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# 1.4: Equations and Inequalities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write equations and inequalities.
• Check solutions to equations.
• Check solutions to inequalities.
• Solve real-world problems using an equation.

## Introduction

In algebra, an equation is a mathematical expression that contains an equal sign. It tells us that two expressions represent the same number. For example, y=12x\begin{align*}y=12x\end{align*} is an equation. An inequality is a mathematical expression that contains inequality signs. For example \begin{align*} y \leq 12x\end{align*} is an inequality. Inequalities are used to tell us that an expression is either larger or smaller than another expression. Equations and inequalities can contain variables and constants.

• Variables are usually given a letter and they are used to represent unknown values. These quantities can change because they depend on other numbers in the problem.
• Constants are quantities that remain unchanged.

Equations and inequalities are used as a short hand notation for situations that involve numerical data. They are very useful because most problems require several steps to arrive at a solution, and it becomes tedious to repeatedly write out the situation in words.

## Write Equations and Inequalities

Here are some examples of equations.

a) \begin{align*}3x-2=5\end{align*}

b) \begin{align*}x+9=2x+5\end{align*}

c) \begin{align*} \frac{x} {3} = 15\end{align*}

d) \begin{align*} x^2+1=10\end{align*}

To write an inequality, we use the following symbols.

\begin{align*} > \end{align*} greater than

\begin{align*}\geq\end{align*} greater than or equal to

\begin{align*}< \end{align*} less than

\begin{align*} \leq \end{align*} less than or equal to

\begin{align*} \neq \end{align*} not equal to

Here are some examples of inequalities.

a) \begin{align*} 3x < 5\end{align*}

b) \begin{align*}4-x \leq 2x\end{align*}

c) \begin{align*} x^2+2x-1 >0 \end{align*}

d) \begin{align*} \frac{3x} {4} \ge \frac{x} {2} - 3\end{align*}

The most important skill in algebra is the ability to translate a word problem into the correct equation or inequality so you can find the solution easily. Going from a word problem to the solution involves several steps. Two of the initial steps are defining the variables and translating the word problem into a mathematical equation.

Defining the variables means that we assign letters to any unknown quantities in the problem.

Translating means that we change the word expression into a mathematical expression containing variables and mathematical operations with an equal sign or an inequality sign.

Example 1

Define the variables and translate the following expressions into equations.

a) A number plus 12 is 20.

b) 9 less than twice a number is 33.

c) Five more than four times a number is 21.

d) 20 was one quarter of the money spent on the pizza. Solution a) Define Let \begin{align*} n= \end{align*} the number we are seeking Translate A number plus 2 is 20 \begin{align*} n+12=20\end{align*} Answer The equation is: \begin{align*} n+12=20\end{align*} b) Define: Let \begin{align*} n=\end{align*} the number we are seeking Translate 9 less than twice a number is 33 This means that twice a number minus 9 is 33 \begin{align*} 2\times n-9=33\end{align*} Answer The equation is: \begin{align*} 2n-9=33\end{align*} c) Define Let \begin{align*} n= \end{align*} the number we are seeking Translate Five more than four times a number is 21. This means that four times a number plus five is 21. \begin{align*} 4 \times n +5 =21\end{align*} Answer The equation is: \begin{align*} 4n +5 =21\end{align*} d) Define Let \begin{align*} m=\end{align*} the money spent on the pizza Translate20 was one quarter of the money spent on the pizza.

Translate

\begin{align*}20= \frac{1} {4} \times m\end{align*}

The equation is: \begin{align*}\frac{1} {4} m = 20\end{align*}

Often word problems need to be reworded before you can write an equation.

Example 2

Find the solution to the following problems.

a) Shyam worked for two hours and packed 24 boxes. How much time did he spend on packing one box?

b) After a 20% discount, a book costs 12. How much was the book before the discount? Solution a) Define Let \begin{align*} t=\end{align*} time it take to pack one box Translate Shyam worked for two hours and packed 24 boxes. This means that two hours is 24 times the time it takes to pack one book. \begin{align*} 2=24 \times t \end{align*} Solve \begin{align*} t = \frac{2} {24}\end{align*} so \begin{align*} t = \frac{1} {12} \ hours\end{align*} or \begin{align*} t = \frac{1} {12} \times 60 \ minutes = 5 \ minutes\end{align*} Answer Shyam takes 5 minutes to pack a box. b) Define: Let \begin{align*}p=\end{align*} the price of the book before the discount. Translate After a 20% discount, a book costs12.

This means that the price -20% of price is 12 \begin{align*}p-0.20p=12\end{align*} Solve \begin{align*}0.8p=12\end{align*} so \begin{align*} p = \frac{12} {0.8}\end{align*} and \begin{align*}p=15\end{align*} Answer The price of the book before the discount was15.

Check

20% discount means: \begin{align*}0.20\times \15 = \3\end{align*}

Price after discount: \begin{align*}\18-\3 = \15\end{align*}

Example 3

Define the variables and translate the following expressions into inequalities.

a) The sum of 5 and a number is less than or equal to 2.

b) The distance from San Diego to Los Angeles is less than 150 miles.

c) Diego needs to earn more than an 82 on his test to receive a B in his algebra class.

d) A child needs to be 42 inches or more to go on the roller coaster.

Solution

a) Define

Let \begin{align*}n=\end{align*} the unknown number.

Translate

\begin{align*} 5+n\leq 2\end{align*}

b) Define

Let \begin{align*}d=\end{align*} the distance from San Diego to Los Angeles in miles.

Translate

\begin{align*} d <150 \end{align*}

c) Define

Let \begin{align*}x=\end{align*} Diego’s test grade.

Translate

\begin{align*} x >82 \end{align*}

d) Define

Let \begin{align*}h=\end{align*} the height of child in inches.

Translate:

\begin{align*} h \geq 42\end{align*}

## Check Solutions to Equations

You will often need to check solutions to equations in order to check your work. In a math class, checking that you arrived at the correct solution is very good practice. We check the solution to an equation by replacing the variable in an equation with the value of the solution. A solution should result in a true statement when plugged into the equation.

Example 4

Check that \begin{align*}x=5\end{align*} is the solution to the equation \begin{align*}3x+2=-2x+27\end{align*}.

Solution

To check that \begin{align*}x=5\end{align*} is the solution to the equation, we “plug in” the value of 5 for the variable, \begin{align*}x\end{align*}:

\begin{align*}3x+2 & =-2x+27\\ 3 \cdot x +2 & = -2 \cdot x +27\\ 3\cdot 5 +2 & = -2 \cdot 5 +27\\ 15 +2& = -10 +27\\ 17 &= 17\end{align*}

This is a true statement.

This means that \begin{align*}x=5\end{align*} is the solution to equation \begin{align*}3x+2=-2x+27\end{align*}.

Example 5

Check that the given number is a solution to the corresponding equation.

a) \begin{align*}y=-1; 3y+5=-2y\end{align*}

b) \begin{align*}z=3; z^2+2z=8\end{align*}

c) \begin{align*}x=-\frac{1}{2}; 3x+1=x\end{align*}

Solution

Replace the variable in each equation with the given value.

a) \begin{align*}3(-1)+5 & = -2(-1)\\ -3+5 & = 2\\ 2 & = 2\end{align*}

This is a true statement. This means that \begin{align*} y=-1\end{align*} is a solution to \begin{align*}3y+5=-2y\end{align*}.

b) \begin{align*}3^2+2(3)&= 8\\ 9+6 &= 8\\ 15 &= 8\end{align*}

This is not a true statement. This means that \begin{align*}z=3\end{align*} is not a solution to \begin{align*}z^2+2z=8\end{align*}.

c) \begin{align*}3\left( -\frac{1}{2} \right) +1&= -\frac{1}{2}\\ \left( -\frac{3}{2} \right) +1&= -\frac{1}{2}\\ -\frac{1}{2} &= -\frac{1}{2}\end{align*}

This is a true statement. This means that \begin{align*}x=\frac{1}{2}\end{align*} is a solution to \begin{align*}3x+1=x\end{align*}.

## Check Solutions to Inequalities

To check the solution to an inequality, we replace the variable in the inequality with the value of the solution. A solution to an inequality produces a true statement when substituted into the inequality.

Example 6

Check that the given number is a solution to the corresponding inequality.

a) \begin{align*} a = 10; 20a \le 250\end{align*}

b) \begin{align*} b = -2; \frac{3 - b} {b} > -4\end{align*}

c) \begin{align*} x = \frac{3}{4}; 4x + 5 \le 8\end{align*}

d) \begin{align*} z = 25; \frac{z} {5} + 1 < z - 20\end{align*}

Solution

Replace the variable in each inequality with the given value.

a) \begin{align*}&20(10) \leq 250\\ &200 \leq 250 \end{align*}

This statement is true. This means that \begin{align*}a=10\end{align*} is a solution to the inequality \begin{align*}20a \leq 250 \end{align*}. Note that \begin{align*} a=10 \end{align*} is not the only solution to this inequality. If we divide both sides of the inequality by 20 we can write that

\begin{align*} a \leq 12.5.\end{align*}

So any number equal to or less than 12.5 is going to be a solution to this inequality.

b)

\begin{align*}\frac{3 - (-2)} {(-2)} > & -4\\ \frac{3 + 2} {-2} > & -4\\ -\frac{5} {2} > & -4\\ -2.5 > & -4 \end{align*}

This statement is true. This means that \begin{align*} b=-2\end{align*} is a solution to the inequality \begin{align*} \frac{3 - b} {b} > -4\end{align*}.

c)

\begin{align*}4 \left(\frac{3} {4} \right) + 5 & \ge 8\\ 3 + 5 & \ge 8\\ 8 & \ge 8\end{align*}

This statement is true. It is true because the equal sign is included in the inequality. This means that \begin{align*} x = \frac{3} {4}\end{align*} is a solution to the inequality \begin{align*}4x+5 \geq 8 \end{align*}.

d)

\begin{align*} \frac{25} {5} + 1 & < 25 - 20\\ 5 + 1 & < 5\\ 6 & < 5\end{align*}

This statement is not true. This means that \begin{align*}z=25\end{align*} is not a solution to \begin{align*} \frac{z} {5} + 1 < z - 20\end{align*}.

## Solve Real-World Problems Using an Equation

Let’s use what we have learned about defining variables, writing equations and writing inequalities to solve some real-world problems.

Example 7

Tomatoes cost $0.50 each and avocados cost$2.00 each. Anne buys six more tomatoes than avocados. Her total bill is 8. How many tomatoes and how many avocados did Anne buy? Solution Define Let \begin{align*}a=\end{align*} number of avocados Anne buys Translate Anne buys six more tomatoes than avocados This means that \begin{align*}a+6=\end{align*} number of tomatoes Translate Tomatoes cost0.50 each and avocados cost $2.00 each. Her total bill is$8.

This means that $0.50 times the number of tomatoes plus$2 times the number of avocados equals 8 \begin{align*}0.5\times (a + 6) + 2 \times a & = 8\\ 0.5a+0.5\times 6+ 2a & = 8 & & \text{THIS IS OUR EQUATION}. \\ 2.5a + 3 & = 8 & & \text{Simplify}\\ 2.5a & = 5\\ a & =2\end{align*} Remember that \begin{align*}a=\end{align*} the number of avocados, so Anne buys two avocados. We also know that the number of tomatoes is given by \begin{align*}a + 6 = 2 + 6 = 8 \end{align*} Answer Anne bought 2 avocados and 8 tomatoes. Check If Anne bought two avocados and eight tomatoes, the total cost is: \begin{align*}2 \times \2 + 8 \times \0.50 = \4 + \4 = \8 \end{align*} The answer checks out. Example 8 To organize a picnic Peter needs at least two times as many hamburgers as hotdogs. He has 24 hotdogs. What is the possible number of hamburgers Peter has? Solution Define Let \begin{align*}x=\end{align*} number of hamburgers Translate Peter needs at least two times as many hamburgers as hot dogs. He has 24 hot dogs. This means that twice the number of hot dogs is less than or equal to the number of hamburgers. \begin{align*}2 \times 24 \leq x\end{align*} Simplify \begin{align*}48 \leq x\end{align*} Answer Peter needs at least 48 hamburgers Check We found \begin{align*}x=48\end{align*}. 48 hamburgers is twice the number of hot dogs. So more than 48 hamburgers is more than twice the number of hot dogs. The answer checks out. ## Review Questions 1. Define the variables and translate the following expressions into equations. 1. Peter’s Lawn Mowing Service charges10 per job and $0.20 per square yard. Peter earns$25 for a job.
2. Renting the ice-skating rink for a birthday party costs $200 plus$4 per person. The rental costs $324 in total. 3. Renting a car costs$55 per day plus $0.45 per mile. The cost of the rental is$100.
2. Define the variables and translate the following expressions into inequalities.
1. A bus can seat 65 passengers or fewer.
2. The sum of two consecutive integers is less than 54.
3. An amount of money is invested at 5% annual interest. The interest earned at the end of the year is greater than or equal to $250. 4. You buy hamburgers at a fast food restaurant. A hamburger costs$0.49. You have at most 3 to spend. Write an inequality for the number of hamburgers you can buy. 3. Check that the given number is a solution to the corresponding equation. 1. \begin{align*} a = -3; 4a + 3 = -9\end{align*} 2. \begin{align*} x = \frac{4} {3}; \frac{3} {4} x + \frac{1} {2} = \frac{3} {2}\end{align*} 3. \begin{align*} y =2; 2.5y - 10.0 = -5.0\end{align*} 4. \begin{align*} z = -5; 2(5 - 2z) = 20 - 2 (z - 1)\end{align*} 4. Check that the given number is a solution to the corresponding inequality. 1. \begin{align*} x = 12; 2(x + 6) \le 8x\end{align*} 2. \begin{align*} z = -9; 1.4z + 5.2 > 0.4z\end{align*} 3. \begin{align*} y = 40; - \frac{5} {2}y + \frac{1} {2} < -18\end{align*} 4. \begin{align*} t = 0.4; 80 \ge 10(3t + 2)\end{align*} 5. The cost of a Ford Focus is 27% of the price of a Lexus GS 450h. If the price of the Ford is15000, what is the price of the Lexus?
6. On your new job you can be paid in one of two ways. You can either be paid $1000 per month plus 6% commission of total sales or be paid$1200 per month plus 5% commission on sales over $2000. For what amount of sales is the first option better than the second option? Assume there are always sales over$2000.

1. \begin{align*}x =\end{align*} number of square yards of lawn; \begin{align*}25 = 10 + 0.2x\end{align*}
2. \begin{align*}p=\end{align*} number of people at the party; \begin{align*}324 = 200 + 4p\end{align*}
3. \begin{align*}m=\end{align*} number of miles; \begin{align*}55 + 0.45m = 100 \end{align*}
4. \begin{align*}n=\end{align*} number of blocks; \begin{align*}n = 4 + 7\end{align*}
1. \begin{align*}x =\end{align*} number of passengers; \begin{align*}x \leq 65\end{align*}
2. \begin{align*}n=\end{align*} the first integer; \begin{align*}2n + 1 < 54\end{align*}
3. \begin{align*}P=\end{align*} amount of money invested; \begin{align*}0.05P \geq 250 \end{align*}
4. \begin{align*}n =\end{align*} number of hamburgers; \begin{align*}0.49n \leq 3\end{align*}
1. \begin{align*}4(-3) + 3 = -9 \end{align*} so \begin{align*}-12 + 3 = -9\end{align*} so \begin{align*}-9 = -9\end{align*}. This is a true statement.
2. \begin{align*} \frac{3} {4} \left (\frac{4} {3} \right) + \frac{1} {2} = \frac{3} {2}\end{align*} so \begin{align*} 1 + \frac{1} {2} = \frac{3} {2}\end{align*} so \begin{align*} \frac{3} {2} = \frac{3} {2}\end{align*} This is a true statement.
3. \begin{align*}2.5(2) -10.0 = - 5.0\end{align*} so \begin{align*}5.0-10.0 = -5.0\end{align*} so \begin{align*}-5.0 = -5.0\end{align*}. This is a true statement.
4. \begin{align*}2(5 -2(-5)) = 20 -2((-5) - 1)\end{align*} so \begin{align*}2(5 + 10) = 20 - 2(-6)\end{align*} so \begin{align*}2(15) = 20 + 12\end{align*} so \begin{align*}30 = 32\end{align*}. This is not a true statement.
1. \begin{align*} 2(12 + 6) \leq 8(12)\end{align*} so \begin{align*}2(18) \leq 96\end{align*} so \begin{align*}36 \leq 96\end{align*}. This is true statement.
2. \begin{align*}1.4(-9) + 5.2 > 0.4(-9)\end{align*} so \begin{align*}-12.6 + 5.2 > -3.6\end{align*} so \begin{align*}-7.4 > -3.6\end{align*}. This is not a true statement.
3. \begin{align*}-\frac{5} {2} (40) < -18\end{align*} so \begin{align*} -100 + \frac{1} {2} < -18\end{align*} so \begin{align*}-99.5 < -18\end{align*}. This is a true statement.
4. \begin{align*}80 \geq 10(3(0.4) + 2)\end{align*} so \begin{align*}80 \geq 10(1.2 + 2)\end{align*} so \begin{align*}80 \geq 10(3.2)\end{align*} so \begin{align*}80 \geq 32\end{align*}. This is a true statement.
1. \begin{align*}x=\end{align*} price of a Lexus; \begin{align*}0.27x = 15000; x = \55556\end{align*}
2. \begin{align*}x=\end{align*} total sales; \begin{align*}1000 + 0.06x > 1200 + 0.05(x-2000)\end{align*} so \begin{align*}x >10000.\end{align*}

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