Solve for l 15-28i=3l+(4m)i

$15 - 28 i = 3 l + (4 m) i$

Step 1

Rewrite the equation as $3 l + 4 m i = 15 - 28 i$ .

$3 l + 4 m i = 15 - 28 i$

Step 2

Subtract $4 m i$ from both sides of the equation.

$3 l = 15 - 28 i - 4 m i$

Step 3

Divide each term in $3 l = 15 - 28 i - 4 m i$ by $3$ and simplify.

Divide each term in $3 l = 15 - 28 i - 4 m i$ by $3$ .

$\frac{3 l}{3} = \frac{15}{3} + \frac{- 28 i}{3} + \frac{- 4 m i}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 l}{3} = \frac{15}{3} + \frac{- 28 i}{3} + \frac{- 4 m i}{3}$

Divide $l$ by $1$ .

$l = \frac{15}{3} + \frac{- 28 i}{3} + \frac{- 4 m i}{3}$

Simplify the right side.

Simplify each term.

Divide $15$ by $3$ .

$l = 5 + \frac{- 28 i}{3} + \frac{- 4 m i}{3}$

Move the negative in front of the fraction.

$l = 5 - \frac{(28) i}{3} + \frac{- 4 m i}{3}$

Move the negative in front of the fraction.

$l = 5 - \frac{28 i}{3} - \frac{4 m i}{3}$