The Addition Principle
After studying this section, you will be able to:
1. Solve equations of the form x + b = c using the addition principle.
2. Using the Addition Principle
When we use the equals sign (=), we indicate that two expressions are equal in value. This is called an equation. For example, x + 5 = 23 is an equation. By choosing certain procedures, you can go step by step from a given equation to the equation x = some number. The number is the solution to the equation.
One of the first procedures used in solving equations has an application in our everyday world. Suppose that we place a 10-kilogram box on one side of a seesaw and a 10-kilogram stone on the other side. If the center of the box is the same distance from the balance point as the center of the stone, we would expect the seesaw to balance. The box and the stone do not look the same, but they have the same value in weight. If we add a 2-kilogram lead weight to the center of weight of each object at the same time, the seesaw should still balance. The results are equal.
There is a similar principle in mathematics. We can state it in words like this.
The Addition Principle
If the same number is added to both sides of an equation, the results on each side are equal in value.
We can restate it in symbols this way.
For real numbers a, b, c if a=b thenat+tc=b+ec
Here is an example.
If 3=6/2, then 3+5=6/2+5
Since we added the same amount 5 to both sides, each side has an equal value.
3+5=6/2+5
8 =6/2+10/2
8 = 16/2
8=8
We can use the addition principle to solve an equation.
EXAMPLE 1 Solve for x. x + 16 = 20
x + 16 + (-16) = 20 + (-16) Use the addition principle to add -16 to both sides.
x+0=4 Simplify.
x=4 The value of x is 4.
We have just found the solution of the equation. The solution is a value for the variable that makes the equation true. We then say that the value, 4, in our example, satisfies the equation. We can easily verify that 4 is a solution by substituting this value in the original equation. This step is called checking the solution.
Check. x + 16 = 20
4+16 ≟ 20
20 = 20 ✔
When the same value appears on both sides of the equals sign, we call the equation an identity. Because the two sides of the equation in our check have the same value, we know that the original equation has been correctly solved. We have found the solution.
When you are trying to solve these types of equations, you notice that you must add a particular number to both sides of the equation. What is the number to choose? Look at the number that is on the same side of the equation with x, that is, the number added to x. Then think of the number that is opposite in sign. This is called the additive inverse of the number. The additive inverse of 16 is -16. The additive inverse of -3 is 3. The number to add to both sides of the equation is precisely this additive inverse.
It does not matter which side of the equation contains the variable. The x term may be on the right or left. In the next example the x term will be on the right.
EXAMPLE 2 Solve for x. 14 =x- 3
14+3=x-3 +3 Add 3 to both sides, since 3 is the additive inverse of -3. This will eliminate the -3 on the right and isolate x.
17 =x+0 Simplify.
17=x The value of x is 17.
Check. 14 = x-3
14 ≟ 17-3 Replace x by 17.
14 =14 ✔ Simplify. It checks. The solution is x = 17.
Before you add a number to both sides, you should always simplify the equation. The following example shows how combining numbers by addition separately, on both sides of the equation—simplifies the equation.
EXAMPLE 3 Solve for x. 15 +2=3+x+2
17=x+5 Simplify by adding.
17+ (-5) =x+5+(-5) Add the value -5 to both sides, since -5 is the additive inverse of 5.
12=x Simplify. The value of x is 12.
Check. 15+2 = 3+x+2
15+2 ≟ 3+12+2 Replace x by 12 in the original equation.
17=17 ✔ It checks.
In Example 3 we added -5 to each side. You could subtract 5 from each side and get the same result. In earlier lesson we discussed how subtracting a 5 is the same as adding a negative 5. Do you see why?
We can determine if a value is the solution to an equation by following the same steps used to check an answer. Substitute the value to be tested for the variable in the original equation. We will obtain an identity if the value is the solution.
EXAMPLE 4 Is x = 10 the solution to the equation -15 + 2 = x-3? If it is not, find the solution.
We substitute 10 for x in the equation and see if we obtain an identity.
-15+2=x-3
-15+2=10-3
-13 ≠ 7 This is not true. It is not an identity.
Thus, x = 10 is not the solution. Now we take the original equation and solve to find the solution.
-15+2=x-3
-13=x-3 Simplify by adding.
-13+3=x-3+3 Add 3 to both sides. 3 is the additive inverse of -3.
-10=x
Check to see if x = -10 is the solution. The value x = 10 was incorrect because of a sign error. We must be especially careful to write the correct sign for each number when solving equations.
EXAMPLE 5 Find the value of x that satisfies the equation 1/5+x = −1/10+1/2
To combine the fractions, the fractions must have common denominators. The least common denominator (LCD) of the fractions is 10.
(1*2)/(5*2)+x = −1/10+(1*5)/(2*5) Change each fraction to an equivalent fraction with a denominator of 10.
2/10 + x = −1/10+5/10 This is an equivalent equation.
2/10+x = 4/10 Simplify by adding.
2/10+(-2/10)+x = 4/10+(-2/10) Add the additive inverse of 2/10 to each side
x=2/10 Add the fractions.
x= 1/5 Simplify the answer.
Check. We substitute 1/5 for x in the original equation and see if we obtain an identity.
1/5+x = −1/10+1/2
1/5+1/5 ≟ −1/10+1/2 Substitute 1/5 for x
2/5 ≟ −1/10+1/2
2/5 = 4/10
2/5 = 2/5 ✔ It checks.
The Multiplication Principle
After studying this section, you will be able to:
1. Solve equations of the form 1/ax=b.
2. Solve equations of the form ax = b.
Solving Equations of the Form 1/ax=b
The addition principle allows us to add the same number to both sides of an equation. What would happen if we multiplied each side of an equation by the same number? For example, what would happen if we multiplied each side of an equation by 3?
To answer this question, let’s return to our simple example of the box and the stone on a balanced seesaw. If we triple the number of weights on each side (we are multiplying each side by 3), the seesaw should still balance. The ‘‘weight value’’ of each side remains equal.
In words we can state this principle thus.
Multiplication Principle
If both sides of an equation are multiplied by the same number, the results on each
side are equal in value.
In symbols we can restate the multiplication principle this way.
|
For real numbers a,b,c wihc #0 ifa@=b thenca=cb |
Let us look at an equation where it would be helpful to multiply each side of the equation by 3.
EXAMPLE 1 Solve for x. 1/3x=-15
We know that (3)(1/3) = 1. We will multiply each side of the equation by 3, because we want to isolate the variable x.
(3)(1/3x)=3(-15) Multiply each side of the equation by 3 since (3)(1/3) = 1.
(3/1)(1/3)(x)=-45
1x=-45 Simplify.
x= -45
Check. 1/3(-45) ≟ -15 Substitute -45 for x in the original equation.
-15=-15 ✔ It checks.
Note that 1/5x can be written as x/5. To solve the equation x/5=3, we could multiply each side of the equation by 5. Try it. Then check your solution.
Solving Equations of the Form ax = b
We can see that using the multiplication principle to multiply each side of an equation by 1/2 is the same as dividing each side of the equation by 2. Thus, it would seem that the multiplication principle would allow us to divide each side of the equation by any nonzero real number. Is there a real-world example of this idea?
Let’s return to our simple example of the box and the stone on a balanced seesaw. Suppose that we were to cut the two objects in half (so that the amount of weight of each was divided by 2). We then return the objects to the same places on the seesaw. The seesaw would still balance. The ‘‘weight value’’ of each side remains equal.
In words we can state this principle thus:
Division Principle
If both sides of an equation are divided by the same nonzero number, the results on
each side are equal in value.
Note: We put a restriction on the number by which we are dividing. We cannot divide by zero. We say that expressions like 2/0 are not defined. Thus we restrict our divisor to nonzero numbers. We can restate the division principle this way.
a b
For real numbers a, b, c where c ~ 0 ifa=b then — = —
coc
EXAMPLE 2 Solve for x. 5x = 125
(5x)/5=125/5 Divide both sides by 5.
x = 25 Simplify. The solution is 25.
Check. 5x = 125
5(25) ≟ 125 Replace x by 25.
125 = 125 ✔ It checks.
For equations of the form ax = b (a number multiplied by x equals another number), we solve the equation by choosing to divide both sides by a particular number. What is the number to choose? We look at the side of the equation that contains x. We notice the number that is multiplied by x. We divide by that number. The division principle tells us that we can still have a true equation provided that we divide by that number on both sides of the equation.
The solution to an equation may be a proper fraction or an improper fraction.
EXAMPLE 3 Solve for x. 4x = 38
(4x)/4= 38/4 Divide both sides by 4.
x=19/2 Simplify. The solution is 19/2.
If you leave the solution as a fraction, it will be easier to check that solution in the original equation.
Check: 4x = 38 Replace x by 19/2.
4(19/2) ≟ 38
38 = 38 ✔ It checks.
In examples 2 and 3 we divided by the number multiplied by x (the coefficient of x). This procedure is followed regardless of whether the sign of that number is positive or negative.
EXAMPLE 4 Solve for x. -3x = 48
(-3x)/-3=48/-3 Divide both sides by -3.
x=-16 The solution is -16.
The coefficient of x may be 1 or -1. You may have to rewrite the equation so that the coefficient of 1 or -1 is obvious. With practice you may be able to “see’’ the coefficient without actually rewriting the equation.
EXAMPLE 5 Solve for x. -x = -24
-1x = -24 Rewrite the equation. -1x is the same as -x. Now the coefficient of -1 is obvious.
(-1x)/-1=-24/-1 Divide both sides by -1
x= 24 The solution is 24.
Use the Addition and Multiplication Principles Together
After studying this section, you will be able to:
1. Solve equations of the form ax + b =c.
2. Solve equations in which the variable appears on both sides of the equation.
3. Solve equations with parentheses.
Solving Equations of the Form ax +b=c
To solve many equations, we must use both the addition principle and the multiplication principle.
EXAMPLE 1 Solve for x and check your solution. 5x +3 = 18
5x + 3 + (-3)= 18+ (-3) Add -3 to both sides, using the addition principle.
5x = 15 Simplify.
(5x)/5=15/5 Divide both sides by 5, using the division principle.
x=3 The solution is 3.
Check. 5(3)+3 ≟ 18
Check. 15+3 ≟ 18
Check. 18=18 ✔ It checks.
A Variable on Both Sides of the Equation
In some cases the variable appears on both sides of the equation. We would like to rewrite the equation so that all the terms containing the variable appear on one side. To do this, we apply the addition principle to the variable term.
EXAMPLE 2 Solve for x. 9x = 6x + 15
9x + (-6x) = 6x + (-6x) + 15 Add -6x to both sides. Notice 6x + (-6x) eliminates the variable on the right side.
3x = 15 Collect like terms.
(3x)/3=15/3 Divide both sides by 3.
x=5 The solution is 5.
Many problems have variable terms and constant terms on both sides of the equation. You will want to get all the variable terms on one side and all the constant terms on the other side.
EXAMPLE 3 Solve for x and check your solution. 9x + 3 = 7x -2.
9x + (-7x) + 3 = 7x + (-7x) - 2 Add -7x to both sides of the equation.
2x+3=-2 Combine like terms.
2x + 3+ (-3) = -2 + (-3) Add -3 to both sides.
2x = -5 Simplify.
(2x)/2=-5/2 Divide both sides by 2.
x = −5/2 The solution is −5/2.
Check. 9x + 3 = 7x -2
Check. 9(-5/2)+3 ≟ 7(-5/2)-2 Replace x by −5/2.
Check. −45/2+3 ≟ −35/2-2 Simplify.
Check. −45/2+6/2 ≟ −35/2-4/2 Change to equivalent fractions with a common denominator.
Check. −39/2 = −39/2 ✔ It checks. x = −5/2 is the solution.
In our next example we will study equations that need simplifying before any other steps are taken. Where it is possible, you should first collect like terms on one or both sides of the equation. The variable terms can be collected on the right side or the left side. In this example we will collect all the x terms on the right side.
EXAMPLE 4 Solve for x. 5x + 26 -6 = 9x + 12x
5x + 20 = 21x Combine like terms.
5x + (-5x) + 20 = 21x + (-5x) Add -5x to both sides.
20 = 16x Combine like terms.
20/16 =(16x)/16 Divide both sides by 16
5/4=x (Don’t forget to reduce the resulting fraction.)
All the equations we have been studying so far are called first-degree equations.
This means the variable terms are not squared (such as x^2 or y^2) or some higher power. It is possible to solve equations with x^2 and y^2 terms by the same methods we have used so far. If x^2 or y^2 terms appear, try to collect them on one side of the equation. If the squared term drops out, you may solve it as a first-degree equation using the methods discussed in this section.
EXAMPLE 5 Solve for y. 5y^2 + 6y -2 = -y + 5y^2 + 12
5y^2-5y^2 + 6y -2 = -y + 5y^2-5y^2 + 12 Subtract 5y^2 from both sides.
6y -2=-y+12 Combine, since 5y^2-5y^2 = 0.
6y+y-2=-y+y+12 Add y to each side.
7y-2= 12 Simplify.
7y-2+2=12+2 Add 2 to each side.
7y=14 Simplify.
(7y)/7 = 14/7 Divide each side by 7.
y=2 Simplify. The solution is2.
Solving Equations with Parentheses
The equations that you just solved are simpler versions of equations that we will now discuss. These equations contain parentheses. If the parentheses are first removed, the problems then become just like those encountered previously. We use the distributive property to remove the parentheses.
EXAMPLE 6 Solve for x and check your solution. 4(x + 1)- 3(x-3) = 25
4(x + 1)- 3(x-3) = 25
4x +4-3x+9 = 25 Multiply by 4 and -3 to remove parentheses. Be careful of the signs. Remember (-3)(-3) = 9.
After removing the parentheses, it is important to collect like terms on each side of the equation. Do this before going on to isolate the variable.
x + 13 = 25 Collect like terms.
x+ 13-13 = 25-13 Add -13 to both sides to isolate the variable.
x = 12 The solution is 12.
Check. 4(12+1)-3(12-3) ≟ 25 Replace x by 12.
4(13)-3(9) ≟ 25 Combine numbers inside parentheses.
52-27 ≟ 25 Multiply.
25=25 ✔ Simplify. It checks.
In problems that involve decimals, great care should be taken. In some steps you will be multiplying decimal quantities, and in other steps you will be adding them.
EXAMPLE 7 Solve for x. 0.3(1.2x-3.6) = 4.2x-16.44
0.36x-1.08 = 4.2x -16.44 Remove parentheses.
0.36x-0.36x-1.08 = 4.2x-0.36x-16.44 Subtract 0.36x from both sides.
-1.08 = 3.84x -16.44 Collect like terms.
-1.08 + 16.44 = 3.84x-16.44 + 16.44 Add 16.44 to both sides.
15.36 = 3.84x Simplify.
15.36/3.84=(3.84x)/3.84 Divide both sides by 3.84.
4=x The solution is x = 4.
EXAMPLE 8 Solve for z and check. 2(3z-5) + 2 = 4z -3(2z + 8)
6z-10 + 2 = 4z-6z-24 Remove parentheses.
6z- 8 = -2z-24 Collect like terms.
6z-8 + 2z = -2z + 2z-24 Add 2z to each side.
8z-8 = -24 Simplify.
8z-8 +8 = -24+ 8 Add 8 to each side.
8z =-16 Simplify.
(8z)/8=-16/8 Divide each side by 8.
z=-2 Simplify. The solution is -2.
Check. 2[3(-2)-5] +2 ≟ 4(-2) -3[2(-2) + 8] Replace z by -2.
2[-6-5] +2 ≟ -8 -3[-4 + 8] Multiply.
2[-11] +2 ≟ -8 -3[4] Simplify.
-22 +2 ≟ -8 -12
-20 = -20 ✔ It checks.
Equations with Fractions
After studying this section, you will be able to:
1. Solve equations with fractions.
Solving Equations with Fractions
Equations with fractions can be rather difficult to solve. This difficulty is simply due to the extra care we usually have to use when computing with fractions. The actual equation solving procedures are the same, with fractions or without. To avoid unnecessary work, we transform the given equation with fractions to an equivalent equation that does not contain fractions. How do we do this? We multiply each side of the equation by the lowest common denominator of all the fractions contained in the equation. We then use the distributive property so that the LCD is multiplied by each term of the equation.
EXAMPLE 1 Solve for x. 1/4x-2/3=5/12x
First we find that the LCD = 12.
12(1/4x-2/3)=12(5/12x) Multiply each side by 12
(12/1)(1/4)(x)-(12/1)(2/3)=(12/1)(5/12)(x) Use the distributive property.
3x-8 = 5x Simplify.
3x + (-3x)-8 = 5x + (-3x) Add -3x to each side.
-8 = 2x Simplify.
−8/2=(2x)/2 Divide each side by 2.
-4= x Simplify.
Check. 1/4(-4)-2/3 ≟ 5/12(-4)
-1-2/3 ≟ −5/3
−3/3-2/3 ≟ −5/3
−5/3 = −5/3 ✔ It checks.
In Example 1 we multiplied each side of the equation by the LCD. It is common practice to immediately go to the second Step and multiply each term by the LCD, rather
EXAMPLE 2 Solve for x. (x+5)/7=x/4+1/2
x/7+5/7=x/4+1/2 First we write as separate fractions
(28)(x/7)+(28)(5/7)=(28)(x/4)+(28)(1/2) We observe that the LCD is 28, so we multiply each term by 28.
4x + 20 = 7x + 14 Simplify.
4x-4x + 20 = 7x-4x + 14 Add -4x to both sides.
20 = 3x + 14 Collect like terms.
20-14=3x + 14- 14 Add -14 to both sides.
6 = 3x Collect like terms.
6/3=(3x)/3 Divide both sides by 3.
2=x The solution is x = 2.
If a problem contains both parentheses and fractions, it is best to remove the parentheses first. Many students find it is helpful to have a written procedure to follow in solving these more involved equations.
Procedure to Solve Linear Equations
1. Remove any parentheses.
2. If fractions exist, multiply all terms on both sides by the lowest common denominator of all the fractions.
3. Collect like terms if possible. Simplify numerical work if possible.
4. Add or subtract terms on both sides of the equation to get all terms with the variable on one side of the equation.
5. Add or subtract a value on both sides of the equation to get all terms not containing the variable on the other side of the equation.
6. Divide both sides of the equation by the coefficient of the variable.
7. Simplify the solution (if possible).
8. Check your solution.
Let’s use each step in solving this example.
EXAMPLE 3 Solve for x and check your solution. 1/3(x-2)= 1/5(x+4)+2
Step 1 x/3-2/3=x/5+4/5+2 Remove parentheses.
Step 2 15(x/3)-15(2/3) = 15(x/5) +15(4/5) +15(2) Multiply by the LCD, 15.
Step 3 5x-10 = 3x + 12 + 30 Simplify.
5x-10 = 3x + 42 Simplify.
Step 4 5x-3x-10 = 3x-3x + 42 Add -3x to both sides.
2x-10 = 42 Simplify.
Step 5 2x-10+ 10 = 42+ 10 Add 10 to both sides.
2x = 52 Simplify.
Step 6 (2x)/2=52/2 Divide both sides by 2.
Step 7 x = 26 Simplify the solution.
Step 8 Check. 1/3(26-2) ≟ 1/5(26 +4)+2 Replace x by 26.
1/3(24) ≟ 1/5(30)+2 Combine values within parentheses.
8 ≟ 6+2 Simplify.
8 = 8 ✔ x=26 is the solution.
It should be remembered that not every step will be needed in each problem. You can combine some steps as well, as long as you are consistently obtaining the correct solution. However, you are encouraged to write out every step as a way of helping you to avoid careless errors.
It is important to remember that when we write decimals these numbers are really fractions written in a special way. Thus, 0.3 = 7 and 0.07 = 745. It is possible to take a linear equation containing decimals and to multiply each term by the appropriate value to obtain integer coefficients.
Formulas
After studying this section, you will be able to:
1. Solve formulas for a specified variable.
Solving for a Specified Variable in a Formula
Formulas are equations with one or more variables that are used to describe real world situations. The formula describes the relationship that exists among the variables. For example, in the formula d = rt, distance (d) is related to the rate of speed (r) and to time (t). We can use this formula to find distance if we know the rate and time. Sometimes, however, we are given the distance and the rate, and we are asked to find the time.
EXAMPLE 1 Joseph drove a distance of 156 miles at an average speed of 52 miles per hour. How long did it take Joseph to make the trip?
d= rt Use the distance formula.
156 = 52t Substitute the known values for the variables.
156/52=52/52t Divide both sides of the equation by 52 to solve for t.
3=t We have solved for t.
It took Joseph 3 hours to drive 156 miles at 52 miles per hour.
If we have many problems that ask us to find the time given the distance and rate, it may be worthwhile to rewrite the formula in terms of time.
EXAMPLE 2 Solve for t. d=rt
d/r=(rt)/r We want to isolate t. Therefore, we are dividing both sides of the equation by the coefficient of t, which is r.
d/r=t You have solved for the variable indicated.
A simple first degree equation with two variables can be thought of as the equation of a line. It is often useful to solve for y in order to make graphing the line easier.
EXAMPLE 3 Solve for y. 3x-2y = 6
-2y = 6-3x We want to isolate the term containing y, so we subtract 3x from both sides.
(-2y)/(-2)= (6-3x)/(-2) Divide both sides by the coefficient of y.
y=6/-2+(-3x)/-2 Rewrite the fraction.
y= 3/2x-3 Simplify and regroup.
This is known as the slope-intercept form of the equation of a line.
Our procedure for solving a first-degree equation can be rewritten to give us a procedure for solving a formula for a specified variable.
Procedure to Solve a Formula for a Specified Variable
1. Remove any parentheses.
2. If fractions exist, multiply all terms on both sides by the LCD of all the fractions.
3. Collect like terms or simplify if possible.
4. Add or subtract terms on both sides of the equation to get all terms with the desired variable on one side of the equation.
5. Add or subtract the appropriate quantity to get all terms that do not have the desired variable on the other side of the equation.
6. Divide both sides of the equation by the coefficient of the desired variable.
7. Simplify if possible.
EXAMPLE 4 A trapezoid is a four-sided figure with two parallel sides. If the parallel sides are a and b and the altitude is h, the area is given by
A=h/2(a+b)
Solve this equation for a.
A=h/2(a+b)
A=(ha)/2+(hb)/2 Remove the parentheses.
2(A) = 2((ha)/2)+2((hb)/2) Multiply all terms by LCD of 2.
2A = ha + hb Simplify.
2A-hb = ha We want to isolate the term containing a. Therefore, we subtract hb from both sides.
(2A-hb)/h= (ha)/h Divide both sides by h (the coefficient of a).
(2A-hb)/h=a The solution is obtained.
Note: Although the solution is in simple form, it could be written in an alternative way. Since
(2A-hb)/h=(2A)/h-(hb)/h=(2A)/h-b
we could have (2A)/h-b = a as an alternative way of writing the answer.
Write and Graph Inequalities
After studying this section, you will be able to:
1. Interpret an inequality statement.
2. Graph an inequality on a number line.
Inequality Statements
We frequently speak of one value being greater than or less than another value. We say that ‘‘5 is less than 7’’ or ‘‘9 is greater than 4.’’ These relationships are called inequalities. We can write inequalities in mathematics by using symbols. We use the symbol < to represent the words ‘‘is less than.’’ We use the symbol > to represent the words ‘‘is greater than.’’
Statement in Words Statement in Algebra
5 is less than 7. 5<7
9 is greater than 4. 9>4
Note. ‘‘5 is less than 7’’ and ‘‘7 is greater than 5’’ have the same meaning. Similarly, 5 <7 and 7 > 5 have the same meaning. They represent two equivalent ways of describing the same relationship between the two numbers 5 and 7.
We can illustrate the concept of inequality graphically if we examine a number line.
+++ +—_ +++ +_+_+—_—_+_¢_ _ +>
-5 -4 -3 -—2 -] 0 I 2 3 4 5 6 7 8
We say that one number is greater than another if it is to the right of the other on the number line. Thus 7 > 5, since 7 is to the right of 5.
What about negative numbers? We can say ‘‘-1 is greater than -3” and write it in symbols -1 > -3 because we know that -1 lies to the right of -3 on the number line.
EXAMPLE 1 Replace the question mark with the symbol < or > in each statement.
(a) 3 ? -1 (b) -2 ? 1 (c) -3 ? -4 (d) 0 ? 3
(a) 3>-1 Use >, since 3 is to the right of -1.
(b) -2< 1 Use <, since -2 is to the left of 1. (Or equivalently, we could say that 1 is to the right of -2.)
(c) -3 > -4 Since -3 is to the right of -4.
(d) 0<3
Graphing an Inequality on a Number Line
Sometimes we will use an inequality to express the relationship between a variable and a number. x > 3 means that x could have the value of any number greater than 3. This can be pictured on the number line in a graph as follows:
-5 -4 -3 -2 -1 0 l 2 3 4 5
Note that the open circle at 3 suggests that we do not include the point for the number 3.
Similarly, we could represent graphically x < -2 as follows:
$$ —} fj —_ + —_;_+_+_+_ + > x
-5 -4 -3 -2 -l 0 I 2 3 #4 = =# §
Sometimes a variable will be either greater than or equal to a certain number. In the statement “x is greater than or equal to 3,’’ we are implying that x could have the value of 3 or any number greater than 3. We write this as x >= 3. We represent it graphically as follows:
—+—. >! 4H_{_AH_ ee ee et
—2 -1 0 l 2 3 4 5 6 7
Note that the closed circle at 3 suggests that we do include the point for the number 3.
Similarly, we could represent graphically x <= -2 as follows:
—— t+. et Ht HH HH
-5 -4 -3 -2 -1 0 1 2 3 4 5
EXAMPLE 2 State each mathematical relationship in words and then illustrate it graphically.
(a) x< -2 (b) -3 <x (c) x<= -6
(a) We state that ‘‘x is less than -2.”
x<-20 ——+-+—O—_1+—_+—_+—_++—
-4 -3 -2 -!1 0 ] 2
(b) We can state that ‘‘-3 is less than x’’ or, an equivalent statement, ‘‘x is greater than -3.’’ Be sure you see that -3 < x is equivalent to x > -3. Although both ways are correct, we usually write the variable first in a simple linear inequality containing a variable and a numerical value.
(c) We state that ‘‘x is less than or equal to -6.”
_-—od—_tH¥!_t—_t*—_+—_ +t +
-7 -6 -5 -4 -3 -2 -1 0 l 2
There are many everyday situations involving an unknown value and an inequality. We can translate these situations into algebraic statements. This is the first step in solving word problems using inequalities.
EXAMPLE 3 Translate each English statement into an algebraic statement.
(a) The police on the scene said that the car was traveling greater than 80 miles per hour (use the variable s for speed).
(b) The owner of the trucking company said that the payload of a truck must never exceed 4500 pounds (use the variable p for payload).
(a) Since the speed must be greater than 80 we have s > 80.
(b) If the payload of the truck can never exceed 4500 pounds, then the payload must be always less than or equal to 4500 pounds. Thus we write p <= 4500.
Solve Inequalities
After studying this section, you will be able to:
1. Solve an inequality.
Solving Inequalities
The possible values that make an inequality true are called its solutions. Thus, when we solve an inequality, we are finding all the values that make it true. To solve an inequality, we simplify it to the point where we can clearly see the possible values for the variable. We’ve solved equations by adding, subtracting, multiplying, and dividing a particular value on both sides of the equation. Here we do similar operations with inequalities, with one important exception. We’ll show some examples so that you can see the operations we can do with inequalities just as with equations.
We will first examine the pattern that takes place when we perform a given operation on both sides of an inequality.
EXAMPLE 1
Original Inequality New Inequality
(a) 3<5 → Multiply both sides by 2 → 6< 10
(b) -2<-1 → Add -3 to both sides. → -5<-4
(c) 0>-4 → Divide both sides by 2. → 0>-2
(d) 8 >4 → Subtract 6 from both sides. → 2>-2
Note that we avoided multiplying or dividing by a negative number !
Now let us examine what would happen if we did multiply or divide by a negative number. We start with an original, true inequality. We want to get a new, also true inequality.
Original Inequality New Inequality
3<5 → Multiply by -2. → 6 ? -10
What is the correct inequality sign? Since -6 is to the right of -10, we know the new inequality should be -6 > -10, if we wish the statement to remain true. Notice how we reverse the direction of the inequality from < (less than) to > (greater than). We would thus obtain the new inequality -6 > -10. Thus
3<5 → Multiply by -2. → -6>-10
The < sign we started with (3 < 5) is reversed to > (-6 >-10). A similar reversal takes place in the following example.
EXAMPLE 2
Original Inequality New Inequality
(a) -2<-1 → Multiply by -3. → 6>3
(b) 0>-4 → Divide both sides by -2. → 0<2
(c) 8 >4 → Divide both sides by -4. → -2<-1
Notice that we do the arithmetic with signed numbers just as we always do. But the new inequality has its inequality sign reversed (from that of the original inequality). Whenever both sides of an inequality are multiplied or divided by a negative quantity, the direction of the inequality is reversed.
Procedure for Solving Inequalities
| You may use the same procedures to solve inequalities that you did to solve equa-
tions except that the direction of an inequality is reversed if you multiply or divide
both sides by a negative number.
EXAMPLE 3 Solve and graph 3x + 7 >= 13.
3x +7-7>=13-7 Subtract 7 from both sides.
3x>= 6 Simplify.
(3x)/3>=6/3 Divide both sides by 3.
x>=2 Simplify. Note the direction of the inequality is not changed, since we have divided by a positive number.
The graphical representation is en ee
—2 -Il 0 I 2 3 4
EXAMPLE 4 Solve and graph 5- 3x > 7.
5-5-3x>7-5 Subtract 5 from both sides.
-3x>2 Simplify.
(-3x)/-3<2/-3 Divide by -3, and reverse the inequality, since both sides are divided by negative 3.
x< -2/3 Note the direction of the inequality.
The graphical representation is Ht Ht
-1_2_1 0 l
3 3
Just like equations, some inequalities contain parentheses and fractions. The initial steps to solve these inequalities will be the same as those used to solve equations with parentheses and fractions. When the variable appears on both sides of the inequality, it is advisable to collect the x terms on the left side of the inequality symbol.
EXAMPLE 5 Solve and graph −(13x)/2<=x/2-15/8
(8)((-13x)/2)<=(8)(x/2)-(8)(15/8) Multiply all terms by LCD = 8. We do not reverse the direction of the inequality symbol since we are multiplying by a positive number, 8.
-52x <= 4x-5 Simplify.
-52x-4x <= 4x-15-4x Add -4x to both sides.
-56x <= -15 Combine like terms.
(-56x)/56>= -15/-56 Divide both sides by -56. We reverse the direction of the inequality when we divide both sides by a negative number.
x>=15/56
The graphical representation is 0 15 28 l
56 56
. ]
The most common error students make in solving inequalities is forgetting to reverse the direction of the inequality symbol when multiplying or dividing by a negative number.