Solve for x 8ix=-5

$8 i x = - 5$

Step 1

Divide each term in $8 i x = - 5$ by $8 i$ .

$\frac{8 i x}{8 i} = \frac{- 5}{8 i}$

Step 2

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 i x}{8 i} = \frac{- 5}{8 i}$

Rewrite the expression.

$\frac{i x}{i} = \frac{- 5}{8 i}$

Cancel the common factor of $i$ .

Cancel the common factor.

$\frac{i x}{i} = \frac{- 5}{8 i}$

Divide $x$ by $1$ .

$x = \frac{- 5}{8 i}$

Step 3

Simplify the right side.

Multiply the numerator and denominator of $\frac{- 5}{8 i}$ by the conjugate of $8 i$ to make the denominator real.

$x = \frac{- 5}{8 i} \cdot \frac{i}{i}$

Multiply.

Combine.

$x = \frac{- 5 i}{8 i i}$

Simplify the denominator.

Add parentheses.

$x = \frac{- 5 i}{8 (i i)}$

Raise $i$ to the power of $1$ .

$x = \frac{- 5 i}{8 (i^{1} i)}$

Raise $i$ to the power of $1$ .

$x = \frac{- 5 i}{8 (i^{1} i^{1})}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \frac{- 5 i}{8 i^{1 + 1}}$

Add $1$ and $1$ .

$x = \frac{- 5 i}{8 i^{2}}$

Rewrite $i^{2}$ as $- 1$ .

$x = \frac{- 5 i}{8 \cdot - 1}$

Multiply $8$ by $- 1$ .

$x = \frac{- 5 i}{- 8}$

Dividing two negative values results in a positive value.

$x = \frac{5 i}{8}$