lesson 3-5 systems of linear equations in two variables
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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.
CByAx
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 = CStep1: rewrite both equations in the form Ax + By = C
x – y = 1 3x + y = 11
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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.
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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-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
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add the equationsadd the equations
243 yx
2x 2x
Thus the Solution is (1, 2)
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
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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 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.
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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
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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.
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Home Work (1)
)8 ,10 ,12(
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8) 5x + 6y +8= 0 3x – 2y +16= 0
Solve each systemWritten Exercises .. page 129Written Exercises .. page 129
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10) 8x – 3y= 3 3x – 2y + 5= 0
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12) 3p + 2q= -29p – q= -6
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Substitution method
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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
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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
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. 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
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Home Work (2)
)18 ,20 ,24 ,26 ,28(
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18) 3x – 2y = 6 5x + 3y + 9= 0
Solve each systemWritten Exercises .. page 129Written 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 + 2x2)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 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
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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(
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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
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Home Work (3)
)14 ,16 ,30 ,32(
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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)
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Step2: Graph both equations on the same coordinate plane.Step2: Graph both equations on the same coordinate plane.
Step3: Estimate the coordinates of the point where the lines intersect.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 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.Step2: Graph both equations on the same coordinate plane.
Step3: Estimate the coordinates of the point where the lines intersect.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 + 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
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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)