In this chapter, we will investigate matrices and their inverses, and various ways to use matrices to solve systems of equations. First, however, we will study systems of equations on their own: linear and nonlinear, and then partial fractions.
 9.0: Prelude to Systems of Equations and Inequalities
 In this chapter, we will investigate matrices and their inverses, and various ways to use matrices to solve systems of equations. First, however, we will study systems of equations on their own: linear and nonlinear, and then partial fractions. We will not be breaking any secret codes here, but we will lay the foundation for future courses.
 9.1: Systems of Linear Equations: Two Variables
 A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
 9.2: Systems of Linear Equations: Three Variables
 https://math.libretexts.org/TextMaps/Algebra_Textmaps/Map%3A_Elementary_Algebra_(OpenStax)/11%3A_Systems_of_Equations_and_Inequalities/11.3%3A_Systems_of_Linear_Equations%3A_Three_Variables
 9.3: Systems of Nonlinear Equations and Inequalities  Two Variables
 In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations.
 9.4: Partial Fractions
 Decompose a ratio of polynomials by writing the partial fractions. Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. The decomposition with repeated linear factors must account for the factors of the denominator in increasing powers. The decomposition with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor.
 9.5: Matrices and Matrix Operations
 To solve a systems of equations, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters.
 9.6: Solving Systems with Gaussian Elimination
 A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix.
 9.7: Solving Systems with Inverses
 A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as reversibility is a requirement. Not all square matrices have an inverse. We will look at two methods for finding the inverse of a 2×2 matrix and a third method that can be used on both 2×2 and 3×3 matrices.
 9.8: Solving Systems with Cramer's Rule
 In this section, we will study two more strategies for solving systems of equations. A determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a square matrix to determine whether there is a solution to the system of equations. Cramer’s Rule to solve a system of equations in two & three variables.
 9.E: Systems of Equations and Inequalities (Exercises)
9: Systems of Equations and Inequalities  Mathematics
Solve the system of two linear equations and check the solution :
Solve the system of two linear equations with variables in numerator and denominator, check the solution and determine the conditions of solvability :
Solve the system of three linear equations and check the solution :
Solve the system of four linear equations and check the solution :
Solve the system of linear and quadratic equation :
Solve the system of linear inequalities with one variable :
Solve the system of linear inequalities with two variables :
9: Systems of Equations and Inequalities  Mathematics
In this chapter we will look at one of the standard topics in any Algebra class. The ability to solve equations and/or inequalities is very important and is used time and again both in this class and in later classes. We will cover a wide variety of solving topics in this chapter that should cover most of the basic equations/inequalities/techniques that are involved in solving.
Here is a brief listing of the material covered in this chapter.
Solutions and Solution Sets – In this section we introduce some of the basic notation and ideas involved in solving equations and inequalities. We define solutions for equations and inequalities and solution sets.
Linear Equations – In this section we give a process for solving linear equations, including equations with rational expressions, and we illustrate the process with several examples. In addition, we discuss a subtlety involved in solving equations that students often overlook.
Applications of Linear Equations – In this section we discuss a process for solving applications in general although we will focus only on linear equations here. We will work applications in pricing, distance/rate problems, work rate problems and mixing problems.
Equations With More Than One Variable – In this section we will look at solving equations with more than one variable in them. These equations will have multiple variables in them and we will be asked to solve the equation for one of the variables. This is something that we will be asked to do on a fairly regular basis.
Quadratic Equations, Part I – In this section we will start looking at solving quadratic equations. Specifically, we will concentrate on solving quadratic equations by factoring and the square root property in this section.
Quadratic Equations, Part II – In this section we will continue solving quadratic equations. We will use completing the square to solve quadratic equations in this section and use that to derive the quadratic formula. The quadratic formula is a quick way that will allow us to quickly solve any quadratic equation.
Quadratic Equations : A Summary – In this section we will summarize the topics from the last two sections. We will give a procedure for determining which method to use in solving quadratic equations and we will define the discriminant which will allow us to quickly determine what kind of solutions we will get from solving a quadratic equation.
Applications of Quadratic Equations – In this section we will revisit some of the applications we saw in the linear application section, only this time they will involve solving a quadratic equation. Included are examples in distance/rate problems and work rate problems.
Equations Reducible to Quadratic Form – Not all equations are in what we generally consider quadratic equations. However, some equations, with a proper substitution can be turned into a quadratic equation. These types of equations are called quadratic in form. In this section we will solve this type of equation.
Equations with Radicals – In this section we will discuss how to solve equations with square roots in them. As we will see we will need to be very careful with the potential solutions we get as the process used in solving these equations can lead to values that are not, in fact, solutions to the equation.
Linear Inequalities – In this section we will start solving inequalities. We will concentrate on solving linear inequalities in this section (both single and double inequalities). We will also introduce interval notation.
Polynomial Inequalities – In this section we will continue solving inequalities. However, in this section we move away from linear inequalities and move on to solving inequalities that involve polynomials of degree at least 2.
Rational Inequalities – We continue solving inequalities in this section. We now will solve inequalities that involve rational expressions, although as we’ll see the process here is pretty much identical to the process used when solving inequalities with polynomials.
Absolute Value Equations – In this section we will give a geometric as well as a mathematical definition of absolute value. We will then proceed to solve equations that involve an absolute value. We will also work an example that involved two absolute values.
Systems of linear inequalities
A system of linear inequalities in two variables consists of at least two linear inequalities in the same variables. The solution of a linear inequality is the ordered pair that is a solution to all inequalities in the system and the graph of the linear inequality is the graph of all solutions of the system.
Graph the system of inequalities
Graph one line at the time in the same coordinate plane and shade the halfplane that satisfies the inequality.
The solution region which is the intersection of the halfplanes is shown in a darker shade
Usually only the solution region is shaded which makes it easier to see which region is the solution region
Equations and inequalities.
An equation is a statement of equality between two expressions, often viewed as a question asking for which values of the variables the expressions on either side are in fact equal. These values are the solutions to the equation. An identity, in contrast, is true for all values of the variables identities are often developed by rewriting an expression in an equivalent form.
The solutions of an equation in one variable form a set of numbers the solutions of an equation in two variables form a set of ordered pairs of numbers, which can be plotted in the coordinate plane. Two or more equations and/or inequalities form a system. A solution for such a system must satisfy every equation and inequality in the system.
An equation can often be solved by successively deducing from it one or more simpler equations. For example, one can add the same constant to both sides without changing the solutions, but squaring both sides might lead to extraneous solutions. Strategic competence in solving includes looking ahead for productive manipulations and anticipating the nature and number of solutions.
Some equations have no solutions in a given number system, but have a solution in a larger system. For example, the solution of x + 1 = 0 is an integer, not a whole number the solution of 2x + 1 = 0 is a rational number, not an integer the solutions of x 2  2 = 0 are real numbers, not rational numbers and the solutions of x 2 + 2 = 0 are complex numbers, not real numbers.
The same solution techniques used to solve equations can be used to rearrange formulas. For example, the formula for the area of a trapezoid, A = ((b_{1}+b_{2})/2)h, can be solved for h using the same deductive process. Inequalities can be solved by reasoning about the properties of inequality. Many, but not all, of the properties of equality continue to hold for inequalities and can be useful in solving them.
Connections to Functions and Modeling. Expressions can define functions, and equivalent expressions define the same function. Asking when two functions have the same value for the same input leads to an equation graphing the two functions allows for finding approximate solutions of the equation. Converting a verbal description to an equation, inequality, or system of these is an essential skill in modeling.
Solution of Equations and Inequalities
The solution of equations and inequalities involves finding the value (or values) of the variable(s) that make the mathematical statements true. The addition and multiplication properties of equations and inequalities, as well as the properties of real numbers, are used to simplify the equation/inequality as much as possible, prior to formulating the solution set for the variable in question.
Solving linear equations in one variable, i.e., equations dealing with algebraic expressions having one variable with an exponent of one, are the simplest to solve for.
EX.
2x + 3 = 7
2x + 3 + (3) = 7 + (3)
2x = 4
(0.5)2x = (0.5) 4
x = 2
EX.
4x + 5 = 15
4x + 5 + (5) = 15 + (5)
4x = 20
¼(4x) = ¼(20)
x = 5
When the equation involves fractions, it is helpful to multiply both sides of the equation by the least common denominator of all fractions:
where the LCD (lowest common denominator) of 4, 5 and 8 is 40
30x + 32 = 35
30x + 32  32 = 35  32
30x = 3
x = 3 / 30 = 1/10
When the equation involves decimals, it is helpful to multiply both sides of the equation by the lowest power of 10 that converts all decimal numbers to integers:
EX.
2.5x  3.2 = 12.5
2.5x  3.2 + 3.2 = 12.5 + 3.2
2.5x = 15.7
By multiplying each side by 10, we get
The process of solving inequalities is similar to that of solving equations, except that different additive and multiplicative properties apply.
EX.
2x  3 > 5
2x  3 + 3 > 5 + 3
2x > 8
½ (2x) > ½ (8)
x > 4
The solution set is < x  x > 4>
EX.
3x  4 < 8
3x  4 + 4 < 8 + 4
3x < 12
1/3 (3x) > 1/3 (12)
(note the change in the inequality from < to >)
x > 4
EX.
1/3 x 3/4 < 1/2
1/3 x (12) 3/4 (12) < 1/2 (12)
where the LCD of 2, 3 and 4 is 12
4x  9 < 6
4x  9 + 9 < 6 + 9
4x < 15
x < 15/4
If the equation or inequality involves absolute values, then its solution will involve solving two equations or inequalities.
(1)  ax + b  = k is equivalent to ax + b = k or ax + b = k.
EX.
2x + 1 = 5
2x + 1 = 5 or 2x + 1 = 5
2x = 4
2x = 6
x = 2 or x = 3
the solution set is <2 , 3>.
(2)  ax + b  < k is equal to k < ax + b < k, where k > 0.
the solution set is the interval (3, 5/3).
(3)  ax + b  > k is equal to ax + b > k or ax + b < k, where k > 0.
EX.  2x  1  > 3
2x  1 > 3 or 2x  1 < 3
2x > 4 or 2x < 2
x > 2 or x < 1
Systems of Equations and Inequalities
It's raining (systems of) cats and dogs! The fifth unit in a ninepart course presents systems of equations and inequalities within the context of pets. Scholars use systems of inequalities to represent constraints within situations and then create and solve world problems leading to systems of equations.
Concepts
Additional Tags
Instructional Ideas
 Review graphing linear equations
 Make a connection between the graphs of one and twovariable inequalities
Classroom Considerations
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Worked example 14: Nature of roots
For which value(s) of (k) will the roots of (6x^2 + 6 = 4kx) be real and equal?
Interpret the question
For roots to be real and equal, we need to solve for the value(s) of (k) such that (Delta =0).
Check that the equation is in standard form (ax^2+bx+c=0)
Identify the coefficients to substitute into the formula for the discriminant
[a = 6 qquad b = 4k qquad c = 6]
Write down the formula and substitute values
For roots to be real and equal, (Delta =0).
Check both answers by substituting back into the original equation
For (k = 3): egin
We see that for (k = 3) the quadratic equation has real, equal roots (x = 1).
For (k =  3): egin
We see that for (k = 3) the quadratic equation has real, equal roots (x = 1).
Write the final answer
For the roots of the quadratic equation to be real and equal, (k = 3) or (k = 3). ()
Cooperative Activity
In this cooperative activity, I use a map off of this website with my instruction sheet. I changed some of the inequalities being used in the activity, and the questions for students to complete. I modified the worksheet to just use the Treasure map only. I give each pair of students or group of 3 if necessary, a treasure map. I give every student Instructions for the map activity, and state that my expectation is for every student to show the work and turn in the answered questions from the instruction sheet. Every pair or group is to put all members names on the map and complete solving the inequalities by graphing with colored pencils on the treasure map. One student example of the treasure activity is available in the resources of this lesson.
In the cooperative activity, all of the inequalities have greater than or less than results, therefore none of the points on the lines are solutions. Some of my students used markers, but the intersection was difficult to determine, so colored pencils are recommended.
Some common mistakes during this cooperative activity are the following:
 The slope is graphed incorrectly.
 The y intercept is graphed incorrectly.
 The shading is on the wrong side of the line.
 The algebra is done incorrectly when using inverse operations to solve for y, or the direction of the sign is not changed when multiplying or dividing by a negative.
 The math is done incorrectly to check the point as a solution.
Students will critique the findings of other students and exchange reasoning in order to form a consensus about the correct location of the treasure in the Peer Feedback. Students will join other groups and provide feedback and make corrections based on the skills listed above. I will assist students that are unable to agree, and use questioning to help them continue moving forward in their productive struggle.
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9: Systems of Equations and Inequalities  Mathematics
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.
&emsp&emspOne 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.
&emsp&emspThere is a similar principle in mathematics. We can state it in words like this.
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
Since we added the same amount 5 to both sides, each side has an equal value.
We can use the addition principle to solve an equation.
EXAMPLE 1 Solve for x . &emsp&emsp x + 16 = 20
x + 16 + (16) = 20 + (16) &emsp&emsp Use the addition principle to add 16 to both sides.
&emsp&emspWe 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. &emsp&emsp x + 16 = 20
&emsp&emspWhen 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.
&emsp&emspWhen 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.
&emsp&emspIt 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 . &emsp&emsp 14 =x 3
14+3=x3 +3 &emsp&emsp Add 3 to both sides, since 3 is the additive inverse of 3 . This will eliminate the 3 on the right and isolate x .
Check. &emsp&emsp 14 = x3
&emsp&emspBefore 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&mdashsimplifies the equation.
EXAMPLE 3 Solve for x . &emsp&emsp 15 +2=3+x+2
17+ (5) =x+5+(5) &emsp&emsp Add the value 5 to both sides, since 5 is the additive inverse of 5 .
12=x &emsp&emsp Simplify. The value of x is 12 .
Check. &emsp&emsp 15+2 = 3+x+2
&emsp&emsp &emsp&emsp 15+2 ≟ 3+12+2 &emsp&emspReplace x by 12 in the original equation.
&emsp&emspIn 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?
&emsp&emspWe 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 = x3 ? If it is not, find the solution.
We substitute 10 for x in the equation and see if we obtain an identity.
13 &ne 7 &emsp&emsp 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.
13+3=x3+3 &emsp&emsp Add 3 to both sides. 3 is the additive inverse of 3 .
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 = &minus 1/10+1/2
&emsp&emspTo combine the fractions, the fractions must have common denominators. The least common denominator (LCD) of the fractions is 10 .
(1*2)/(5*2)+x = &minus 1/10+(1*5)/(2*5) &emsp&emspChange each fraction to an equivalent fraction with a denominator of 10 .
2/10 + x = &minus 1/10+5/10 &emsp&emspThis is an equivalent equation.
2/10+x = 4/10 &emsp&emspSimplify by adding.
2/10+(2/10)+x = 4/10+(2/10) &emsp&emspAdd the additive inverse of 2/10 to each side
x= 1/5 &emsp&emsp Simplify the answer.
Check. We substitute 1/5 for x in the original equation and see if we obtain an identity.
1/5+1/5 ≟ &minus 1/10+1/2 &emsp&emspSubstitute 1/5 for x
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 ?
&emsp&emspTo answer this question, let&rsquos 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 &lsquo&lsquoweight value&rsquo&rsquo 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 [email protected]=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 . &emsp&emsp 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) &emsp&emspMultiply each side of the equation by 3 since (3)(1/3) = 1 .
Check. &emsp&emsp 1/3(45) ≟ 15 &emsp&emspSubstitute 45 for x in the original equation.
&emsp&emspNote 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 realworld example of this idea?
&emsp&emspLet&rsquos 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 &lsquo&lsquoweight value&rsquo&rsquo 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
EXAMPLE 2 Solve for x . &emsp&emsp 5x = 125
(5x)/5=125/5 &emsp&emspDivide both sides by 5 .
x = 25 &emsp&emsp Simplify. &emsp&emspThe solution is 25 .
Check. &emsp&emsp 5x = 125
&emsp&emspFor 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.
&emsp&emspThe solution to an equation may be a proper fraction or an improper fraction.
EXAMPLE 3 Solve for x . &emsp&emsp 4x = 38
(4x)/4= 38/4 &emsp&emspDivide both sides by 4 .
x=19/2 &emsp&emspSimplify. The solution is 19/2 .
&emsp&emspIf you leave the solution as a fraction, it will be easier to check that solution in the original equation.
Check: &emsp&emsp 4x = 38 &emsp&emsp Replace x by 19/2 .
&emsp&emspIn 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 . &emsp&emsp 3x = 48
(3x)/3=48/3 &emsp&emspDivide both sides by 3 .
&emsp&emspThe 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 &ldquosee&rsquo&rsquo the coefficient without actually rewriting the equation.
EXAMPLE 5 Solve for x . &emsp&emsp x = 24
1x = 24 &emsp&emsp Rewrite the equation. 1x is the same as x . Now the coefficient of 1 is obvious.
(1x)/1=24/1 &emsp&emspDivide both sides by 1
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.&emsp&emsp 5x +3 = 18
5x + 3 + (3)= 18+ (3) &emsp&emsp Add 3 to both sides, using the addition principle.
(5x)/5=15/5 &emsp&emspDivide both sides by 5 , using the division principle.
Check.&emsp&emsp 5(3)+3 ≟ 18
Check.&emsp&emsp 15+3 ≟ 18
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 . &emsp&emsp 9x = 6x + 15
9x + (6x) = 6x + (6x) + 15 &emsp&emsp Add 6x to both sides. Notice 6x + (6x) eliminates the variable on the right side.
3x = 15 &emsp&emsp Collect like terms.
(3x)/3=15/3 &emsp&emspDivide both sides by 3 .
&emsp&emspMany 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 &emsp&emsp Add 7x to both sides of the equation.
2x + 3+ (3) = 2 + (3) &emsp&emsp Add 3 to both sides.
(2x)/2=5/2 &emsp&emspDivide both sides by 2 .
Check. &emsp&emsp 9x + 3 = 7x 2
Check. &emsp&emsp 9(5/2)+3 ≟ 7(5/2)2 &emsp&emspReplace x by &minus 5/2 .
Check. &emsp&emsp &minus 45/2+6/2 ≟ &minus 35/24/2 &emsp&emspChange to equivalent fractions with a common denominator.
&emsp&emspIn 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. &emsp&emsp 5x + 26 6 = 9x + 12x
5x + 20 = 21x &emsp&emsp Combine like terms.
5x + (5x) + 20 = 21x + (5x) &emsp&emsp Add 5x to both sides.
20 = 16x &emsp&emsp Combine like terms.
20/16 =(16x)/16 &emsp&emsp Divide both sides by 16
5/4=x &emsp&emsp(Don&rsquot forget to reduce the resulting fraction.)
&emsp&emspAll the equations we have been studying so far are called firstdegree 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 firstdegree equation using the methods discussed in this section.
EXAMPLE 5 Solve for y . &emsp&emsp 5y^2 + 6y 2 = y + 5y^2 + 12
5y^25y^2 + 6y 2 = y + 5y^25y^2 + 12 &emsp&emspSubtract 5y^2 from both sides.
6y 2=y+12 &emsp&emsp Combine, since 5y^25y^2 = 0 .
6y+y2=y+y+12 &emsp&emsp Add y to each side.
7y2+2=12+2 &emsp&emsp Add 2 to each side.
(7y)/7 = 14/7 &emsp&emsp Divide each side by 7.
y=2 &emsp&emsp Simplify. &emsp&emspThe solution is 2 .
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(x3) = 25
4x +43x+9 = 25 &emsp&emsp Multiply by 4 and 3 to remove parentheses. Be careful of the signs. Remember (3)(3) = 9 .
&emsp&emspAfter 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 &emsp&emsp Collect like terms.
x+ 1313 = 2513 &emsp&emspAdd 13 to both sides to isolate the variable.
x = 12 &emsp&emsp The solution is 12 .
Check. &emsp&emsp 4(12+1)3(123) ≟ 25 &emsp&emsp Replace x by 12 .
&emsp&emsp &emsp&emsp 4(13)3(9) ≟ 25 &emsp&emspCombine numbers inside parentheses.
&emsp&emspIn 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(1.2x3.6) = 4.2x16.44
0.36x1.08 = 4.2x 16.44 &emsp&emsp Remove parentheses.
0.36x0.36x1.08 = 4.2x0.36x16.44 &emsp&emsp Subtract 0.36x from both sides.
1.08 = 3.84x 16.44 &emsp&emsp Collect like terms.
1.08 + 16.44 = 3.84x16.44 + 16.44 &emsp&emsp Add 16.44 to both sides.
15.36/3.84=(3.84x)/3.84 &emsp&emspDivide both sides by 3.84 .
EXAMPLE 8 Solve for z and check. 2(3z5) + 2 = 4z 3(2z + 8)
6z10 + 2 = 4z6z24 &emsp&emsp Remove parentheses.
6z 8 = 2z24 &emsp&emsp Collect like terms.
6z8 + 2z = 2z + 2z24 &emsp&emsp Add 2z to each side.
8z8 +8 = 24+ 8 &emsp&emsp Add 8 to each side.
(8z)/8=16/8 &emsp&emsp Divide each side by 8 .
z=2 &emsp&emsp Simplify. The solution is 2 .
Check. &emsp&emsp 2[3(2)5] +2 ≟ 4(2) 3[2(2) + 8] &emsp&emsp Replace z by 2 .
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 . &emsp&emsp 1/4x2/3=5/12x
First we find that the LCD = 12 .
12(1/4x2/3)=12(5/12x) &emsp&emspMultiply each side by 12
(12/1)(1/4)(x)(12/1)(2/3)=(12/1)(5/12)(x) &emsp&emsp Use the distributive property.
3x + (3x)8 = 5x + (3x) &emsp&emsp Add 3x to each side.
&minus 8/2=(2x)/2 &emsp&emspDivide each side by 2 .
Check.&emsp&emsp 1/4(4)2/3 ≟ 5/12(4)
&emsp&emspIn 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 . &emsp&emsp (x+5)/7=x/4+1/2
x/7+5/7=x/4+1/2 &emsp&emspFirst we write as separate fractions
(28)(x/7)+(28)(5/7)=(28)(x/4)+(28)(1/2) &emsp&emspWe observe that the LCD is 28 , so we multiply each term by 28 .
4x4x + 20 = 7x4x + 14 &emsp&emsp Add 4x to both sides.
20 = 3x + 14 &emsp&emsp Collect like terms.
2014=3x + 14 14 &emsp&emspAdd 14 to both sides.
6/3=(3x)/3 &emsp&emsp Divide both sides by 3 .
&emsp&emspIf 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
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).
Let&rsquos use each step in solving this example.
EXAMPLE 3 Solve for x and check your solution. 1/3(x2)= 1/5(x+4)+2
Step 1 &emsp&emsp x/32/3=x/5+4/5+2 &emsp&emspRemove parentheses.
Step 2 &emsp&emsp 15(x/3)15(2/3) = 15(x/5) +15(4/5) +15(2) &emsp&emsp Multiply by the LCD, 15 .
Step 3 &emsp&emsp 5x10 = 3x + 12 + 30 &emsp&emsp Simplify.
Step 4 &emsp&emsp 5x3x10 = 3x3x + 42 &emsp&emsp Add 3x to both sides.
Step 5 &emsp&emsp 2x10+ 10 = 42+ 10 &emsp&emsp Add 10 to both sides.
Step 6 &emsp&emsp (2x)/2=52/2 &emsp&emsp Divide both sides by 2 .
Step 7 &emsp&emsp x = 26 &emsp&emsp Simplify the solution.
Step 8 Check. &emsp&emsp 1/3(262) ≟ 1/5(26 +4)+2 &emsp&emsp Replace x by 26 .
&emsp&emspIt 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.
&emsp&emspIt 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 &emsp&emsp Use the distance formula.
156 = 52t &emsp&emsp Substitute the known values for the variables.
156/52=52/52t &emsp&emspDivide both sides of the equation by 52 to solve for t.
It took Joseph 3 hours to drive 156 miles at 52 miles per hour.
&emsp&emspIf 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 . &emsp&emsp d=rt
d/r=(rt)/r &emsp&emspWe want to isolate t . Therefore, we are dividing both sides of the equation by the coefficient of t , which is r .
d/r=t &emsp&emsp You have solved for the variable indicated.
&emsp&emspA 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 . &emsp&emsp 3x2y = 6
2y = 63x &emsp&emsp We want to isolate the term containing y , so we subtract 3x from both sides.
(2y)/(2)= (63x)/(2) &emsp&emspDivide both sides by the coefficient of y .
y=6/2+(3x)/2 &emsp&emspRewrite the fraction.
y= 3/2x3 &emsp&emsp Simplify and regroup.
This is known as the slopeintercept form of the equation of a line.
&emsp&emspOur procedure for solving a firstdegree 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
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.
EXAMPLE 4 A trapezoid is a foursided figure with two parallel sides. If the parallel sides are a and b and the altitude is h , the area is given by
Solve this equation for a .
A=(ha)/2+(hb)/2 &emsp&emspRemove the parentheses.
2(A) = 2((ha)/2)+2((hb)/2) &emsp&emsp Multiply all terms by LCD of 2 .
2Ahb = ha &emsp&emsp We want to isolate the term containing a . Therefore, we subtract hb from both sides.
(2Ahb)/h= (ha)/h &emsp&emspDivide both sides by h (the coefficient of a ).
(2Ahb)/h=a &emsp&emspThe solution is obtained.
Note: Although the solution is in simple form, it could be written in an alternative way. Since
we could have (2A)/hb = 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 &lsquo&lsquo 5 is less than 7 &rsquo&rsquo or &lsquo&lsquo 9 is greater than 4 .&rsquo&rsquo These relationships are called inequalities. We can write inequalities in mathematics by using symbols. We use the symbol < to represent the words &lsquo&lsquois less than.&rsquo&rsquo We use the symbol > to represent the words &lsquo&lsquois greater than.&rsquo&rsquo
Note. &lsquo&lsquo 5 is less than 7 &rsquo&rsquo and &lsquo&lsquo 7 is greater than 5 &rsquo&rsquo 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 .
&emsp&emspWe can illustrate the concept of inequality graphically if we examine a number line.
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 .
&emsp&emspWhat about negative numbers? We can say &lsquo&lsquo 1 is greater than 3 &rdquo 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 &emsp&emsp Use >, since 3 is to the right of 1 .
(b) 2< 1 &emsp&emspUse <, since 2 is to the left of 1 . (Or equivalently, we could say that 1 is to the right of 2 .)
(c) 3 > 4 &emsp&emspSince 3 is to the right of 4 .
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:
Note that the open circle at 3 suggests that we do not include the point for the number 3 .
&emsp&emspSimilarly, we could represent graphically x < 2 as follows:
&emsp&emspSometimes a variable will be either greater than or equal to a certain number. In the statement &ldquo x is greater than or equal to 3 ,&rsquo&rsquo 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:
Note that the closed circle at 3 suggests that we do include the point for the number 3 .
&emsp&emspSimilarly, we could represent graphically x <= 2 as follows:
&mdash&mdash 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) We state that &lsquo&lsquo x is less than 2 .&rdquo
(b) We can state that &lsquo&lsquo 3 is less than x &rsquo&rsquo or, an equivalent statement, &lsquo&lsquo x is greater than 3 .&rsquo&rsquo 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 &lsquo&lsquo x is less than or equal to 6 .&rdquo
&emsp&emspThere 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:
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&rsquove 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&rsquoll show some examples so that you can see the operations we can do with inequalities just as with equations.
&emsp&emspWe will first examine the pattern that takes place when we perform a given operation on both sides of an inequality.
Note that we avoided multiplying or dividing by a negative number !
&emsp&emspNow 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.
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
The < sign we started with ( 3 < 5 ) is reversed to > ( 6 >10 ). A similar reversal takes place in the following example.
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 +77>=137 &emsp&emsp Subtract 7 from both sides.
(3x)/3>=6/3 &emsp&emsp Divide both sides by 3 .
x>=2 &emsp&emsp Simplify. Note the direction of the inequality is not changed, since we have divided by a positive number.
The graphical representation is en ee
&mdash2 Il 0 I 2 3 4
EXAMPLE 4 Solve and graph 5 3x > 7 .
553x>75 &emsp&emsp Subtract 5 from both sides.
(3x)/3<2/3 &emsp&emspDivide by 3 , and reverse the inequality, since both sides are divided by negative 3 .
x< 2/3 &emsp&emsp 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 &minus (13x)/2<=x/215/8
(8)((13x)/2)<=(8)(x/2)(8)(15/8) &emsp&emspMultiply all terms by LCD = 8 . We do not reverse the direction of the inequality symbol since we are multiplying by a positive number, 8.
52x4x <= 4x154x &emsp&emspAdd 4x to both sides.
56x <= 15 &emsp&emspCombine like terms.
(56x)/56>= 15/56 &emsp&emspDivide both sides by 56 . We reverse the direction of the inequality when we divide both sides by a negative number.
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.