Solve for m 4m+3-m=3(m+1)

$4 m + 3 - m = 3 (m + 1)$

Subtract $m$ from $4 m$ .

$3 m + 3 = 3 (m + 1)$

Simplify $3 (m + 1)$ .

Apply the distributive property.

$3 m + 3 = 3 m + 3 \cdot 1$

Multiply $3$ by $1$ .

$3 m + 3 = 3 m + 3$

Move all terms containing $m$ to the left side of the equation.

Subtract $3 m$ from both sides of the equation.

$3 m + 3 - 3 m = 3$

Combine the opposite terms in $3 m + 3 - 3 m$ .

Subtract $3 m$ from $3 m$ .

$0 + 3 = 3$

Add $0$ and $3$ .

$3 = 3$

Since $3 = 3$ , the equation will always be true for any value of $m$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$