EXAMPLE 4 Solve linear systems with many or no solutions Solve the linear system. a.x – 2y = 4 3x – 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.

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 5.12x – 3y = – 9 –4x + y = 3 infinitely many solutions ANSWER 6.6x + 15y = – 12 – 2x – 5y = 9 ANSWER no solutions

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 8.12x – 2y = 21 3x + 12y = – 4 The solution is (, ) ANSWER – x + 3y = 20 – x – y = – infinitely many solutions ANSWER

GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 9.8x + 9y = 15 5x – 2y = 17 5x + 5y = x + 3y = 4.2 (3, –1) ANSWER (0.6, 0.4) ANSWER

HW- Assignment Section 4-2 Homework Assignment From Section 4-2 page 166 and 167 Even Problems 2-26

EXAMPLE 2 Solve a system of equations by graphing Solve the system. 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 of equations by graphing Solve the system. 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= x 2.5 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 = = 2.5(20) ANSWER The total costs are equal after 20 rides. The correct answer is B.

GUIDED PRACTICE for Examples 2,3, and 4 Solve the system. 4.2x + 5y = 6 4x + 10y = 12 ANSWER Infinitely many solutions

GUIDED PRACTICE for Examples 2,3, and 4 Solve the system 5.3x – 2y = 10 3x – 2y = 2 ANSWER no solutiononsistent

GUIDED PRACTICE for Examples 2,3, and 4 Solve the system. 6. –2x + y = 5 y = –x + 2 ANSWER (–1, 3)

HW- Assignment Section 4-1 Homework Assignment Section 4-2 page #161 Even problems 2-16

EXAMPLE 1 Graph a system of two inequalities Graph the system of inequalities. y > –2x – 5 Inequality 1 y < x + 3 Inequality 2

EXAMPLE 1 Graph a system of two inequalities STEP 2 Identify the region that is common to both graphs. It is the region that is shaded purple. SOLUTION STEP 1 Graph each inequality in the system. Use red for y > –2x – 5 and blue for y ≤ x + 3.

EXAMPLE 2 Graph a system with no solution Graph the system of inequalities. 2x + 3y < 6 Inequality 1 y < – x Inequality 2

EXAMPLE 2 Graph a system with no solution STEP 2 Identify the region that is common to both graphs. There is no region shaded both red and blue. So, the system has no solution. SOLUTION STEP Graph each inequality in the system. Use red for 2x + 3y – x + 4.

EXAMPLE 3 Graph a system with an absolute value inequality Graph the system of inequalities. y < 3 Inequality 1 y > x + 4 Inequality 2

GUIDED PRACTICE for Examples 1, 2 and 3 Graph the system of inequalities. 1. y < 3x – 2 y > – x + 4

GUIDED PRACTICE for Examples 1, 2 and x – y > x – y < 5

GUIDED PRACTICE for Examples 1, 2 and 3 3. x + y > – 3 –6x + y < 1

GUIDED PRACTICE for Examples 1, 2 and 3 6. y > 2 x + 1 y < x + 1 This has no solution.

HW- Assignment Worksheet Homework Assignment Skills Practice 6-8 Worksheet Problems 1-12

EXAMPLE 1 Use the elimination method Solve the system. 4x + 2y + 3z = 1 Equation 1 2x – 3y + 5z = –14 Equation 2 6x – y + 4z = –1 Equation 3 SOLUTION STEP 1 Rewrite the system as a linear system in two variables. 4x + 2y + 3z = 1 12x – 2y + 8z = –2 Add 2 times Equation 3 to Equation 1. 16x + 11z = –1 New Equation 1

EXAMPLE 1 2x – 3y + 5z = –14 –18x + 3y –12z = 3 Add – 3 times Equation 3 to Equation 2. –16x – 7z = –11 New Equation 2 STEP 2 Solve the new linear system for both of its variables. 16x + 11z = –1 Add new Equation 1 and new Equation 2. –16x – 7z = –11 4z = –12 z = –3 Solve for z. x = 2 Substitute into new Equation 1 or 2 to find x. Use the elimination method

6x – y + 4z = –1 EXAMPLE 1 Use the elimination method STEP 3 Substitute x = 2 and z = – 3 into an original equation and solve for y. Write original Equation 3. 6(2) – y + 4(–3) = –1 Substitute 2 for x and –3 for z. y = 1 Solve for y.

EXAMPLE 2 Solve a three-variable system with no solution Solve the system.x + y + z = 3 Equation 1 4x + 4y + 4z = 7 Equation 2 3x – y + 2z = 5 Equation 3 SOLUTION When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation. Add – 4 times Equation 1 to Equation 2. –4x – 4y – 4z = –12 4x + 4y + 4z = 7 0 = –5 New Equation 1

EXAMPLE 2 Solve a three-variable system with no solution Because you obtain a false equation, you can conclude that the original system has no solution.

EXAMPLE 3 Solve a three-variable system with many solutions Solve the system. x + y + z = 4 Equation 1 x + y – z = 4 Equation 2 3x + 3y + z = 12 Equation 3 SOLUTION STEP 1 Rewrite the system as a linear system in two variables. Add Equation 1 to Equation 2. x + y + z = 4 x + y – z = 4 2x + 2y = 8 New Equation 1

EXAMPLE 3 Solve a three-variable system with many solutions x + y – z = 4 Add Equation 2 3x + 3y + z = 12 to Equation 3. 4x + 4y = 16 New Equation 2 Solve the new linear system for both of its variables. STEP 2 Add –2 times new Equation 1 to new Equation 2. Because you obtain the identity 0 = 0, the system has infinitely many solutions. –4x – 4y = –16 4x + 4y = 16

HW- Assignment Section 4-4 and 4-5 Homework Assignments Section 4-4 page #178 Problems Section 4-5 page #181 and 182 Problems1-18

EXAMPLE 1 Evaluate determinants Evaluate the determinant of the matrix. a SOLUTION b – – – –

GUIDED PRACTICE for Examples 1 and – 15 Evaluate the determinant of the matrix. ANSWER – – – – –21 ANSWER – – – – 470 ANSWER

EXAMPLE 3 Use Cramer’s rule to solve this system: 3x 5y = 21 9x + 4y = 6 – – – SOLUTION STEP 1 Evaluate the determinant of the coefficient matrix. 94 3–5 = –45 – 12 = –57 Use Cramer’s rule for a 2 X 2 system

EXAMPLE 3 STEP 2 Apply Cramer’s rule because the determinant is not 0. = –57 –171 = 3 y = 9–6 3–21 –57 = –189 – (–18) ANSWER The solution is ( 2, 3). –x = –64 –21 –5 –57 = 30 – (–84) = – = –2 Use Cramer’s rule for a 2 X 2 system

EXAMPLE 3 CHECK Check this solution in the original equations. –21 = –21 Use Cramer’s rule for a 2 X 2 system 9x + 4y = –6 9(–2) + 4(3) = –6 ? – = –6 ? –6 = ? 3(–2) – 5(3) = –21 3x – 5y = –21 –6 – 15 = –21 ?

GUIDED PRACTICE for Examples 3 and 4 Use Cramer’s rule to solve the linear system. 5. 3x – 4y = –15 2x + 5y = 13 (–1, 3). ANSWER 6. 4x + 7y = 2 –3x – 2y = 28 (4, –2). ANSWER 7. 3x – 4y + 2z = 18 4x + y – 5z = –13 2x – 3y + z = 11 (2, –1, 4). ANSWER

HW- Assignment Section 13-3 Homework Assignment Section 13-3 page 580 Problems #2-26