Solve for x 3+(x-2)/(x-3)<=4

$3 + \frac{x - 2}{x - 3} \leq 4$

Step 1

Subtract $4$ from both sides of the inequality.

$3 + \frac{x - 2}{x - 3} - 4 \leq 0$

Step 2

Simplify $3 + \frac{x - 2}{x - 3} - 4$ .

Find the common denominator.

Write $3$ as a fraction with denominator $1$ .

$\frac{3}{1} + \frac{x - 2}{x - 3} - 4 \leq 0$

Multiply $\frac{3}{1}$ by $\frac{x - 3}{x - 3}$ .

$\frac{3}{1} \cdot \frac{x - 3}{x - 3} + \frac{x - 2}{x - 3} - 4 \leq 0$

Multiply $\frac{3}{1}$ by $\frac{x - 3}{x - 3}$ .

$\frac{3 (x - 3)}{x - 3} + \frac{x - 2}{x - 3} - 4 \leq 0$

Write $- 4$ as a fraction with denominator $1$ .

$\frac{3 (x - 3)}{x - 3} + \frac{x - 2}{x - 3} + \frac{- 4}{1} \leq 0$

Multiply $\frac{- 4}{1}$ by $\frac{x - 3}{x - 3}$ .

$\frac{3 (x - 3)}{x - 3} + \frac{x - 2}{x - 3} + \frac{- 4}{1} \cdot \frac{x - 3}{x - 3} \leq 0$

Multiply $\frac{- 4}{1}$ by $\frac{x - 3}{x - 3}$ .

$\frac{3 (x - 3)}{x - 3} + \frac{x - 2}{x - 3} + \frac{- 4 (x - 3)}{x - 3} \leq 0$

Combine the numerators over the common denominator.

$\frac{3 (x - 3) + x - 2 - 4 (x - 3)}{x - 3} \leq 0$

Simplify each term.

Apply the distributive property.

$\frac{3 x + 3 \cdot - 3 + x - 2 - 4 (x - 3)}{x - 3} \leq 0$

Multiply $3$ by $- 3$ .

$\frac{3 x - 9 + x - 2 - 4 (x - 3)}{x - 3} \leq 0$

Apply the distributive property.

$\frac{3 x - 9 + x - 2 - 4 x - 4 \cdot - 3}{x - 3} \leq 0$

Multiply $- 4$ by $- 3$ .

$\frac{3 x - 9 + x - 2 - 4 x + 12}{x - 3} \leq 0$

Add $3 x$ and $x$ .

$\frac{4 x - 9 - 2 - 4 x + 12}{x - 3} \leq 0$

Combine the opposite terms in $4 x - 9 - 2 - 4 x + 12$ .

Subtract $4 x$ from $4 x$ .

$\frac{0 - 9 - 2 + 12}{x - 3} \leq 0$

Subtract $9$ from $0$ .

$\frac{- 9 - 2 + 12}{x - 3} \leq 0$

Subtract $2$ from $- 9$ .

$\frac{- 11 + 12}{x - 3} \leq 0$

Add $- 11$ and $12$ .

$\frac{1}{x - 3} \leq 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 3 = 0$

Step 4

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Find the domain of $\frac{1}{x - 3}$ .

Set the denominator in $\frac{1}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The domain is all values of $x$ that make the expression defined.

$(- \infty, 3) \cup (3, \infty)$

Step 6

The solution consists of all of the true intervals.

$x < 3$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x < 3$

Interval Notation:

$(- \infty, 3)$

Step 8

Solve for x 3+(x-2)/(x-3)<=4

Related Questions

Support LCMGCF.com by disabling your adblocker.