Solve the Inequality for t 8(t+2)+7(t+2)-3(t-2)<0

$8 (t + 2) + 7 (t + 2) - 3 (t - 2) < 0$

Step 1

Simplify $8 (t + 2) + 7 (t + 2) - 3 (t - 2)$ .

Simplify each term.

Apply the distributive property.

$8 t + 8 \cdot 2 + 7 (t + 2) - 3 (t - 2) < 0$

Multiply $8$ by $2$ .

$8 t + 16 + 7 (t + 2) - 3 (t - 2) < 0$

Apply the distributive property.

$8 t + 16 + 7 t + 7 \cdot 2 - 3 (t - 2) < 0$

Multiply $7$ by $2$ .

$8 t + 16 + 7 t + 14 - 3 (t - 2) < 0$

Apply the distributive property.

$8 t + 16 + 7 t + 14 - 3 t - 3 \cdot - 2 < 0$

Multiply $- 3$ by $- 2$ .

$8 t + 16 + 7 t + 14 - 3 t + 6 < 0$

$8 t + 16 + 7 t + 14 - 3 t + 6 < 0$

Simplify by adding terms.

Add $8 t$ and $7 t$ .

$15 t + 16 + 14 - 3 t + 6 < 0$

Subtract $3 t$ from $15 t$ .

$12 t + 16 + 14 + 6 < 0$

Simplify by adding numbers.

Add $16$ and $14$ .

$12 t + 30 + 6 < 0$

Add $30$ and $6$ .

$12 t + 36 < 0$

$12 t + 36 < 0$

$12 t + 36 < 0$

$12 t + 36 < 0$

Step 2

Subtract $36$ from both sides of the inequality.

$12 t < - 36$

Step 3

Divide each term in $12 t < - 36$ by $12$ and simplify.

Divide each term in $12 t < - 36$ by $12$ .

$\frac{12 t}{12} < \frac{- 36}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 t}{12} < \frac{- 36}{12}$

Divide $t$ by $1$ .

$t < \frac{- 36}{12}$

$t < \frac{- 36}{12}$

$t < \frac{- 36}{12}$

Simplify the right side.

Divide $- 36$ by $12$ .

$t < - 3$

$t < - 3$

$t < - 3$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$t < - 3$

Interval Notation:

$(- \infty, - 3)$

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