Copyright © 2011 Pearson, Inc. 7.1 Solving Systems of Two Equations.

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