Question

2 u + 12 < 23

0,1 d + 8 > 0
3 - 4 r > 7
13 < 2 z + 3
6 ≥ 9 - 0, 15 i

Answer 1

Part 1)
`u< 5.5`

Part 2)
`d > -80`

Part 3)
`r < -1`

Part 4)
`z>5`

Part 5)
`i\geq 20`

Explanation:

The question is

Solve each inequality for the indicated variable

Part 1) we have

`2u+12<23`

subtract 12 both sides

`2u< 23-12\\2u<11`

Divide by 2 both sides

`u< 5.5`

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

`0.1d+8 >0`

subtract 8 both sides

`0.1d > -8`

Divide by 0.1 both sides

`d > -80`

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

`3-4r> 7`

Subtract 3 both sides

`-4r>7-3\\-4r>4`

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

`r < -1`

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

`13< 2z+3`

Subtract 3 both sides

`13-3<2z\\10 <2z`

Divide by 2 both sides

`5<z`

Rewrite

`z>5`

The solution is the interval (5,∞)

In a number line the solution is the shaded area at right of z=5 (open circle)

The number 5 is not included in the solution

Part 5) we have

`6\geq 9-0.15i`

Subtract 9 both sides

`6-9\geq -0.15i\\-3\geq -0.15i`

Divide by -0.15 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

`20\leq i`

Rewrite

`i\geq 20`

The solution is the interval [20,∞)

In a number line the solution is the shaded area at right of i=20 (closed circle)

The number 20 is included in the solution