Solving Systems of Linear Equations by Elimination; Applications Solve systems of linear equations using elimination. 2.Solve applications using elimination.

x + y = y = 9 y = 5 The solution is (4, 5). Solve the system of equations using elimination. One equation with one unknown. Look for additive inverses. Solve for y.

The solution is (4, 0). Solve the system of equations using elimination. Look for additive inverses.

-3 Solve the system of equations. Solve for y. x + y = y = 8 y = 5 The solution is (3, 5). Look for additive inverses. 4

Solve the system of equations. The solution is (4, 6). 4

-3 4 Solve the system of equations. Multiply by 2. Multiply by  5. Look for additive inverses. 4x – 5y = 19 4(1) – 5y = 19 4 – 5y = 19 –5y = 15 y =  3 The solution is (1, –3). Solve for y. 2 -5

Slide 5- 7 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution 9.3

Slide 5- 8 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution 9.3

Solve the system of equations. Multiply by 100. Multiply by 5. To clear the decimals in Equation 1, multiply by 100. To clear the fractions in Equation 2, multiply by 5.

continued Substitute to find y. The solution is (  1,  3).

Copyright © 2011 Pearson Education, Inc. Solving Systems of Two Linear Equations Using Elimination 1. Write the equations in standard form (Ax + By = C). 2. Use the multiplication principle to clear fractions or decimals (optional). 3. If necessary, multiply one or both equations by a number so that they have a pair of terms that are additive inverses. 4. Add the equations. The result should be an equation in one variable. 5. Solve the equation from step Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable. 7. Check your solution in the original equations.

Slide Copyright © 2011 Pearson Education, Inc. Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5) 9.3

Slide Copyright © 2011 Pearson Education, Inc. Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5) 9.3

Inconsistent Systems and Dependent Equations When both variables have been eliminated and the resulting equation is false, such as 0 = 5, there is no solution. The system is inconsistent. When both variables have been eliminated and the resulting equation is true, such as 0 = 0, the equations are dependent. There are an infinite number of solutions.

Solve the system of equations. False statement. The system is inconsistent and has no solution.

Solve the system of equations. True statement. The equations are dependent. There are an infinite number of solutions. Multiply by 2. To eliminate y, multiply the first equation by 2. 2

Slide Copyright © 2011 Pearson Education, Inc. Solve the system. a) (0, 10) b) (-2, -3) c) infinite number of solutions d) no solution 9.3

Slide Copyright © 2011 Pearson Education, Inc. Solve the system. a) (0, 10) b) (-2, -3) c) infinite number of solutions d) no solution 9.3