Lesson 3-5 Systems of linear Equations in Two variables

Lesson 3-5

Systems of linear Equations in Two variables

A system of two linear equations consists of two equations that can be written in the form:

A solution of a system of linear equations is an ordered pair (x, y) that satisfies each equation.

CByAx

There are several methods of solving systems of linear equations

- Substitution Method

-Linear combination Method

-Graphing Method

Linear combination Method

Example 1 Solve by the linear combination method

The two equations are already in the form Ax + By = C, so we go to the step2

Solution

Step2: Adjust the coefficients of the variables so that the x-terms or the y-terms will cancel out. Step2: Adjust the coefficients of the variables so that the x-terms or the y-terms will cancel out.

Note that the coefficients of the y-terms are already adjusted and will cancel out when the two equations are added. so we go to step3

Step1: rewrite both equations in the form Ax + By = CStep1: rewrite both equations in the form Ax + By = C

x – y = 1 3x + y = 11

Step3: add the equations and solveStep3: add the equations and solve

1 yx

2y 2y

Thus the Solution is (3, 2)

4x = 12 Divide both sides of the equation by 4

3x 3x

13 y

2 y

x – y = 1 3x + y = 11

Step4: Back-substitute and find the other variable.Step4: Back-substitute and find the other variable.


3x – 4y = 2 x – 2y = 0

Solution

Adjust the coefficients of the variables so that the x-terms or the y-termswill cancel out. Adjust the coefficients of the variables so that the x-terms or the y-termswill cancel out.

Multiply the second equation by (-3), This will set up the x-terms to cancel.

3x – 4y = 2 -3x +6y = 0

add the equationsadd the equations

243 yx

2x 2x


3x – 4y = 2 -3x +6y = 0

2y = 2 Divide both sides of the equation by 2

1y 1y

Back-substitute and find the other variable.Back-substitute and find the other variable.

2)1(43 x

243 x

63 x


Solution

2x – 4y = -6 5x + 3y = 11

Neither variable is the obvious choice for cancellation. However I can multiply toconvert the x-terms to 10x or the y-terms to 12y. Since I'm lazy and 10 is smaller than 12, I'll multiply to cancel the x-terms. I will multiply the first equation by (-5)and the second row by (2); then I'll add down and solve

Adjust the coefficients of the variables so that the x-terms or the y-termswill cancel out. Adjust the coefficients of the variables so that the x-terms or the y-termswill cancel out.

add the equations and solveadd the equations and solve

Multiply by (-5)2x – 4y = -6 5x + 3y = 11 Multiply by (2)

-10x +20y = 30 10x + 6y = 22

26y = 52

2y 2y

Divide both sides of the equation by 26

642 yx

Thus the Solution is (1, 2)682 x

6)2(42 x

22 x

1x 1x

Back-substitute and find the other variable.Back-substitute and find the other variable.

Home Work (1)

)8 ,10 ,12(

Page 129

8) 5x + 6y +8= 0 3x – 2y +16= 0

Solve each systemWritten Exercises .. page 129Written Exercises .. page 129

10) 8x – 3y= 3 3x – 2y + 5= 0

12) 3p + 2q= -29p – q= -6

Substitution method

2y 2y

Example Solve by the substitution method

To avoid having fractions in the substitution process, let’s choose the 2nd equation and add (2y) to both sides of the equation.

3x – 4y = 2 x – 2y = 0

Solution

Step1: Rewrite either equation for one variable in terms of the other. Step1: Rewrite either equation for one variable in terms of the other.

02 yx yx 2

Step2: Substitute into the other equation and solve Step2: Substitute into the other equation and solve

243 yx 24)2(3 yy

24)2(3 yy simplify

246 yy Simplify like terms

22 y Divide both sides of the equation by 2

1y 1y

Step3: Back-substitute the value found into the other equation Step3: Back-substitute the value found into the other equation

yx 2

)1(2x

2x 2x


Your Turn Solve by the substitution method

x – y = 1 3x + y = 11

Step1: Rewrite either equation for one variable in terms of the other. Step1: Rewrite either equation for one variable in terms of the other.

Step2: Substitute into the other equation and solveStep2: Substitute into the other equation and solve

Step3: Back-substitute the value found into the other equation Step3: Back-substitute the value found into the other equation

Home Work (2)

)18 ,20 ,24 ,26 ,28(

Page 129

18) 3x – 2y = 6 5x + 3y + 9= 0

Solve each systemWritten Exercises .. page 129Written Exercises .. page 129

20) 6x = 4y + 5 6y = 9x – 5

24) 2x + y = 2 – x x + 2y = 2 + y

26) x + y = 4(y + 2)x – y = 2(y + 4)

28) 2)y – x = (5 + 2x2)y + x = (5 – 2y

Graphing method

Example Solve by Graphing

x + 2y = 4 -x + y = -1

Solution

Step1: Put both equations in slope-intercept form: y = mx + b Step1: Put both equations in slope-intercept form: y = mx + b

x + 2y = 4

2y = -x + 4

-x + y = -1

y = x – 1

22

x

y

2

1m

2

1m

2b 2b

1

1m

1

1m 1b 1b

-x -x

2 2 2

x x

Step2: Graph both equations on the same coordinate plane.Step2: Graph both equations on the same coordinate plane.

2

1m

2

1m

2b 2b

1

1m

1

1m

1b 1b

Graph b: on the y-axis

Use m: rise then run and graph a second point

Draw a line: it should pass through the two points.

Step3: Estimate the coordinates of the point where the lines intersect.Step3: Estimate the coordinates of the point where the lines intersect.

b

b

)1,2(

system of linear equations

CONCEPT

SUMMARY

y

x

y

x

Lines intersectone solution

Lines are parallelno solution

y

x

Lines coincideinfinitely many solutions

Consistent system Inconsistent system dependent system

Home Work (3)

)14 ,16 ,30 ,32(

Page 129

2x + y = -2 2x – 3y = 15

Step1: Put both equations in slope-intercept form y = mx + b Step1: Put both equations in slope-intercept form y = mx + b

Graph both equations in the same coordinate system, then estimate the solution.

Written Exercises .. page 129Written Exercises .. page 129

14)






3x + 5y = 15 x – y = 4

Step1: Put both equations in slope-intercept form y = mx + b Step1: Put both equations in slope-intercept form y = mx + b

16)






3x = 4y + 83y = 4x + 8

write both equations in slope-intercept form y = mx + b write both equations in slope-intercept form y = mx + b

Write each system in slope-intercept form. By Comparing the slopes and the y-intercepts, determine whether the equations are consistent or inconsistent.

30)

3x = 4y + 8

3x – 8 = 4y

3y = 4x + 8

24

3

xy

4

3m

4

3m

2b 2b

3

4m

3

4m

3

8b

3

8b

-8 -8

4 4 4 3

8

3

4

xy

3 3 3

The Slopes are not equal so the two lines will intersect at one point, thus the system is Consistent.The Slopes are not equal so the two lines will intersect at one point, thus the system is Consistent.

yx

24

3

3x – 6y = 9 4x – 3y = 12

Put both equations in slope-intercept form y = mx + b Put both equations in slope-intercept form y = mx + b

32)

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Lesson 3-5 Systems of linear Equations in Two variables