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Copyright © 2011 Pearson, Inc.
7.1Solving
Systems of Two Equations
Slide 7.1 - 2 Copyright © 2011 Pearson, Inc.
What you’ll learn about
The Method of Substitution Solving Systems Graphically The Method of Elimination Applications
… and whyMany applications in business and science can be modeled using systems of equations.
Slide 7.1 - 3 Copyright © 2011 Pearson, Inc.
Solution of a System
A solution of a system of two equations in two
variables is an ordered pair of real numbers that
is a solution of each equation.
Slide 7.1 - 4 Copyright © 2011 Pearson, Inc.
Example Using the Substitution Method
Solve the system using the substitution method.
2x −y=106x+ 4y=1
Slide 7.1 - 5 Copyright © 2011 Pearson, Inc.
Example Using the Substitution Method
Solve the first equation for y.
2x −y=10y=2x−10Substitute the expression for y into the second equation:6x+ 4(2x−10) =1
x=4114
y=2x−10
=24114⎛
⎝⎜⎞
⎠⎟−10
=−297
Solve the system using the substitution method.
2x −y=106x+ 4y=1
Slide 7.1 - 6 Copyright © 2011 Pearson, Inc.
Example Using the Substitution Method
6x + 4(2x−10) =1
x=4114
y=2x−10
=24114⎛
⎝⎜⎞
⎠⎟−10
=−297
The solution is the ordered pair
41
14,−
297
⎛
⎝⎜⎞
⎠⎟.
Slide 7.1 - 7 Copyright © 2011 Pearson, Inc.
Example Solving a Nonlinear System Algebraically
Solve the system algebraically.
y =x2 +6xy=8x
Slide 7.1 - 8 Copyright © 2011 Pearson, Inc.
Example Solving a Nonlinear System Algebraically
Substitute the values of y from the first equation into
the second equation:
8x =x2 +6x
0 =x2 −2xx=0, x=2.If x=0, then y=0. If x=2, then y=16.The system of equations has two solutions: (0,0) and (2,16).
y =x2 +6xy=8x
Slide 7.1 - 9 Copyright © 2011 Pearson, Inc.
Example Using the Elimination Method
Solve the system using the elimination method.
3x + 2y=124x−3y=33
Slide 7.1 - 10 Copyright © 2011 Pearson, Inc.
Example Using the Elimination Method
Multiply the first equation by 3 and the second
equation by 2 to obtain:
9x +6y=368x−6y=66Add the two equations to eliminate the variable y.
17x=102 so x=6
Solve the system using the elimination method.
3x + 2y=124x−3y=33
Slide 7.1 - 11 Copyright © 2011 Pearson, Inc.
Example Using the Elimination Method
Substitue x =6 into either of the two original equations:3(6) + 2y=12
2y=−6y=−3
The solution of the original system is (6,−3).
Solve the system using the elimination method.
3x + 2y=124x−3y=33
Slide 7.1 - 12 Copyright © 2011 Pearson, Inc.
Example Finding No Solution
Solve the system:
3x + 2y=5−6x−4y=10
Slide 7.1 - 13 Copyright © 2011 Pearson, Inc.
Example Finding No Solution
Multiply the first equation by 2.
6x + 4y=10−6x−4y=10Add the equations:
0 =20The last equation is true for no values of x and y.The equation has no solution.
Solve the system:
3x + 2y=5−6x−4y=10
Slide 7.1 - 14 Copyright © 2011 Pearson, Inc.
Example Finding Infinitely Many Solutions
Solve the system.
3x +6y=−109x+18y=−30
Slide 7.1 - 15 Copyright © 2011 Pearson, Inc.
Example Finding Infinitely Many Solutions
Multiply the first equation by −3.−9x−18y=309x+18y=−30Add the two equations.
0 =0The last equation is true for all values of x and y.The system has infinitely many solutions.
Solve the system.
3x +6y=−109x+18y=−30
Slide 7.1 - 16 Copyright © 2011 Pearson, Inc.
Quick Review
1. Solve for y in terms of x. 2x +3y=6 Solve the equation algebraically.
2. x3 =9x 3. x2 +5x=64. Write the equation of the line that contains the point
(1,1) and is perpendicular to the line 2x+3y=6.5. Write an equation equivalent to x+ y=5 withcoefficient of x equal to −2.
Slide 7.1 - 17 Copyright © 2011 Pearson, Inc.
Quick Review
1. Solve for y in terms of x. 2x +3y=6 y=−23x+ 2
Solve the equation algebraically.
2. x3 =9x 0,±3 3. x2 +5x=6 −6,14. Write the equation of the line that contains the point
(1,1) and is perpendicular to the line 2x+3y=6.
y−1=32(x−1)
5. Write an equation equivalent to x+ y=5 withcoefficient of x equal to −2. −2x−2y=−10
