Solve the System of @WORD 12<=5x+2<=27

$12 \leq 5 x + 2 \leq 27$

Step 1

Solve $12 \leq 5 x + 2$ for $x$ .

Rewrite so $x$ is on the left side of the inequality.

$5 x + 2 \geq 12$

Move all terms not containing $x$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$5 x \geq 12 - 2$

Subtract $2$ from $12$ .

$5 x \geq 10$

Divide each term in $5 x \geq 10$ by $5$ and simplify.

Divide each term in $5 x \geq 10$ by $5$ .

$\frac{5 x}{5} \geq \frac{10}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} \geq \frac{10}{5}$

Divide $x$ by $1$ .

$x \geq \frac{10}{5}$

Simplify the right side.

Divide $10$ by $5$ .

$x \geq 2$

Step 2

Solve $5 x + 2 \leq 27$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$5 x \leq 27 - 2$

Subtract $2$ from $27$ .

$5 x \leq 25$

Divide each term in $5 x \leq 25$ by $5$ and simplify.

Divide each term in $5 x \leq 25$ by $5$ .

$\frac{5 x}{5} \leq \frac{25}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} \leq \frac{25}{5}$

Divide $x$ by $1$ .

$x \leq \frac{25}{5}$

Simplify the right side.

Divide $25$ by $5$ .

$x \leq 5$

Step 3

Find the intersection of $x \geq 2$ and $x \leq 5$ .

$2 \leq x \leq 5$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$2 \leq x \leq 5$

Interval Notation:

$[2, 5]$

Step 5

Solve the System of @WORD 12<=5x+2<=27

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