Solve the Inequality for y 6(5y-(3y-1))>=4(3y-7)

$6 (5 y - (3 y - 1)) \geq 4 (3 y - 7)$

Simplify $6 (5 y - (3 y - 1))$ .

Start simplifying.

$0 + 0 + 6 (5 y - (3 y - 1)) \geq 4 (3 y - 7)$

Simplify by adding zeros.

$6 (5 y - (3 y - 1)) \geq 4 (3 y - 7)$

Simplify each term.

Apply the distributive property.

$6 (5 y - (3 y) - - 1) \geq 4 (3 y - 7)$

Multiply $3$ by $- 1$ .

$6 (5 y - 3 y - - 1) \geq 4 (3 y - 7)$

Multiply $- 1$ by $- 1$ .

$6 (5 y - 3 y + 1) \geq 4 (3 y - 7)$

Simplify terms.

Subtract $3 y$ from $5 y$ .

$6 (2 y + 1) \geq 4 (3 y - 7)$

Apply the distributive property.

$6 (2 y) + 6 \cdot 1 \geq 4 (3 y - 7)$

Multiply.

Multiply $2$ by $6$ .

$12 y + 6 \cdot 1 \geq 4 (3 y - 7)$

Multiply $6$ by $1$ .

$12 y + 6 \geq 4 (3 y - 7)$

Simplify $4 (3 y - 7)$ .

Apply the distributive property.

$12 y + 6 \geq 4 (3 y) + 4 \cdot - 7$

Multiply.

Multiply $3$ by $4$ .

$12 y + 6 \geq 12 y + 4 \cdot - 7$

Multiply $4$ by $- 7$ .

$12 y + 6 \geq 12 y - 28$

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

Subtract $12 y$ from both sides of the inequality.

$12 y + 6 - 12 y \geq - 28$

Combine the opposite terms in $12 y + 6 - 12 y$ .

Subtract $12 y$ from $12 y$ .

$0 + 6 \geq - 28$

Add $0$ and $6$ .

$6 \geq - 28$

Since $6 \geq - 28$ , the inequality will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$