EXAMPLE 1 Solve a system graphically

Graph the linear system and estimate the solution. Then check the solution algebraically.

4x + y = 8

2x – 3y = 18

Equation 1

Equation 2

SOLUTION

Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, – 4). You can check this algebraically as follows.

EXAMPLE 1 Solve a system graphically

Equation 1 Equation 2

4x + y = 8

4(3) + (– 4) 8=?

=?12 – 4 8

8 = 8

2x – 3y = 18

=?2(3) – 3( – 4) 18

=?6 + 12 18

18 = 18

The solution is (3, – 4).

EXAMPLE 2 Solve a system with many solutions

Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.

4x – 3y = 8

8x – 6y = 16

Equation 1

Equation 2

SOLUTION

The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent.

EXAMPLE 3 Solve a system with no solution

Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.

2x + y = 4

2x + y = 1

Equation 1

Equation 2

SOLUTION

The graphs of the equations are two parallel lines.Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent.

EXAMPLE 4 Standardized Test Practice

SOLUTION

Equation 1 (Option A)

y = 1 x + 30

Equation 2 (Option B)

EXAMPLE 4 Standardized Test Practice

y = x2.5

To solve the system, graph the equations y = x + 30 and y = 2.5x, as shown at the right.

EXAMPLE 4 Standardized Test Practice

Notice that you need to graph the equations only in the first quadrant because only nonnegative values of x and y make sense in this situation.

The lines appear to intersect at about the point (20, 50). You can check this algebraically as follows.

Equation 1 checks.

Equation 2 checks.

50 = 20 + 30

50 = 2.5(20)

ANSWERThe total costs are equal after 20 rides.

The correct answer is B.

EXAMPLE 1 Use the substitution method

Solve the system using the substitution method.

2x + 5y = – 5

x + 3y = 3

Equation 1

Equation 2

SOLUTION

STEP 1 Solve Equation 2 for x.

x = – 3y + 3 Revised Equation 2

EXAMPLE 1 Use the substitution method

STEP 2Substitute the expression for x into Equation 1 and solve for y.

2x +5y = – 5

2(– 3y + 3) + 5y = – 5

y = 11

Write Equation 1.

Substitute – 3y + 3 for x.

Solve for y.

STEP 3Substitute the value of y into revised Equation 2 and solve for x.

x = – 3y + 3

x = – 3(11) + 3

x = – 30

Write revised Equation 2.

Substitute 11 for y.

Simplify.

EXAMPLE 1 Use the substitution method

CHECK Check the solution by substituting into the original equations.

2(– 30) + 5(11) – 5=? Substitute for x and y. =?– 30 + 3(11) 3

Solution checks. 3 = 3– 5 = – 5

The solution is (– 30, 11).

ANSWER

EXAMPLE 2 Use the elimination method

Solve the system using the elimination method.

3x – 7y = 10

6x – 8y = 8

Equation 1

Equation 2

SOLUTION

Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign.

STEP 1

3x – 7y = 10

6x – 8y = 8

– 6x + 14y = 20

6x – 8y = 8

EXAMPLE 2 Use the elimination method

STEP 2Add the revised equations and solve for y. 6y = – 12

y = – 2STEP 3Substitute the value of y into one of the original equations. Solve for x.

3x – 7y = 10

3x – 7( – 2) = 10

3x + 14 = 10

x = 43

– Solve for x.

Simplify.

Substitute – 2 for y.

Write Equation 1.

EXAMPLE 2 Use the elimination method

The solution is ( , – 2)43

–

ANSWER

CHECK

You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.

EXAMPLE 3 Standardized Test Practice

SOLUTION

Write verbal models for this situation.

Equation 1

EXAMPLE 3 Standardized Test Practice

Equation 2

EXAMPLE 3 Standardized Test Practice

STEP 2 Write a system of equations.

Equation 1

Equation 2

5x + 7y = 2500

8x + 12y = 4200

Total cost for all T-shirts

Total revenue from all T-shirts sold

STEP 3Solve the system using the elimination method.

Multiply Equation 1 by – 8 and Equation 2 by 5 so that the coefficients of x differ only in sign.

5x + 7y = 2500

8x + 12y = 4200

– 40x – 56y = – 20,000

40x + 60y = 21,000

Add the revised equations and solve for y. 4y = 1000y = 250

EXAMPLE 3 Standardized Test Practice

Substitute the value of y into one of the original equations and solve for x.

5x + 7y = 2500

5x + 7(250) = 2500

5x + 1750 = 2500x = 150

Write Equation 1.

Substitute 250 for y.

Simplify.

Solve for x.

The school sold 150 short sleeve T-shirts and 250 long sleeve T-shirts.

ANSWERThe correct answer is C.

EXAMPLE 4 Solve linear systems with many or no solutions

Solve the linear system.

a. x – 2y = 43x – 6y = 8

b. 4x – 10y = 8– 14x + 35y = – 28

SOLUTION

a. Because the coefficient of x in the first equation is 1, use the substitution method.

Solve the first equation for x.

x – 2y = 4

x = 2y + 4

Write first equation.

Solve for x.

EXAMPLE 4 Solve linear systems with many or no solutions

Substitute the expression for x into the second equation.

3x – 6y = 8

3(2y + 4) – 6y = 8

12 = 8

Write second equation.

Substitute 2y + 4 for x.

Simplify.

Because the statement 12 = 8 is never true, there is no solution.

ANSWER

EXAMPLE 4 Solve linear systems with many or no solutions

b.Because no coefficient is 1 or – 1, use the elimination method.

Multiply the first equation by 7 and the second equation by 2.

4x – 10y = 8

– 14x + 35y = – 28

28x – 70y = 56

– 28x + 70y = – 56

Add the revised equations. 0 = 0

ANSWER

Because the equation 0 = 0 is always true, there are infinitely many solutions.