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. There are several methods of solving systems of linear equations - Substitution Method - Linear combination Method - Graphing Method

Linear combination Method

x – y = 1 3x + y = 11 Example 1 Solution Solve by the linear combination method x – y = 1 3x + y = 11 Solution Step1: rewrite both equations in the form Ax + By = C The two equations are already in the form Ax + By = C, so we go to the step2 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

x – y = 1 3x + y = 11 4x = 12 Thus the Solution is (3, 2) Step3: add the equations and solve x – y = 1 3x + y = 11 4x = 12 Divide both sides of the equation by 4 Step4: Back-substitute and find the other variable. Thus the Solution is (3, 2)

3x – 4y = 2 x – 2y = 0 3x – 4y = 2 -3x +6y = 0 Example 2 Solution Solve by the linear combination method 3x – 4y = 2 x – 2y = 0 Solution Adjust the coefficients of the variables so that the x-terms or the y-terms will cancel out. Multiply the second equation by (-3), This will set up the x-terms to cancel. 3x – 4y = 2 -3x +6y = 0

3x – 4y = 2 -3x +6y = 0 2y = 2 Thus the Solution is (1, 2) add the equations 3x – 4y = 2 -3x +6y = 0 2y = 2 Divide both sides of the equation by 2 Back-substitute and find the other variable. Thus the Solution is (1, 2)

2x – 4y = -6 5x + 3y = 11 Example 3 Solution Solve by the linear combination method 2x – 4y = -6 5x + 3y = 11 Solution Adjust the coefficients of the variables so that the x-terms or the y-terms will cancel out. Neither variable is the obvious choice for cancellation. However I can multiply to convert 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

2x – 4y = -6 5x + 3y = 11 -10x +20y = 30 10x + 6y = 22 26y = 52 add the equations and solve 2x – 4y = -6 5x + 3y = 11 Multiply by (-5) Multiply by (2) -10x +20y = 30 10x + 6y = 22 26y = 52 Divide both sides of the equation by 26

Back-substitute and find the other variable. Thus the Solution is (1, 2)

Home Work (1) (8, 10, 12) Page 129

Written Exercises .. page 129 Solve each system 8) 5x + 6y +8= 0 3x – 2y +16= 0

10) 8x – 3y= 3 3x – 2y + 5= 0

12) 3p + 2q= -2 9p – q= -6

Substitution method

3x – 4y = 2 x – 2y = 0 Example Solution Solve by the substitution method 3x – 4y = 2 x – 2y = 0 Solution Step1: Rewrite either equation for one variable in terms of the other. To avoid having fractions in the substitution process, let’s choose the 2nd equation and add (2y) to both sides of the equation. 2y Step2: Substitute into the other equation and solve

Thus the Solution is (1, 2) simplify Simplify like terms Divide both sides of the equation by 2 Step3: Back-substitute the value found into the other equation Thus the Solution is (1, 2)

x – y = 1 3x + y = 11 Your Turn Solve by the substitution method Step1: Rewrite either equation for one variable in terms of the other. Step2: Substitute into the other equation and solve Step3: Back-substitute the value found into the other equation

Home Work (2) (18, 20, 24, 26, 28) Page 129

Written Exercises .. page 129 Solve each system 18) 3x – 2y = 6 5x + 3y + 9= 0

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 + 2x 2(y + x) = 5 – 2y

Graphing method

x + 2y = 4 -x + y = -1 x + 2y = 4 -x + y = -1 2y = -x + 4 y = x – 1 Example Solve by Graphing x + 2y = 4 -x + y = -1 Solution Step1: Put both equations in slope-intercept form: y = mx + b x + 2y = 4 -x + y = -1 -x x 2y = -x + 4 y = x – 1 2

Step2: Graph both equations on the same coordinate plane. Graph b: on the y-axis Use m: rise then run and graph a second point b Draw a line: it should pass through the two points. Step3: Estimate the coordinates of the point where the lines intersect.

infinitely many solutions CONCEPT SUMMARY system of linear equations y x y x y x Lines intersect one solution Lines are parallel no solution Lines coincide infinitely many solutions Consistent system dependent system Inconsistent system

Home Work (3) (14, 16, 30, 32) Page 129

Written Exercises .. page 129 Graph both equations in the same coordinate system, then estimate the solution. 14) 2x + y = -2 2x – 3y = 15 Step1: Put both equations in slope-intercept form y = mx + b

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

16) 3x + 5y = 15 x – y = 4 Step1: Put both equations in slope-intercept form y = mx + b

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

30) 3x = 4y + 8 3y = 4x + 8 3x = 4y + 8 3y = 4x + 8 3x – 8 = 4y 3 4 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 3y = 4x + 8 write both equations in slope-intercept form y = mx + b 3x = 4y + 8 3y = 4x + 8 -8 3 3x – 8 = 4y 4 The Slopes are not equal so the two lines will intersect at one point, thus the system is Consistent.

32) 3x – 6y = 9 4x – 3y = 12 Put both equations in slope-intercept form y = mx + b