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Lesson 3-5 Systems of linear Equations in Two variables

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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

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Linear combination Method

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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 = C x – y = 1 3x + y = 11

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

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Example 2 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. 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

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

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Example 3 Solve by the linear combination method Solution 2x – 4y = -6 5x + 3y = 11 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 Adjust the coefficients of the variables so that the x-terms or the y-terms will cancel out. Adjust the coefficients of the variables so that the x-terms or the y-terms will cancel out.

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

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

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

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

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10) 8x – 3y= 3 3x – 2y + 5= 0

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12) 3p + 2q= -2 9p – q= -6

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Substitution method

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2y Example Solve by the substitution method To avoid having fractions in the substitution process, lets choose the 2 nd 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. Step2: Substitute into the other equation and solve

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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)

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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. Step2: Substitute into the other equation and solve Step3: Back-substitute the value found into the other equation

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

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

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20) 6x = 4y + 5 6y = 9x – 5

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24) 2x + y = 2 – x x + 2y = 2 + y

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26) x + y = 4(y + 2) x – y = 2(y + 4)

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

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Graphing method

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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 2y = -x + 4 -x + y = -1 y = x – 1 -x 222 x x

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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. bb

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system of linear equations C ONCEPT S UMMARY y x y x Lines intersect one solution Lines are parallel no solution y x Lines coincide infinitely many solutions Consistent system Inconsistent system dependent system

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

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2x + y = -2 2x – 3y = 15 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 )

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Step2: Graph both equations on the same coordinate plane. Step3: Estimate the coordinates of the point where the lines intersect. 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.

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

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Step2: Graph both equations on the same coordinate plane. Step3: Estimate the coordinates of the point where the lines intersect. 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.

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3x = 4y + 8 3y = 4x + 8 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 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.

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

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