Solve linear systems by Multiplication

ObjectiveObjective

To solve linear systems using elimination To solve linear systems using elimination (adding, subtracting, and multiplying).(adding, subtracting, and multiplying).

Today we are focusing on solving systems Today we are focusing on solving systems by elimination when you have to multiply by elimination when you have to multiply firstfirst

Steps are on PAGE 305Steps are on PAGE 305

Equation 1 Equation 1

-2x + 5y = 13-2x + 5y = 13 Equation 2Equation 2

2x + 3y = 11 2x + 3y = 11

BELLWORKBELLWORK

Equation 1 Equation 1 2x + 3y = 11 2x + 3y = 11 Substitute value forSubstitute value fory into either of the y into either of the original equations original equations 2x + 3(3) = 112x + 3(3) = 11

2x + 9 = 112x + 9 = 11

The solution is the point (1,3). The solution is the point (1,3). Substitute (1,3) into both Substitute (1,3) into both

equations to check.equations to check.

-2(1) + 5(3) = 13-2(1) + 5(3) = 1313 = 1313 = 13

2(1) + 3(3) = 112(1) + 3(3) = 1111 = 11 11 = 11

ADDITIONADDITION

+ 8y = 248y = 24 y = 3y = 3

x = 1x = 1

EliminatedEliminated

Solution: (1,3)

Could we solve this system by addition or subtraction?

6x + 5y = 19 6x + 5y = 19 2x + 3y = 52x + 3y = 5

NO!We need to do somethingBefore they cancel

Section 9.4 “Solve Linear Systems Section 9.4 “Solve Linear Systems by Multiplying” p.303by Multiplying” p.303

ELIMINATION-ELIMINATION-adding or subtracting equations to obtain a adding or subtracting equations to obtain a

new equation in one variable. new equation in one variable.

Solving Linear Systems Using EliminationSolving Linear Systems Using Elimination

(1) (1) Multiply Multiply the whole equation by a constant in order to the whole equation by a constant in order to be able to eliminate a variable. be able to eliminate a variable.

(2) (2) Add or SubtractAdd or Subtract the equations to eliminate one the equations to eliminate one variable.variable.

(3) (3) SolveSolve the resulting equation for the other variable.the resulting equation for the other variable.

(4) (4) Substitute Substitute in either original equation to find the value in either original equation to find the value of the eliminated variable. of the eliminated variable.

Equation 1 Equation 1

2x + 3y = 52x + 3y = 5 Equation 2Equation 2

6x + 5y = 19 6x + 5y = 19

““Solve Linear Systems by Elimination Solve Linear Systems by Elimination Multiplying First!!”Multiplying First!!”

Equation 2 Equation 2 2x + 3y = 52x + 3y = 5 Substitute value forSubstitute value fory into either of the y into either of the original equations original equations

2x + 3(-1) = 52x + 3(-1) = 5 2x - 3 = 52x - 3 = 5

The solution is the point (4,-1). The solution is the point (4,-1). Substitute (4,-1) into both Substitute (4,-1) into both

equations to check.equations to check.

6(4) + 5(-1) = 196(4) + 5(-1) = 1919 = 1919 = 19

2(4) + 3(-1) = 52(4) + 3(-1) = 55 = 55 = 5

Multiply Multiply

FirstFirst

++ -4y = 4-4y = 4

y = -1y = -1

x = 4x = 4

EliminatedEliminated

x (-3)x (-3) -6x – 9y = -15-6x – 9y = -15 6x + 5y = 19 6x + 5y = 19

Solution: (4,-1)

Equation 1 Equation 1

3x + 10y = -33x + 10y = -3 Equation 2Equation 2

2x + 5y = 3 2x + 5y = 3

““Solve Linear Systems by Elimination Solve Linear Systems by Elimination Multiplying First!!”Multiplying First!!”

Equation 1 Equation 1 2x + 5y = 32x + 5y = 3 Substitute value forSubstitute value forx into either of the x into either of the original equations original equations 2(9) + 5y = 32(9) + 5y = 3

18 + 5y = 318 + 5y = 3

The solution is the point (9,-3). The solution is the point (9,-3). Substitute (9,-3) into both Substitute (9,-3) into both

equations to check.equations to check.

2(9) + 5(-3) = 32(9) + 5(-3) = 33 = 33 = 3

3(9) + 10(-3) = -33(9) + 10(-3) = -3-3 = -3 -3 = -3

Multiply Multiply

FirstFirst

++ -x = -9-x = -9

x = 9x = 9

y = -3y = -3

EliminatedEliminated

x (-2)x (-2)

3x + 10y = -33x + 10y = -3 -4x - 10y = -6 -4x - 10y = -6

Solution: (9,-3)

Equation 1 Equation 1

-3x + 2y = -9-3x + 2y = -9 Equation 2Equation 2

4x + 5y = 35 4x + 5y = 35

““Solve Linear Systems by Elimination Solve Linear Systems by Elimination Multiplying First!!”Multiplying First!!”

Equation 1 Equation 1 4x + 5y = 354x + 5y = 35 Substitute value forSubstitute value forx into either of the x into either of the original equations original equations 4(5) + 5y = 354(5) + 5y = 35

20 + 5y = 3520 + 5y = 35

The solution is the point (5,3). The solution is the point (5,3). Substitute (5,3) into both Substitute (5,3) into both

equations to check.equations to check.

4(5) + 5(3) = 354(5) + 5(3) = 3535 = 3535 = 35

-3(5) + 2(3) = -9-3(5) + 2(3) = -9-9 = -9-9 = -9

Multiply Multiply

FirstFirst

++ 23x = 11523x = 115

x = 5x = 5

y = 3y = 3

EliminatedEliminated

x (2)x (2)

15x - 10y = 4515x - 10y = 45 8x + 10y = 70 8x + 10y = 70

x (-5)x (-5)

Solution: (5,3)

Equation 1 Equation 1

6x + 13y = -96x + 13y = -9 Equation 2Equation 2

9x + 2y = 39 9x + 2y = 39

““Solve Linear Systems by Elimination Solve Linear Systems by Elimination Multiplying First!!”Multiplying First!!”

Equation 1 Equation 1 9x + 2y = 399x + 2y = 39 Substitute value forSubstitute value fory into either of the y into either of the original equations original equations 9x + 2(-3) = 399x + 2(-3) = 39

9x - 6 = 399x - 6 = 39

The solution is the point (5,-3). The solution is the point (5,-3). Substitute (5,-3) into both Substitute (5,-3) into both

equations to check.equations to check.

9(5) + 2(-3) = 399(5) + 2(-3) = 3939 = 3939 = 39

6(5) + 13(-3) = -96(5) + 13(-3) = -9-9 = -9-9 = -9

Multiply Multiply

FirstFirst

++ -35y = 105-35y = 105

y = -3y = -3

x = 5x = 5

EliminatedEliminated

x (2)x (2)

-18x - 39y = 27-18x - 39y = 27 18x + 4y = 78 18x + 4y = 78

x (-3)x (-3)

Solution: (5,-3)

Guided PracticeGuided Practice

x + y = 2x + y = 2

2x + 7y = 92x + 7y = 9

6x – 2y = 16x – 2y = 1

-2x + 3y = -5-2x + 3y = -5

(1,1)(1,1) (-0.5, -2)(-0.5, -2)

3x - 7y = 53x - 7y = 5

9y = 5x + 59y = 5x + 5

(-10,-5)(-10,-5)

HomeworkHomework

Workbook page 309-310Workbook page 309-310