Question

this series of problems will build your fluency with solving equations and inequalities. for each question below, write an equation or inequality describing the scenario and define the meaning of your variable. then, solve.

use the answer key to determine that you accurately modeled the scenario as an equation or inequality and solved correctly.

if necessary, revise your work.

1. i am thinking of a number. 3 more than twice my number is less than 17 but greater than -1. if my number is an integer, what could it be?

1. in january of 2006, the population of new hope, mi, was 18,324. the town grew at a constant monthly rate for several years. in september of 2008, the town had 21,926 residents. after how many months from january 2006 did the town have at least 25,000 residents? (hint: you may need to write multiple equations.)

Explanation:

Step1: Define variable and write inequality for problem 1

Let the number be $x$. Twice the number is $2x$, and 3 more than twice the number is $2x + 3$. The inequality is $- 1<2x + 3<17$.

Step2: Solve the left - hand side of the inequality

Subtract 3 from all parts: $-1-3<2x+3 - 3<17 - 3$, which simplifies to $-4<2x<14$.

Step3: Solve for $x$

Divide all parts by 2: $\frac{-4}{2}<\frac{2x}{2}<\frac{14}{2}$, so $-2 < x<7$. Since $x$ is an integer, $x$ can be $-1,0,1,2,3,4,5,6$.

Step4: Define variables and find growth rate for problem 2

Let $t$ be the number of months since January 2006, and the population $P$ has a linear growth model $P=P_0+rt$, where $P_0 = 18324$. The number of months from January 2006 to September 2008 is $t_1=32$ months, and $P_1 = 21926$. Substitute into the equation: $21926=18324+r\times32$.

Step5: Solve for the growth rate $r$

First, subtract 18324 from both sides: $21926 - 18324=r\times32$, so $3602 = 32r$. Then $r=\frac{3602}{32}=112.5625$.

Step6: Write an equation to find when $P\geq25000$

The equation is $25000\leq18324 + 112.5625t$.

Step7: Solve the inequality for $t$

Subtract 18324 from both sides: $25000-18324\leq112.5625t$, so $6676\leq112.5625t$. Then $t\geq\frac{6676}{112.5625}\approx59.3$. Since $t$ represents months, $t = 60$ months.

Answer:

1. The possible integer values of the number are $-1,0,1,2,3,4,5,6$.
2. After 60 months from January 2006, the town had at least 25000 residents.