Solving One-Step Inequalities by Adding or Subtracting

Solving One-Step Inequalities by Adding or Subtracting exercise

• Determine the maximum weight of Carol's second suitcase.

  Hints

  Use Opposite Operations:

  • Multiplication ($\times~\longleftrightarrow~\div$)
  • Division ($\div~\longleftrightarrow~\times$)
  • Addition ($+~\longleftrightarrow~-$)
  • Subtraction ($-~\longleftrightarrow~+$)

  Adding or subtracting doesn't change the sign of the inequality.

  When multiplying or dividing by a negative number, you must remember to flip the sign of the inequality.

  Solution

  Consider the facts then write an inequality to calculate the weight limit of the second suitcase:

  • A suitcase weighing $43$ lbs. is already on the scale
  • the unknown weight of her second piece of luggage is $x$
  • the total weight is $x+43$
  • the weight limit is 50 lbs.

  This is our inequality: $x+43\le50$

  We can solve it using Opposite Operations:

  $\begin{array}{rcl} x+43&\le&~~~50\\ \color{#669900}{-43} && \color{#669900}{-43}\\ x&\le&~~~~~7 \end{array}$

  Carol's second suitcase can weigh $7$ lbs. or less.

  Remember: if we add or subtract on both sides of the inequality sign, the sign of the inequality doesn't change directions.

• Describe how to represent solutions of an inequality on a number line.

  Hints

  You can transform the inequality using opposite operations.

  Use Opposite Operations:

  • Multiplication ($\times~\longleftrightarrow~\div$)
  • Division ($\div~\longleftrightarrow~\times$)
  • Addition ($+~\longleftrightarrow~-$)
  • Subtraction ($-~\longleftrightarrow~+$)

  Remember that opposite operations must be carried out on both sides of the inequality.

  Solution

  Let's have a look at the inequality $x+43\le50$.

  Subtracting $43$ on both sides, we get $x\le 7$.

  The number line shown on the right models the solution set for this inequality.

  • Because of the sign $\le$, which means less than or equal to, the circle at $7$ is filled in.
  • The arrow is pointing to the left and ends at zero because there's no such thing as negative weights.

• Find the correct inequalites for the number lines.

  Hints

  Transform each inequality so that you isolate the variable $x$.

  When graphing inequalities on the number line, the inequality signs $\le$ or $\ge$ both indicate closed circles while $<$ or $>$ indicate open circles.

  Solution

  Top Number Line

  $\mathbf{x+27\le38}$

  • Subtracting $27$ on both sides gives us $x\le 11$.
  • Because of $\le$ sign, the circle at $11$ is closed.
  • The arrow points to the left.

  Middle Number Line

  $\mathbf{x-5<3}$

  • Adding $5$ on both sides of the inequality sign gives us $x<8$.
  • Because of $<$ sign, the circle at $8$ is open.
  • The arrow points to the left.

  Bottom Number Line

  $\mathbf{x+14\le26}$

  • Subtracting $14$ on both sides gives us $x\le 12$.
  • Because of $\le$ sign, the circle at $12$ is closed.
  • The arrow points to the left.

• Figure out how much Sue can spend on ice cream.

  Hints

  All the information needed to solve the problem is included above.

  Isolate the variable $x$ using opposite operations.

  To check solutions, substitute them back into the inequality.

  Solution

  To solve inequalities, we should use the following steps:

  1. Understand the problem
  2. Write the inequality
  3. Solve the inequality
  4. Check our work

  Let's try those steps for the situation described above:

  1. We know that Paula has $\$20$ in total. She spends $\$12$ on a new shirt. $x$ is the amount for the ice cream.
  2. So we can say that $x+12$ represents the money Paula spent, and $20$ is her spending limit. This gives us the inequality $x+12\le 20$.
  3. By subtracting $12$ on both sides of the inequality, we can isolate the variable $x$, giving us $x\le 8$.
  4. Let's look at two solutions from the number line.

  $\mathbf{x=8}$:

  $8+12=20\le20$ $\surd$

  $\mathbf{x=4}$:

  $4+12=16\le20$ $\surd$

  So the girls can spend $\$8$ or less on ice cream.

  Naturally, we don't have negative numbers in mind...no one will PAY you to eat ice cream!

• Decide how much Carol's suitcase can weigh.

  Hints

  This number line models the solution set of the inequality $x+43\le 50$.

  Substitute $x$ with each given value.

  Two of the given values are correct solutions for the inequality.

  Solution

  $x$ represents the possible weight of Carol's remaining suitcase.

  Subtracting $43$ on both sides leads to $x\le 7$.

  All solutions of this inequality can be seen on the right.

  Let's check different values for $x$:

  $\begin{array}{rclcrcl} x&=&7&&7+43&=&50\le 50 \surd\\ x&=&2&&2+43&=&45\le50 \surd\\ x&=&8&&8+43&=&51\not \le 50\\ x&=&10&&10+43&=&53\not \le 50\\ x&=&12&&12+43&=&55\not \le 50 \end{array}$

• Solve the inequalities.

  Hints

  First, if needed, combine like terms by adding or subtracting values.

  Use Opposite Operations:

  • Multiplication ($\times~\longleftrightarrow~\div$)
  • Division ($\div~\longleftrightarrow~\times$)
  • Addition ($+~\longleftrightarrow~-$)
  • Subtraction ($-~\longleftrightarrow~+$)

  Remember to carry out the opposite operations on both sides of the equation. However, you can combine like terms on one or on both sides of the equation.

  Solution

  To solve inequalities, you must isolate the variable $x$. We can do this by Combining Like Terms and using Opposite Operations.

  $\mathbf{x+4-7\ge 12}$

  $\begin{array}{rclcl} x+4-7&\ge& ~12&&|\text{ Combine Like Terms}\\ x-3 &\ge&~12\\ \color{#669900}{+3}&&\color{#669900}{+3}&&| (-~\leftrightarrow~+) \text{ Opposite Operations}\\ x&\ge&~15\\ ~\\ \end{array}$

  $\mathbf{x+12\le 20+14}$

  $\begin{array}{rclcl} x+12&\le& 20+14&&|\text{ Combine Like Terms}\\ x+12 &\le&~~~34\\ \color{#669900}{-12}&&\color{#669900}{-12}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ x&\le&~~~22\\ ~\\ \end{array}$

  $\mathbf{x+5< 22}$

  $\begin{array}{rclcl} x+5&<& ~22\\ \color{#669900}{-5}&&\color{#669900}{-5}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ x&<&~17 \end{array}$