CEMC's Open Courseware - Lesson 12: Solving Two-Step Inequalities Using Algebra

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Answers and Solutions

1. Yes, since $2(9)-7 = 11$ and $11 \gt 10$ is a true statement, $k=9$ is a solution to $2k-7 \gt 10$.
2. No, since $3(4)+8 =20$ and $20 \lt 20$ is a false statement, $t=4$ is not a solution to $20 \lt 3t+8$.
3. No, since $17-3(9)= -10$ and $-10 \geq 26$ is a false statement, $x=9$ is not a solution to $17-3x \geq 26$.
4. Yes, since $\frac{-18}{2} + 20 = 11$ and $-18 \leq 11$ is a true statement, $p=-18$ is a solution to $-18 \leq \frac{p}{2} +20$.
1. $2p$ $-9$ $\lt 21$
  First we would reverse the subtraction, and then reverse the multiplication in the second step.
2. $4a$ $+8$$\gt 15$
  First we would reverse the addition, and then reverse the multiplication in the second step.
3. $16 \leq $ $\dfrac{m}{3}$ $+10$
  First we would reverse the addition, and then reverse the division in the second step.
Notice that there are two terms on the left side of this inequality: $17$ and $-3y$.
We need to reverse the first term, $17$, first. Since $17$ is positive, we subtract $17$ from both sides.
Step 1: Subtract $17$ from both sides.
$\begin{align*} 17-3y &\gt 26 \\ 17-3y \class{hl2}{-17} &\gt 26 \class{hl2}{-17} \\ -3y &\gt 9 \end{align*}$
Step 2: Divide both sides by $-3$, remembering that the $\gt$ must become a $\lt$ when dividing by a negative value.
$\begin{align*} -3y &\gt 9 \\ \dfrac{-3y}{\class{hl1}{-3}} & \class{bg6}{\lt \;} \dfrac{9}{\class{hl1}{-3}} \\ y & \lt -3 \end{align*}$
Verifying: Convince yourself that if you were to:
- choose any value for $y$ that is smaller than $-3$, then the original inequality will be true.
- choose $-3$ as the value for $y$, then the original inequality will be false.
- choose any value for $y$ that is larger than $-3$, then the original inequality will be false.
Therefore, $y \lt -3$ is the solution to the inequality $17-3y \gt 26$.
1. To solve, we first add $9$ to both sides of the inequality, and then divide each side by $15$.
  $\begin{align*} 15x - 9 &\gt 21 \\ 15x - 9 \class{hl2}{+9} & \gt 21 \class{hl2}{+9} \\ 15x &\gt 30 \\ \dfrac{15x}{\class{hl1}{15}} & \gt \dfrac{30}{\class{hl1}{15}} \\ x &\gt 2 \end{align*}$
  This means if we choose any value for $x$ that is greater 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 false.
  - 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 true.
2. To solve, we first subtract $3$ from both sides of the inequality, and then multiply each side by $2$.
  $\begin{align*} 3 + \dfrac{k}{2} &\geq 11 \\ 3 + \dfrac{k}{2} \class{hl2}{-3} &\geq 11 \class{hl2}{-3} \\ \dfrac{k}{2} &\geq 8 \\ \dfrac{k}{2} \class{hl1}{(2)} &\geq 8 \class{hl1}{(2)} \\ k &\geq 16 \end{align*}$
  This means if we choose any value for $k$ that is greater than or equal to $16$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $k$ that is smaller than $16$, then the original inequality will be false.
  - choose $16$ as the value for $k$, then the original inequality will be true.
  - choose any value for $k$ that is larger than $16$, then the original inequality will be true.
3. To solve, we first subtract $33$ from both sides of the inequality, and then divide each side by $-8$.
  $\begin{align*} 33 - 8w &\leq -7 \\ 33 - 8w \class{hl2}{-33} &\leq -7 \class{hl2}{-33} \\ -8w &\leq -40 \\ \frac{-8w}{\class{hl1}{-8}} & \class{bg6}{\geq \;} \frac{-40}{\class{hl1}{-8}} \\ w &\geq 5 \end{align*}$
  This means if we choose any value for $w$ that is greater than or equal to $5$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $w$ that is smaller than $5$, then the original inequality will be false.
  - choose $5$ as the value for $w$, then the original inequality will be true.
  - choose any value for $w$ that is larger than $5$, then the original inequality will be true.
4. To solve, we first subtract $8$ from both sides of the inequality, and then multiply each side by $5$.
  $\begin{align*} 14 &\gt \frac{r}{5} + 8 \\ 14 \class{hl2}{-8} &\gt \frac{r}{5} + 8 \class{hl2}{-8} \\ 6 &\gt \frac{r}{5} \\ 6 \class{hl1}{(5)} &\gt \frac{r}{5} \class{hl1}{(5)} \\ 30 &\gt r \\ \textrm{or }~~~ r &\lt 30 \end{align*}$
  This means if we choose any value for $r$ that is less than $30$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $r$ that is smaller than $30$, then the original inequality will be true.
  - choose $30$ as the value for $r$, then the original inequality will be false.
  - choose any value for $r$ that is larger than $30$, then the original inequality will be false.
5. To solve, we first subtract $12$ from both sides of the inequality, and then multiply each side by $-3$.
  $\begin{align*} 12 - \frac{y}{3} &\lt 16 \\ 12 - \frac{y}{3} \class{hl2}{-12} &\lt 16 \class{hl2}{-12} \\ - \frac{y}{3} &\lt 4 \\ - \frac{y}{3} \class{hl1}{(-3)} &\class{bg6}{\gt \;} 4 \class{hl1}{(-3)} \\ y &\gt -12 \end{align*}$
  This means if we choose any value for $y$ that is greater than $-12$, the original inequality will be true.
  Verifying: Convince yourself that if you were to:
  - choose any value for $y$ that is smaller than $-12$, then the original inequality will be false.
  - choose $-12$ as the value for $y$, then the original inequality will be false.
  - choose any value for $y$ that is larger than $-12$, then the original inequality will be true.
1. To solve the inequality we add $19$ to both sides and then divide each side by $6$.
  $\begin{align*} 6x-19 &\geq -15 \\ 6x-19 \class{hl2}{+19} &\geq -15 \class{hl2}{+19} \\ 6x &\geq 4 \\ \dfrac{6x}{\class{hl1}{6}} &\geq \dfrac{4}{\class{hl1}{6}} \\ x &\geq 0.\overline{6} \end{align*}$
  This means if we choose any value for $x$ that is greater than $0.\overline{6}$, the original inequality will be true. The smallest integer greater than $0.\overline{6}$ is $1$. Therefore, $1$ is the smallest integer value that satisfies the inequality $6x-19 \geq -15$.
2. To solve the inequality we subtract $9$ from both sides and then divide each side by $5$.
  $\begin{align*} 9 + 5x &\gt 3 \\ 9 + 5x \class{hl2}{-9} &\gt 3 \class{hl2}{-9} \\ 5x &\gt -6 \\ \dfrac{5x}{\class{hl1}{5}} & \gt \dfrac{-6}{\class{hl1}{5}} \\ x &\gt -1.2 \end{align*}$
  This means if we choose any value for $x$ that is greater than $-1.2$, the original inequality will be true. The smallest integer greater than $-1.2$ is $-1$. Therefore, $-1$ is the smallest integer value that satisfies the inequality $9 + 5x \gt 3$.
1. To solve the inequality we subtract $8$ from both sides and then divide each side by $3$.
  $\begin{align*} 8 + 3x &\lt -6 \\ 8 + 3x \class{hl2}{-8} &\lt -6 \class{hl2}{-8} \\ 3x &\lt -14 \\ \dfrac{3x}{\class{hl1}{3}} & \lt \dfrac{-14}{\class{hl1}{3}} \\ x &\lt -4.\overline{6} \end{align*}$
  This means if we choose any value for $x$ that is less than $-4.\overline{6}$, the original inequality will be true. The greatest integer smaller than $-4.\overline{6}$ is $-5$. Therefore, $-5$ is the greatest integer value that satisfies the inequality $8+3x \lt -6$.
2. To solve the inequality we add $25$ to both sides and then divide each side by $10$.
  $\begin{align*} 10x - 25 &\leq -3 \\ 10x - 25 \class{hl2}{+25} &\leq -3 \class{hl2}{+25} \\ 10x &\leq 22 \\ \dfrac{10x}{\class{hl1}{10}} &\leq \dfrac{22}{\class{hl1}{10}} \\ x &\leq 2.2 \end{align*}$
  This means if we choose any value for $x$ that is less than $2.2$, the original inequality will be true. The greatest integer smaller than $2.2$ is $2$. Therefore, $2$ is the greatest integer value that satisfies the inequality $10x-25 \leq -3$.
Let $n$ be the number of times Ammar visits the tennis club. Then the equation $C=15 + 2n$ represents the cost of one month of tennis. If Ammar will spend no more than $$50$, then the inequality $15 + 2n \leq 50$ will help us determine the number of times Ammar can visit the club.
$\begin{align*} 15 + 2n &\leq 50 \\ 15 + 2n \class{hl2}{-15} &\leq 50 \class{hl2}{-15} \\ 2n &\leq 35 \\ \dfrac{2n}{\class{hl1}{2}} &\leq \dfrac{35}{\class{hl1}{2}} \\ n &\leq 17.5 \end{align*}$
Since Ammar can only visit the club a whole number of times, he can visit the club at most $17$ times to stay within his budget.
Let $f$ represent the amount of filament used for a print. Then the equation $C_1 = 0.12f$ represents the cost of printing using Option 1 and the equation $C_2 = 5 + 0.06f$ represents the cost of printing using Option 2.
Then, the inequality $5 + 0.06f \lt 0.12f$ will help us determine when the cost of Option 2 is less than the cost of Option 1.
$\begin{align*} 5 + 0.06f &\lt 0.12f \\ 5 + 0.06f \class{hl2}{-5} &\lt 0.12f \class{hl2}{-5} \\ 0.06f &\lt 0.12f - 5 \end{align*}$
Notice that subtracting $5$ from both sides does not help simplify our inequality in this case because we have two like terms, involving variables, on each side of the inequality. Ultimately, we want to be able to collect the like terms, and so to do that, we must instead subtract $0.06f$ from both sides.
$\begin{align*} 5 + 0.06f &\lt 0.12f \\ 5 + 0.06f \class{hl2}{-0.06f} &\lt 0.12f \class{hl2}{-0.06f} \\ 5 &\lt 0.06f \\ \dfrac{5}{\class{hl1}{0.06}} &\lt \dfrac{0.06f}{\class{hl1}{0.06}} \\ 83. \overline{3} &\lt f \end{align*}$
Therefore, in order to make Option 2 cheaper than Option 1, you would need to use more than $83.33$ g of filament per month.
1. $-8s+9 \geq -8t+9$
2. $2a-7 \gt 2b-7$
1. Answers will vary.
  Since $k \lt 5$ it follows that $4k \lt 4(5)$ and so $4k \lt 20$.
  Since $4k \lt 20$ it follows that $4k - 7 \lt 20 - 7$ and so $4k-7 \lt 13$
  Therefore, the inequality $4k-7 \lt 13$ has the solution $k \lt 5$.
2. Answers will vary.
  Since $k \lt 5$ it follows that $\frac{k}{5} \lt \frac{5}{5}$ and so $\frac{k}{5} \lt 1$.
  Since $\frac{k}{5} \lt 1$ it follows that $\frac{k}{5} + 8 \lt 1 + 8$ and so $\frac{k}{5} + 8 \lt 9$.
  Therefore, the inequality $\frac{k}{5} + 8 \lt 9$ has the solution $k \lt 5$.
1. The following image is not drawn to scale. Your image should have the same shape, but will be larger.
2. If $AB=1$ then $BC=8$ and $AC=5$. To show that we cannot draw $\triangle ABC$, start by constructing $BC$. Then draw a circle of radius $1$ with centre $B$. This circle represents all possible locations for $A$ that are $1$ cm from $B$.
  Then, draw a circle of radius $5$ with centre $C$. This circle represents all possible locations for $A$ that are $5$ cm from $C$.
  Since the two circles do not intersect, there is no location for $A$ that is both $1$ cm from $B$ and $5$ cm from $C$. Therefore, we cannot draw $\triangle ABC$.
3. From part (b) we can deduce that if we are able to draw a triangle, then the two circles must intersect. We can use trial and error, on the value of $AB$.
  - If $AB=3$ then $BC=10$ and $AC=7$. We can draw a circle of radius $3$ with centre $B$ and a circle of radius $7$ with centre $C$. The two circles intersect once and the intersection point is on the line $BC$.
    This means there is only one possible location for $A$ such that $AB=3$ and $AC=7$. But, if $A$ is on $BC$ then we do not have a triangle. Therefore, if $AB$ is $3$ cm we cannot draw $\triangle ABC$.
  - If $AB=4$ then $BC=11$ and $AC=8$. We can draw a circle of radius $4$ with centre $B$ and a circle of radius $8$ with centre $C$. The two circles intersect twice, once on each side of $BC$.
    This means that there are two possible locations for $A$ such that $AB=4$ and $AC=8$. Choose one of these intersection points as the point $A$ and you will have constructed $\triangle ABC$.
  Therefore $AB=4$ cm is the smallest possible value for $AB$ for which you can draw $\triangle ABC$.
4. From part (c) we can conclude that in order to be able to draw $\triangle ABC$, the value of $AB + AC$ must be greater than $BC$. That is, $AB + AC \gt BC$.
  In general, given any triangle with side lengths $a, b$ and $c$, the sum of the lengths of any two sides must be greater than the length of the third side. That is,
  - $a + b \gt c$,
  - $b + c \gt a$, and
  - $a + c \gt b$.
  This result is known as the triangle inequality.