}{o\left(h+6\right)}$$

Solve for h

$h=\frac{2\left(2-3co\right)}{co+1}$
$c=0\text{ or }o\neq -\frac{1}{c}$

Steps for Solving Linear Equation

Calculate $i$ to the power of $2$ and get $-1$.

$$5+3\left(2oc-1\right)=6-h\left(oc+1\right)$$

Use the distributive property to multiply $3$ by $2oc-1$.

$$5+6oc-3=6-h\left(oc+1\right)$$

Subtract $3$ from $5$ to get $2$.

$$2+6oc=6-h\left(oc+1\right)$$

Use the distributive property to multiply $h$ by $oc+1$.

$$2+6oc=6-\left(hoc+h\right)$$

To find the opposite of $hoc+h$, find the opposite of each term.

$$2+6oc=6-hoc-h$$

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

$$6-hoc-h=2+6oc$$

Subtract $6$ from both sides.

$$-hoc-h=2+6oc-6$$

Subtract $6$ from $2$ to get $-4$.

$$-hoc-h=-4+6oc$$

Combine all terms containing $h$.

$$\left(-oc-1\right)h=-4+6oc$$

The equation is in standard form.

$$\left(-co-1\right)h=6co-4$$

Divide both sides by $-oc-1$.

$$\frac{\left(-co-1\right)h}{-co-1}=\frac{6co-4}{-co-1}$$

Dividing by $-oc-1$ undoes the multiplication by $-oc-1$.

$$h=\frac{6co-4}{-co-1}$$

Divide $-4+6oc$ by $-oc-1$.

$$h=-\frac{2\left(3co-2\right)}{co+1}$$