Question

mp using tools in exercises 29 and 30, use technology to solve the equation. 29. $0.7x + 0.5 = -0.2x - 1.3$ 30. $2.1x + 0.6 = -1.4x + 6.9$ 31. modeling real life there are about 34 million gallons of water in reservoir a and about 38 million gallons in reservoir b. during a drought, reservoir a loses about 0.8 million gallons per month and reservoir b loses about 1.1 million gallons per month. after how many months will the reservoirs contain the same amount of water? example 3 32. modeling real life your dog is 16 years old in dog years. your cat is 28 years old in cat years. for every human year, your dog ages by 7 dog years and your cat ages by 4 cat years. in how many human years will both pets be the same age in their respective types of years?

Explanation:

Step1: Solve equation 29: isolate x

Add $0.2x$ to both sides, subtract $0.5$:
$0.7x + 0.2x = -1.3 - 0.5$
$0.9x = -1.8$
$x = \frac{-1.8}{0.9} = -2$

Step2: Solve equation 30: isolate x

Add $1.4x$ to both sides, subtract $0.6$:
$2.1x + 1.4x = 6.9 - 0.6$
$3.5x = 6.3$
$x = \frac{6.3}{3.5} = 1.8$

Step3: Solve problem 31: set up equation

Let $m$ = number of months.
Reservoir A: $34 - 0.8m$; Reservoir B: $38 - 1.1m$
Set equal: $34 - 0.8m = 38 - 1.1m$
Add $1.1m$, subtract 34:
$0.3m = 4$
$m = \frac{4}{0.3} = \frac{40}{3} \approx 13.33$

Step4: Solve problem 32: set up equation

Let $h$ = number of human years.
Dog's age: $16 + 7h$; Cat's age: $28 + 4h$
Set equal: $16 + 7h = 28 + 4h$
Subtract $4h$, subtract 16:
$3h = 12$
$h = 4$

Answer:

1. $x = -2$
2. $x = 1.8$
3. $\frac{40}{3}$ (or about 13.33) months
4. 4 human years