Solve the Inequality for x 3x+y<=3

$3 x + y \leq 3$

Step 1

Subtract $y$ from both sides of the inequality.

$3 x \leq 3 - y$

Step 2

Divide each term in $3 x \leq 3 - y$ by $3$ and simplify.

Divide each term in $3 x \leq 3 - y$ by $3$ .

$\frac{3 x}{3} \leq \frac{3}{3} + \frac{- y}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{3}{3} + \frac{- y}{3}$

Divide $x$ by $1$ .

$x \leq \frac{3}{3} + \frac{- y}{3}$

Simplify the right side.

Simplify each term.

Divide $3$ by $3$ .

$x \leq 1 + \frac{- y}{3}$

Move the negative in front of the fraction.

$x \leq 1 - \frac{y}{3}$