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5.1 Systems of Linear Equations Linear Systems Substitution Method Elimination Method Special Systems Applying Systems of Equations Solving Linear Systems with Three Unknowns (Variables) Using Systems of Equations to Model Data

Linear Systems The definition of a linear equation given in Chapter 1 can be extended to more variables; any equation of the form for real numbers a1, a2, …, an (not all of which are 0) and b, is a linear equation or a first-degree equation in n unknowns.

Linear Systems A set of equations is called a system of equations. The solutions of a system of equations must satisfy every equation in the system. If all the equations in a system are linear, the system is a system of linear equations, or a linear system.

Linear Systems The possible graphs of a linear system in two unknowns are as follows. 1. The graphs intersect at exactly one point, which gives the (single) ordered pair solution of the system. The system is consistent and the equations are independent.

Linear Systems 2. The graphs are parallel lines, so there is no solution and the solution set is ø. The system is inconsistent and the equations are independent.

Linear Systems 3. The graphs are the same line, and there is an infinite number of solutions. The system is consistent and the equations are dependent.

Substitution Method In a system of two equations with two variables, the substitution method involves using one equation to find an expression for one variable in terms of the other, and then substituting into the other equation of the system.

Solution SOLVING A SYSTEM BY SUBSTITUTION Example 1 Solve the system. (1) (2) Solution Begin by solving one of the equations for one of the variables. We solve equation (2) for y. (2) Add x.

Note the careful use of parentheses. SOLVING A SYSTEM BY SUBSTITUTION Example 1 Now replace y with x + 3 in equation (1), and solve for x. Note the careful use of parentheses. (1) Let y = x + 3 in (1). Distributive property Combine terms. Subtract.

SOLVING A SYSTEM BY SUBSTITUTION Example 1 Replace x with 1 in equation (3) to obtain y = 1 + 3 = 4. The solution of the system is the ordered pair (1, 4). Check this solution in both equations (1) and (2). Check: (1) (2) ? ? True True Both check; the solution set is {(1, 4)}.

Elimination Method Another way to solve a system of two equations, called the elimination method, uses multiplication and addition to eliminate a variable from one equation. To eliminate a variable, the coefficients of that variable in the two equations must be additive inverses. To achieve this, we use properties of algebra to change the system to an equivalent system, one with the same solution set. The three transformations that produce an equivalent system are listed here.

TransformationS of a Linear System 1. Interchange any two equations of the system. 2. Multiply or divide any equation of the system by a nonzero real number. 3. Replace any equation of the system by the sum of that equation and a multiple of another equation in the system.

Solution SOLVING A SYTEM BY ELIMINATION Example 2 Solve the system. (1) (2) Solution One way to eliminate a variable is to use the second transformation and multiply both sides of equation (2) by – 3, giving the equivalent system (1) Multiply (2) by – 3 (3)

SOLVING A SYTEM BY ELIMINATION Example 2 Now multiply both sides of equation (1) by 2, and use the third transformation to add the result to equation (3), eliminating x. Solve the result for y. Multiply (1) by 2 (3) Add. Solve for y.

SOLVING A SYTEM BY ELIMINATION Example 2 Substitute 2 for y in either of the original equations and solve for x. (1) Let y = 2 in (1). Multiply. Add 8.

SOLVING A SYTEM BY ELIMINATION Example 2 A check shows that (3, 2) satisfies both equations (1) and (2); the solution set is {(3, 2)}.

Solution SOLVING AN INCONSISTENT SYSTEM Example 3 Solve the system. (1) (2) Solution To eliminate the variable x, multiply both sides of equation (1) by 2. Multiply (1) by 2 (2) False.

SOLVING AN INCONSISTENT SYSTEM Example 3 Since 0 = 15 is false, the system is inconsistent and has no solution. As suggested here by the graph, this means that the graphs of the equations of the system never intersect. (The lines are parallel.) The solution set is the empty set.

SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Example 4 Solve the system. (1) (2) Solution Divide both sides of equation (1) by 2, and add the result to equation (2). Divide (1) by 2. (2) True.

SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Example 4 The result, is a true statement, which indicates that the equations of the original system are equivalent. Any ordered pair that satisfies either equation will satisfy the system. From equation (2), (2)

SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS Example 4 The solutions of the system can be written in the form of a set of ordered pairs (x, 2 + 4x), for any real number x. Some ordered pairs in the solution set are (0, 2 + 4 • 0) = (0, 2), (1, 2 + 4 • 1) = (1, 6), (3, 14), and (– 2, – 6). As shown here, the equations of the original system are dependent and lead to the same straight line graph. The solution set is written {(x, 2 + 4x)}.

y = 2 and solving for x to obtain Note In the algebraic solution for Example 4, we wrote the solution set with the variable x arbitrary. We could write the solution set with y arbitrary: By selecting values for y and solving for x in this ordered pair, we can find individual solutions. Verify again that (0, 2) is a solution by letting y = 2 and solving for x to obtain

Applying Systems of Equations Many applied problems involve more than one unknown quantity. Although some problems with two unknowns can be solved using just one variable, it is often easier to use two variables. To solve a problem with two unknowns, we must write two equations that relate the unknown quantities. The system formed by the pair of equations can then be solved using the methods of this chapter.

Solving An Applied Problem By Writing A System of Equations Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Step 3 Write a system of equations that relates the unknowns. Step 4 Solve the system of equations. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem.

USING A LINEAR SYSTEM TO SOLVE AN APPLICATION Example 5 Salaries for the same position can vary depending on the location. In 2006, the average of the salaries for the position of Accountant I in San Diego, California, and Salt Lake City, Utah, was $41,175.50. The salary in San Diego, however, exceeded the salary in Salt Lake City by $3697. Determine the salary for the Accountant I position in San Diego and in Salt Lake City.

Solution USING A LINEAR SYSTEM TO SOLVE AN APPLICATION Example 5 Step 1 Read the problem. We must find the salary of the Accountant I position in San Diego and in Salt Lake City. Step 2 Assign variables. Let x represent the salary of the Accountant I position in San Diego and y represent the salary for the same position in Salt Lake City.

Solution USING A LINEAR SYSTEM TO SOLVE AN APPLICATION Example 5 Step 3 Write a system of equations. Since the average salary for the Accountant I position in San Diego and Salt Lake City was $41,175.50, one equation is Multiply both sides of the equation by 2 to get the equation (1)

USING A LINEAR SYSTEM TO SOLVE AN APPLICATION Example 5 The salary in San Diego exceeded the salary in Salt Lake City by $3697. Thus, which gives the system of equations x – y = 3697, which gives the system of equations (1) (2)

USING A LINEAR SYSTEM TO SOLVE AN APPLICATION Example 5 Step 4 Solve the system. To eliminate y, add the two equations. (1) (2) Add. Solve for x. To find y, substitute 43,024 for x in equation (2). Let x = 43,024 in (2). Subtract 43,024. Multiply by – 1.

USING A LINEAR SYSTEM TO SOLVE AN APPLICATION Example 5 Step 5 State the answer. The salary for the position of Accountant I was $43,024 in San Diego and $39,327 in Salt Lake City. Step 6 Check. The average of $43,024 and $39,327 is Also, $43,024 is $3697 more than $39,327, as required. The answer checks.

Solving Linear Equations with three Unknowns (Variables) Earlier, we saw that the graph of a linear equation in two unknowns is a straight line. The graph of a linear equation in three unknowns requires a three-dimensional coordinate system. The three number lines are placed at right angles. The graph of a linear equation in three unknowns is a plane.

Solving Linear Equations with three Unknowns (Variables) Some possible intersections of planes representing three equations in three variables are shown here.

Solving Linear Equations with three Unknowns (Variables) Some possible intersections of planes representing three equations in three variables are shown here.

Solving Linear Equations with three Unknowns (Variables) Some possible intersections of planes representing three equations in three variables are shown here.

Solving a System To solve a linear system with three unknowns, first eliminate a variable from any two of the equations. Then eliminate the same variable from a different pair of equations. Eliminate a second variable using the resulting two equations in two variables to get an equation with just one variable whose value you can now determine. Find the values of the remaining variables by substitution. The solution of the system is written as an ordered triple.

Solution Example 6 Solve the system. (1) (2) (3) SOLVING A SYSTEM OF EQUATIONS WITH THREE VARIABLES Example 6 Solve the system. (1) (2) (3) Solution Eliminate z by adding equations (2) and (3) to get (4)

Make sure equation (5) has the same two variables as equation (4). SOLVING A SYSTEM OF EQUATIONS WITH THREE VARIABLES Example 6 To eliminate z from another pair of equations, multiply both sides of equation (2) by 6 and add the result to equation (1). Multiply (2) by 6. (1) (5) Make sure equation (5) has the same two variables as equation (4).

SOLVING A SYSTEM OF EQUATIONS WITH THREE VARIABLES Example 6 To eliminate x from equations (4) and (5), multiply both sides of equation (4) by 5 and add the result to equation (5). Solve the resulting equation for y. Multiply (4) by – 5. (5) Add. Divide by 5.

Using y = – 1 , find x from equation (4) by substitution. SOLVING A SYSTEM OF EQUATIONS WITH THREE VARIABLES Example 6 Using y = – 1 , find x from equation (4) by substitution. (4) with y = – 1.

Substitute 2 for x and – 1 for y in equation (3) to find z. SOLVING A SYSTEM OF EQUATIONS WITH THREE VARIABLES Example 6 Substitute 2 for x and – 1 for y in equation (3) to find z. (3) with x = 2, y = – 1. Verify that the ordered triple (2, –1, 1) satisfies all three equations in the original system. The solution set is {(2, –1,1)}.

Caution Be careful not to end up with two equations that still have three variables. Eliminate the same variable from each pair of equations.

Solution Example 7 Solve the system. (1) (2) SOLVING A SYSTEM OF TWO EQUATIONS WITH THREE VARIABLES Example 7 Solve the system. (1) (2) Solution Geometrically, the solution is the intersection of the two planes given by equations (1) and (2). The intersection of two different nonparallel planes is a line. Thus there will be an infinite number of ordered triples in the solution set, representing the points on the line of intersection. To eliminate x, multiply both sides of equation (1) by – 3 and add the result to equation (2). (Either y or z could have been eliminated instead.)

Solve this equation for z. SOLVING A SYSTEM OF TWO EQUATIONS WITH THREE VARIABLES Example 7 Multiply (1) by – 3. (2) (3) Add 7y. Solve this equation for z. Divide each term by – 7.

Use parentheses around SOLVING A SYSTEM OF TWO EQUATIONS WITH THREE VARIABLES Example 7 This gives z in terms of y. Express x also in terms of y by solving equation (1) for x and substituting for z in the result. (1) Solve for x. Substitute (– y + 3) for z. Simplify. Use parentheses around – y + 3. With y arbitrary, the solution set is of the form {(–y +1, y, – y+3)}.

By choosing z = 2, one solution would be Note Had we solved equation (3) in Example 7 for y instead of z, the solution would have had a different form but would have led to the same set of solutions. In that case we would have z arbitrary, and the solution set would be of the form {(– 2 + z, 3 – z, z)}. By choosing z = 2, one solution would be (0, 1, 2), which was found above.

Modeling Data Applications with three unknowns usually require solving a system of three equations. We can find the equation of a parabola in the form by solving a system of three equations with three variables.

USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 Find the equation of the parabola y = ax2 + bx + c that passes through the points (2,4), (– 1, 1), and (– 2, 5). Solution Since the three points lie on the graph of the equation y = ax2 + bx + c, they must satisfy the equation. Substituting each ordered pair into the equation gives three equations with three variables.

USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 (1) (2) (3)

To solve this system, first eliminate c using equations (1) and (2). USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 To solve this system, first eliminate c using equations (1) and (2). (1) Multiply (2) by – 1. (4)

Equation (5) must have the same two variables as equation (4). USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 Now, use equations (2) and (3) to eliminate the same variable, c. (2) Multiply (3) by – 1. (5) Equation (5) must have the same two variables as equation (4).

USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 Solve the system of equations (4) and (5) in two variables by eliminating a. (4) (5) Add. Divide by 4.

Find a by substituting – ¼ for b in equation (4). USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 Find a by substituting – ¼ for b in equation (4). (4) Let b = – ¼. Add ¼ .

Finally, find c by substituting and in equation (2). USING CURVE FITTING TO FIND AN EQUATION THROUGH THREE POINTS Example 8 Finally, find c by substituting and in equation (2). (2) Add. Subtract . The required equation is or

SOLVING AN APPLICATION USING A SYSTEM OF THREE EQUATIONS Example 9 An animal feed is made from three ingredients: corn, soybeans, and cottonseed. One unit of each ingredient provides units of protein, fat, and fiber as shown in the table. How many units of each ingredient should be used to make a feed that contains 22 units of protein, 28 units of fat, and 18 units of fiber? Corn Soybeans Cottonseed Total Protein .25 .4 .2 22 Fat .3 28 Fiber .1 18

SOLVING AN APPLICATION USING A SYSTEM OF THREE EQUATIONS Example 9 Solution Step 1 Read the problem. We must determine the number of units of corn, soybeans, and cottonseed. Step 2 Assign variables. Let x represent the number of units of corn, y the number of units of soybeans, and z the number of units of cottonseed.

Also, for the 28 units of fat, SOLVING AN APPLICATION USING A SYSTEM OF THREE EQUATIONS Example 9 Step 3 Write a system of equations. Since the total amount of protein is to be 22 units, the first row of the table yields (1) Also, for the 28 units of fat, (2) and, for the 18 units of fiber, (3)

SOLVING AN APPLICATION USING A SYSTEM OF THREE EQUATIONS Example 9 Multiply equation (1) on both sides by 100, and equations (2) and (3) by 10 to get the equivalent system (4) (5) (6)

SOLVING AN APPLICATION USING A SYSTEM OF THREE EQUATIONS Example 9 Step 4 Solve the system. Using the methods described earlier in this section, we find that x = 40, y = 15, and z = 30. Step 5 State the answer. The feed should contain 40 units of corn, 15 units of soybeans, and 30 units of cottonseed. Step 6 Check. Show that the ordered triple (40, 15, 30) satisfies the system formed by equations (1), (2), and (3).

Note Notice how the table in Example 9 is used to set up the equations of the system. The coefficients in each equation are read from left to right. This idea is extended in the next section, where we introduce solution of systems by matrices.