Solving Two-VariableSystems of Linear Equations
Digital Lesson
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A solution of such a system is an ordered pair which is a solutionof each equation in the system.
Example: The ordered pair (4, 1) is a solution of the system since 3(4) + 2(1) = 14 and 2(4) – 5(1) = 3.
System of Linear Equations
A set of linear equations in two variables is calleda system of linear equations.
3x + 2y = 14
2x + 5y = 3
Example: The ordered pair (0, 7) is not a solution of the systemsince 3(0) + 2(7) = 14 but 2(0) – 5(7) = – 35, not 3.
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A system of equations with at least one solution is consistent.
A system with no solutions is inconsistent.
Systems of linear equations in two variables have eitherno solutions, one solution, or infinitely many solutions.
Solutions of Linear Equations
y
x
infinitely manysolutions
y
x
no solutions
y
x
unique solution
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The ordered pair (1, 2) is the unique solution.
The system is consistent since it has solutions.
Example: Solve the System
y
x
x – y = –1
2x + y = 4
(1, 2)
To solve the system by the graphingmethod, graph both equations and determine where thegraphs intersect.
x – y = –1
2x + y = 4
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The system has no solutions and is inconsistent.
The lines are parallel and have no point of intersection.
Inconsistent Solutions
y
x
x – 2y = – 4
3x – 6y = 6
Example: Solve the system by thegraphing method.
x – 2y = – 4
3x – 6y = 6
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The graphs of the two equations are the same line and theintersection points are all the points on this line.
The system has infinitely many solutions.
Infinitely Many Solutions
y
x
x – 2y = – 4
3x – 6y = – 12
Example: Solve the system by the graphingmethod.
x – 2y = – 4
3x – 6y = – 12
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To solve a system by the substitution method:
1. Select an equation and solve for one variable in terms of the other.
2. Substitute the expression resulting from Step 1 into the other equation to produce an equation in one variable.
3. Solve the equation produced in Step 2.
4. Substitute the value for the variable obtained in Step 3 into the expression obtained in Step 2.
5. Check the solution.
Definition: Algebraic Methods
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1. From the second equation obtain x = 3y – 3.
3. Solve for y to obtain y = 2.
4. Substitute 2 for y in x = 3y – 3 and conclude x = 3. The solution is (3, 2).
Example: Substitution Method
2. Substitute this expression for x into the first equation.
2(3y –3) + y = 8
5. Check:
2(3) – (2) = 8
(3) – 3(2) = –3
Example: Solve the system by the substitution method.
2x + y = 8
x – 3y = – 3
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1. From the first equation obtain y = 2x – 10.
2. Substitute 2x – 10 for y into the second equation to produce 4x – 2(2x – 10) = 8.
3. Attempt to solve for x.4x – 2(2x – 10) = 84x – 4x + 20 = 820 = 8 False statement
Because there are no values of x and y for which 20 equals 8, thissystem has no solutions.
Example: Substitution Method
Example: Solve the system by the substitutionmethod.
2x – y = 10
4x – 2y = 8
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To solve a system by the addition (or elimination) method:
1. Multiply either or both equations by nonzero constants to obtain opposite coefficients for one of the variables in the system.
2. Add the equations to produce an equation in one variable. Solve this equation.
3. Substitute the value of the variable found in Step 2 into either of the original equations to obtain another equation in one variable. Solve this equation.
4. Check the solution.
Definition: Addition Method
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2. Add the equations to obtain – 7x = –7. Therefore x = 1.
3. Substitute 1 for x in the first equation to produce5(1) + 2y = 112y = 6Therefore y = 3. The solution is (1, 3).
Example: Addition Method
4. Check:
5(1) + 2(3) = 11
3(1) + 4(3) = 15
1. Multiply the first equation by –2 to make the coefficients of y opposites.
– 10 x – 4y = – 22
3x + 4y = 15
Example: Solve the system by the additionmethod.
5x + 2y = 11
3x + 4y = 15
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Let x = speed of the plane in calm air
y = speed of the tailwind
Use the formula Rate × Time = Distance.
Example: A plane with a tailwind flew 1920 mi in 8 hours. On thereturn trip, against the wind, the plane flew the same distance in 12hours. What is the speed of the plane in calm air and the speed ofthe tailwind?
Use the addition method to solve the system.
12(x - y)
12
x - y
Against Wind
8(x + y)
8
x + y
With Wind
Distance
Time
Rate
Example: Application Problem
This yields a system of equations in x and y.
8x + 8y = 1920
12x – 12y = 1920
Example continues
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2. Add the equations to obtain 48 x = 9600.
1. Multiply the first equation by 3 and the second equation by 2.
3. Substitute 200 for x in the first equation. 8(200) + 8y = 1920
In 12 hours with an airspeed of 160 mph the plane willtravel 12 × 160 = 1920 on the return leg of its flight.
4. In 8 hours with an airspeed of 240 mph the plane will travel 8 × 240 = 1920 mi on the first leg of its flight.
Example continued
Therefore x = 200.
y = 40
Example continued: Solve the systemusing the addition method.
8x + 8y = 1920
12x – 12y = 1920
24x + 24y = 5760
24x – 24y = 3840
3(8x + 8y) = 3(1920)
2(12x – 12y) = 2(1920)