CEMC's Open Courseware - Lesson 11: Solving One-Step Inequalities Using Algebra

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  - Inequalities on a Scale
  - Solving Inequalities with Addition and Subtraction
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Answers and Solutions

1. Yes, since $4(1)=4$ and $4 \lt 8$ is a true statement, $x=1$ is a solution to $4x \lt 8$.
2. Yes, since $-3(5) = -15$ and $-15 \geq -15$ is a true statement, $g=5$ is a solution to $-3g \geq -15$.
3. No, since $-2+5=3$ and $1 \gt 3$ is a false statement, $z=5$ is not a solution to $1 \gt -2 + z$.
4. No, since $-2(3) = -6$ and $-6 \lt -6$ is a false statement, $y=3$ is not a solution to $-6 \lt -2y$.
1. To solve, we add $3$ to both sides of the inequality.
  $\begin{align*} x-3 &\gt 8 \\ x - 3 \class{hl2}{+3} &\gt 8 \class{hl2}{+3} \\ x & \gt 11 \end{align*}$
  This means if we choose any value for $x$ that is greater than $11$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $x$ that is smaller than $11$, then the original inequality will be false.
  - choose $11$ as the value for $x$, then the original inequality will be false.
  - choose any value for $x$ that is larger than $11$, then the original inequality will be true.
2. To solve, we subtract $4$ from both sides of the inequality.
  $\begin{align*} h + 4 &\geq 5 \\ h +4 \class{hl2}{-4} &\geq 5 \class{hl2}{-4} \\ h &\geq 1 \end{align*}$
  This means if we choose any value for $h$ that is greater than or equal to $1$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $h$ that is smaller than $1$, then the original inequality will be false.
  - choose $1$ as the value for $h$, then the original inequality will be true.
  - choose any value for $h$ that is larger than $1$, then the original inequality will be true.
3. To solve, we multiply both sides of the inequality by $4$.
  $\begin{align*} \frac{t}{4} &\leq 2 \\ \frac{t}{4} \class{hl2}{(4)} &\leq 2 \class{hl2}{(4)} \\ t & \leq 8 \end{align*}$
  This means if we choose any value for $t$ that is less than or equal to $8$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $t$ that is smaller than $8$, then the original inequality will be true.
  - choose $8$ as the value for $t$, then the original inequality will be true.
  - choose any value for $t$ that is larger than $8$, then the original inequality will be false.
4. To solve, we divide both sides of the inequality by $-7$. Since we are dividing by a negative value, we must remember to reverse $\gt$ to $\lt$.
  $\begin{align*} -7x &\gt 14 \\ \frac{-7x}{\class{hl2}{-7}} & \class{bg6}{\lt \;} \frac{14}{\class{hl2}{-7}} \\ x &\lt -2 \end{align*}$
  This means if we choose any value for $x$ that is less than $-2$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $x$ that is smaller than $-2$, then the original inequality will be true.
  - choose $-2$ as the value for $x$, then the original inequality will be false.
  - choose any value for $x$ that is larger than $-2$, then the original inequality will be false.
5. First, we collect like terms.
  $\begin{align*} 5x - 2x &\lt -24 \\ 3x & \lt -24 \end{align*}$
  To solve, we divide both sides of the inequality by $3$.
  $\begin{align*} 3x & \lt -24 \\ \frac{3x}{\class{hl2}{3}} &\lt \frac{-24}{\class{hl2}{3}} \\ x &\lt -8 \end{align*}$
  This means if we choose any value for $x$ that is less than $-8$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $x$ that is smaller than $-8$, then the original inequality will be true.
  - choose $-8$ as the value for $x$, then the original inequality will be false.
  - choose any value for $x$ that is larger than $-8$, then the original inequality will be false.
6. First, we collect like terms.
  $\begin{align*} \frac{y}{3} - \frac{y}{4} &\leq 1 + \frac{1}{2} \\[1mm] \frac{4y}{12} - \frac{3y}{12} &\leq \frac{2}{2} + \frac{1}{2} \\[1mm] \frac{y}{12} &\leq \frac{3}{2} \end{align*}$
  To solve, we multiply both sides of the inequality by $12$.
  $\begin{align*} \frac{y}{12} &\leq \frac{3}{2} \\[1mm] \frac{y}{12} \class{hl2}{(12)} &\leq \frac{3}{2} \class{hl2}{(12)} \\[1mm] y &\leq \frac{36}{2} \\[1mm] y &\leq 18 \end{align*}$
  This means if we choose any value for $y$ that is less than or equal to $18$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $y$ that is smaller than $18$, then the original inequality will be true.
  - choose $18$ as the value for $y$, then the original inequality will be true.
  - choose any value for $y$ that is larger than $18$, then the original inequality will be false.
1. To solve the inequality we divide both sides by $2$.
  $\begin{align*} 2x &\gt 17 \\ \frac{2x}{\class{hl2}{2}} &\gt \frac{17}{\class{hl2}{2}} \\ x &\gt 8.5 \end{align*}$
  This means if we choose any value for $x$ that is greater than $8.5$, the original inequality will be true. The smallest integer greater than $8.5$ is $9$. Therefore, $9$ is the smallest integer value that satisfies the inequality $2x \gt 17$.
2. To solve the inequality we add $11$ to both sides.
  $\begin{align*} x - 11 \class{hl2}{+11} &\geq \frac{3}{2} \class{hl2}{+11} \\ x &\geq 1.5 + 11 \\ x &\geq 12.5 \end{align*}$
  This means if we choose any value for $x$ that is greater than $12.5$, the original inequality will be true. The smallest integer greater than $12.5$ is $13$. Therefore, $13$ is the smallest integer value that satisfies the inequality $x-11 \geq \frac{3}{2}$.
1. To solve the inequality we subtract $4$ from both sides.
  $\begin{align*} x + 4 &\leq 21.5 \\ x+4 \class{hl2}{-4} & \leq 21.5 \class{hl2}{-4} \\ x &\leq 17.5 \end{align*}$
  This means if we choose any value for $x$ that is less than $17.5$, the original inequality will be true. The greatest integer less than $17.5$ is $17$. Therefore, $17$ is the greatest integer value that satisfies the inequality $x+4 \leq 21.5$.
2. To solve the inequality we divde both sides by $5$.
  $\begin{align*} 5x &\lt -9 \\ \frac{5x}{\class{hl2}{5}} &\lt \frac{-9}{\class{hl2}{5}} \\ x &\lt -1.8 \end{align*}$
  This means if we choose any value for $x$ that is less than $-1.8$, the original inequality will be true. The greatest integer less than $-1.8$ is $-2$. Therefore, $-2$ is the greatest integer value that satisfies the inequality $5x \lt -9$.
1. Let $P$ represent the perimeter of the rectangle. Then, we have
  $\begin{align*} P &= 2w + w + 2w + w \\ &= 6w \end{align*}$
2. If the perimeter of the rectangle is $102$ cm, then we substitute $P=102$ into the above equation and solve for $w$ to find the width.
  $\begin{align*} P &= 6w \\ 102 &= 6w \\ \frac{102}{\class{hl2}{6}} &= \frac{6w}{\class{hl2}{6}} \\ 17 &= w \end{align*}$
  Therefore, if the perimeter is $102$ cm then the width of the rectangle is $17$ cm.
3. If the perimeter of the rectangle is less than $140$ cm, then we write $P \lt 140$ which is equivalent to $6w \lt 140$. Solving the inequality, we have
  $\begin{align*} 6w &\lt 140 \\ \frac{6w}{\class{hl2}{6}} &\lt \frac{140}{\class{hl2}{6}} \\ w &\lt 23 \frac{1}{3} \end{align*}$
  Therefore, if the perimeter of the rectangle is less than $140$ cm then the width can be any value that is less than $23 \frac{1}{3}$. But, the width is a measurement and so it should also be greater than $0$.
1. If $V$ is greater than or equal to $100$ then we write $V \geq 100$. Since $V=7n$ it follows that $7n \geq 100$. Solving this inequality, we have
  $\begin{align*} 7n &\geq 100 \\ \frac{7n}{\class{hl2}{7}} &\geq \frac{100}{\class{hl2}{7}} \\ n &\geq 14 \frac{2}{7} \end{align*}$
  Since $n$ represents the term number, $n$ is an integer. Therefore, $n=15$ is the smallest value for $n$ such that $V$ is greater than or equal to $100$.
2. If $V$ is a 3-digit integer then $V$ is greater than or equal to $100$ and less than $1000$. From part (a) we know that $n=15$ is the smallest value for $n$ such that $V$ is greater than or equal to $100$. We are left to find the largest value for $n$ such that $V$ is less than $1000$.
  $\begin{align*} V &\lt 1000 \\ 7n &\lt 1000 \\[1mm] \frac{7n}{\class{hl2}{7}} &\lt \frac{1000}{\class{hl2}{7}} \\ n &\lt 142 \frac{6}{7} \end{align*}$
  Since $n$ is an integer, $n=142$ is the largest value for $n$ such that $V$ is less than $1000$.
  Therefore, $V$ is a 3-digit integer when $n=15, 16, 17, \ldots, 142$.
  Therefore, $V$ is a 3-digit integer for $128$ values of $n$.
1. $E = 14.65h$
2. Substituting $E = 14.65h$ into the inequality and solving, we have
  $\begin{align*} E &\gt 1573.43 \\ 14.65h &\gt 1573.43 \\ \frac{14.65h}{\class{hl2}{14.65}} &\gt \frac{1573.43}{\class{hl2}{14.65}} \\ h &\gt 107.402 \end{align*}$
  If we assume that Frederick only gets paid for every full hour of work, then $h=108$ is the smallest number of hours he can work so that he has enough money to buy the computer.
  In this case, he would have earned $108 \times $14.65 = $1582.20$.
3. Based on Frederick's estimate, we must solve the inequality $8.75h \gt 1573.43$.
  $\begin{align*} E &\gt 1573.43 \\ 8.75h &\gt 1573.43 \\ \frac{8.75h}{\class{hl2}{8.75}} &\gt \frac{1573.43}{\class{hl2}{8.75}} \\ h &\gt 179.821 \end{align*}$
  Again, assuming that Frederick only gets paid for every full hour of work, then $h=180$ is the smallest number of hours he can work so that he can save enough money to buy the computer.
  In this case, he would have saved $180 \times $8.75 = $1575$.
1. $s+2 \geq t+2$
2. $a-6 \lt b-6$
3. $3g \gt 3h$
4. $-5p \geq -5q$
1. Answers will vary.
  Since $k \gt 3$ it follows that $k+5 \gt 3+5$.
  Therefore, the inequality $k+5 \gt 8$ has the solution $k \gt 3$.
2. Answers will vary.
  Since $k \gt 3$ it follows that $k(10) \gt 3(10)$.
  Therefore, the inequality $10k \gt 30$ has the solution $k \gt 3$.
No, it is not always true. For example, let $a=2$, $b=5$, $c = -10$ and $d=-5$.
Then $a-c = 2-(-10) = 12$ and $b-d = 5 - (-5) = 10$.
It follows that $a-c \lt b-d$ is $12 \lt 10$ which is a false statement.