Solve the Inequality for a 2(7a-1)-11a>=3(a-4)

$2 (7 a - 1) - 11 a \geq 3 (a - 4)$

Simplify $2 (7 a - 1) - 11 a$ .

Simplify each term.

Apply the distributive property.

$2 (7 a) + 2 \cdot - 1 - 11 a \geq 3 (a - 4)$

Multiply $7$ by $2$ .

$14 a + 2 \cdot - 1 - 11 a \geq 3 (a - 4)$

Multiply $2$ by $- 1$ .

$14 a - 2 - 11 a \geq 3 (a - 4)$

Subtract $11 a$ from $14 a$ .

$3 a - 2 \geq 3 (a - 4)$

Simplify $3 (a - 4)$ .

Apply the distributive property.

$3 a - 2 \geq 3 a + 3 \cdot - 4$

Multiply $3$ by $- 4$ .

$3 a - 2 \geq 3 a - 12$

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

Subtract $3 a$ from both sides of the inequality.

$3 a - 2 - 3 a \geq - 12$

Combine the opposite terms in $3 a - 2 - 3 a$ .

Subtract $3 a$ from $3 a$ .

$0 - 2 \geq - 12$

Subtract $2$ from $0$ .

$- 2 \geq - 12$

Since $- 2 \geq - 12$ , the inequality will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$