Question

The Green family is a family of six people. They have used 4,885.78 gallons of water so far this month. They cannot exceed 9,750.05 gallons per month during drought season. Write an inequality to show how much water just one member of the family can use for the remainder of the month, assuming each family member uses the same amount of water every month.

Answer 1

Answer:

Each member can use, x = 810.71 gallons of water for the remaining month.

The equation is 6x + 4885.78 $\leq$ 9750.05

Step-by-step explanation:

Total number of members in Green's family = 6

Let one member uses = x gallons of water

Therefore six member will use = 6x gallons of water

Given the Green's family can use only 9750.05 gallons of water in one month. And they have use so far 4885.78 gallons of water.

Therefore the equation becomes

6x + 4885.78 $\leq$ 9750.05

6x $\leq$ 9750.05 - 4885.78

6x $\leq$ 4864.27

x $\leq$ 4864.27 / 6

x $\leq$ 810.71 gallons

Hence, the equation is 6x + 4885.78 $\leq$ 9750.05

and each member can use 810.71 gallons of water for the remaining days of the month.

Answer 2

Answer:

The inequality is  $6x+4,885.78\leq 9,750.05$

$x\leq 810.71\ gal$

Step-by-step explanation:

Let

x -----> amount of water that can be used by only one family member for the rest of the month

we know that

$6x+4,885.78\leq 9,750.05$

Solve for x

Subtract 4,885.78 both sides

$6x\leq 9,750.05-4,885.78$

$6x\leq 4,864.27$

Divide by 6 both sides

$x\leq 4,864.27/6$

$x\leq 810.71\ gal$