Solve for x 2(4x+3)<5x+21

$2 (4 x + 3) < 5 x + 21$

Step 1

Simplify $2 (4 x + 3)$ .

Rewrite.

$0 + 0 + 2 (4 x + 3) < 5 x + 21$

Simplify by adding zeros.

$2 (4 x + 3) < 5 x + 21$

Apply the distributive property.

$2 (4 x) + 2 \cdot 3 < 5 x + 21$

Multiply.

Multiply $4$ by $2$ .

$8 x + 2 \cdot 3 < 5 x + 21$

Multiply $2$ by $3$ .

$8 x + 6 < 5 x + 21$

Step 2

Move all terms containing $x$ to the left side of the inequality.

Subtract $5 x$ from both sides of the inequality.

$8 x + 6 - 5 x < 21$

Subtract $5 x$ from $8 x$ .

$3 x + 6 < 21$

Step 3

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

Subtract $6$ from both sides of the inequality.

$3 x < 21 - 6$

Subtract $6$ from $21$ .

$3 x < 15$

Step 4

Divide each term in $3 x < 15$ by $3$ and simplify.

Divide each term in $3 x < 15$ by $3$ .

$\frac{3 x}{3} < \frac{15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} < \frac{15}{3}$

Divide $x$ by $1$ .

$x < \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$x < 5$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < 5$

Interval Notation:

$(- \infty, 5)$