Solve the System of Inequalities -y+3<=8 and y+2>9

$- y + 3 \leq 8$ and $y + 2 > 9$

Simplify the first inequality.

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

Subtract $3$ from both sides of the inequality.

$- y \leq 8 - 3$ and $y + 2 > 9$

Subtract $3$ from $8$ .

$- y \leq 5$ and $y + 2 > 9$

Divide each term in $- y \leq 5$ by $- 1$ and simplify.

Divide each term in $- y \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{5}{- 1}$ and $y + 2 > 9$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{5}{- 1}$ and $y + 2 > 9$

Divide $y$ by $1$ .

$y \geq \frac{5}{- 1}$ and $y + 2 > 9$

Simplify the right side.

Divide $5$ by $- 1$ .

$y \geq - 5$ and $y + 2 > 9$

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

Subtract $2$ from both sides of the inequality.

$y \geq - 5$ and $y > 9 - 2$

Subtract $2$ from $9$ .

$y \geq - 5$ and $y > 7$

The intersection consists of the elements that are contained in both intervals.

$y > 7$

The result can be shown in multiple forms.

Inequality Form:

$y > 7$

Interval Notation:

$(7, \infty)$

Solve the System of Inequalities -y+3<=8 and y+2>9

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