bmayer@chabotcollege.edu mth55_lec-34_sec_6-6_rational_equations.ppt 1 bruce mayer, pe chabot...

BMayer@ChabotCollege.edu • MTH55_Lec-34_sec_6-6_Rational_Equations.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

§6.6 Rational§6.6 RationalEquationsEquations

Review §Review §

Any QUESTIONS About• §6.4 → Complex Rational Expressions

Any QUESTIONS About HomeWork• §6.4 → HW-21

6.4 MTH 55

Solving Rational EquationsSolving Rational Equations

In previous Lectures, we learned how to simplify expressions. We now learn to solve a new type of equation. A rational equation is an equation that contains one or more rational expressions. Some examples:

2 7 5 8 6 2 58, , 4.

4 3 3 9

x x x x x x

We want determine the value(s) for x that make these Equations TRUE

To Solve a Rational EquationTo Solve a Rational Equation1. List any restrictions that exist.

Numbers that make a denominator equal 0 canNOT possibly be solutions.

2. CLEAR the equation of FRACTIONS by multiplying both sides by the LCM of ALL the denominators present

3. Solve the resulting equation using the addition principle, the multiplication principle, and the Principle of Zero Products, as needed.

4. Check the possible solution(s) in the original equation.

Example Example Solve Solve

SOLUTION - Because no variable appears in the denominator, no restrictions exist. The LCM of 5, 2, and 4 is 20, so we multiply both sides by 20

20 201

20 204

4 10 5x x 6 5x

Using the multiplication principle to multiply both sides by the LCM. Parentheses are important!

Using the distributive law. Be sure to multiply EACH term by the LCM

Simplifying and solving for x. If fractions remain, we have either made a mistake or have not used the LCM of ALL the denominators.

Checking AnswersChecking Answers

Since a variable expression could represent 0, multiplying both sides of an equation by a variable expression does NOT always produce an Equivalent Equation• COULD be Multiplying by Zero and

Not Know it

Thus checking each solution in the original equation is essential.

SOLUTION - Note that x canNOT equal 0. The Denominator LCM is 15x.

3 15x x

3 1515 15

x xx x

5 15 4x

3x 15 x

CHECKtentative Solution,x = 5

3 15x x

3( ) 15

15 5 151 3 4

15 15 154 4

The Solutionx = 5 CHECKS

SOLUTION - Note that x canNOT equal 0. The Denom LCM is x

12(7)x

x xx 12

2 12 7x x 2 7 12 0x x

( 3)( 4) 0x x ( 3) 0 or ( 4) 0x x

Thus by Zero Products:

x = 3 or

CHK: For x = 3 For x = 4

Both of these check, so there are two solutions; 3 and 4

SOLUTION Note that y canNOT equal 3 or −3. We multiply both sides of the equation by the Denom LCM.

( 3)( 3( 3)( 3) ( 3 ( 3

) 3 3) )y

y y y yy y y

( 3y )( 3y )5

( 3y )( 3y )

)( 3)3

( 3)( 3y y

5 3( 3) 2( 3)y y

3 9 25 6y y

9 3 3y y y

SOLUTION - Note that x canNOT equal 1 or −1. Multiply both sides of the eqn by the LCM

( 1)( 1) ( 1)( 12 5 4

1 1 ( 1)( 1)

)x x x x

x x x x

2( 1) 5( 1) 4x x

2 2 5 5 4x x

3 7 4x

Because of the restriction above, 1 must be rejected as a solution. This equation has NO solution.

SOLUTION: Because the left side of this equation is undefined when x is 0, we state at the outset that x 0.

Next, we multiply both sides of the equation by the LCD, 4x:

2 7 58.

Multiplying by the LCD to clear fractions

2 784 4

x xx x

SOLN cont.

2 7 58.

2 7 544 84

(2 7) 4 ( 5)32

Using the distributive law

Locating factors equal to 1

Removing factors equal to 1

Using the distributive law

(2 7) 4( 5) 32x x x

2 7 4 20 32x x x

SOLN cont.

2 7 58.

This should check since x 0.

6 13 32x x

13 26x12 x

CHECK 2 7 58

1 12 2

Rational Eqn Rational Eqn CAUTIONCAUTION

When solving rational equations, be sure to list any Division-by-Zero restrictions as part of the first step.

Refer to the restriction(s) as you proceed

SOLUTION: To find all restrictions and to assist in finding the LCD, we factor:

Note that to prevent division by zero x 3 and x −3.

Next multiply by the LCD, (x + 3)(x – 3), and then use the distributive law

8 6 2.

3 3 9x x x

8 6 2.

3 3 ( 3)( 3)x x x x

SOLUTION: By LCD Multiplication

Remove factors Equal to One and solve the resulting Eqn• Keep in Mind any restrictions

8 6 2.

3 3 9x x x

3 3 ( 3)( 3)( 3) ( 3)( 3)

( 3)x x x xx x x x

( 3)( 3) ( 3)8 6 ( 3)( 3)2

( 3)3 3 ( 3)( 3)

x x x xx x

x x x x

SOLN cont.: Multiply and Collect Similar terms

A check will confirm that 22 is the solution

8 6 2.

3 3 9x x x

8( 3) 6( 3) 2x x

8 24 6 18 2x x

2 42 2x

Example Example Eqn with NO Soln Eqn with NO Soln

To avoid division by zero, exclude from the expression domain

1 and –1, since these values make one or more of the denominators in the

equation equal 0.

Distributive property

Solve .3x – 1

=2x + 1

x2 – 1

x – 12

x + 1– 6

x2 – 1(x – 1)(x + 1)(x – 1)(x + 1)

x – 12

x + 1– 6

x2 – 1(x – 1)(x + 1)(x – 1)(x + 1) (x – 1)(x + 1)

=– 63(x + 1) 2(x – 1)

=– 63x + 3 2x + 2

= 6x + 5

Multiply each side by the LCD, (x –1)(x + 1).

Multiply.

Distributive property

Combine terms.

Subtract 5.

Example Example Eqn with NO Soln Eqn with NO SolnSolve .3

x – 1=2

x + 1– 6

x2 – 1

Since 1 is not in the domain, it cannot be a solution of the equation.

Substituting 1 in the original equation shows why.

Check: =3

x – 12

x + 1– 6

x2 – 1

1 – 12

1 + 1– 6

12 – 1

– 60

Since division by 0 is undefined, the given equation has no solution,

and the solution set is ∅.

Example Example Fcn to Eqn Fcn to Eqn

Given Function:

Find all values of a for which

5( ) .f x x

( ) 4.f a

SOLUTION On BoardOn Board By Function Notation:

5( )f a a

Thus Need to find all values of a for which f(a) = 4

Solve for a:

First note that a 0. To solve for a, multiply both sides of the equation by the LCD, a:

Multiplying both sides by a. Parentheses are important.

aa a Using the distributive law

CarryOutSolution

Simplifying2 5 4a a

2 4 5 0a a

( 5)( 1) 0a a

5 1a or a

Getting 0 on one side

Factoring

Using the principle of zero products

5( ) 5 1 4;

5( ) 1

1 1 41

STATE: The solutions are 5 and −1. For a = 5 or a = −1, we have f(a) = 4.

Rational Equations and GraphsRational Equations and Graphs

One way to visualize the solution to the last example is to make a graph. This can be done by graphing; e.g., Given

5( ) .f x x

Find x such that f(x) = 4

Graph the function, and on the same grid graph y = g(x) = 4

We then inspect the graph for any x-values that are paired with 4. It appears from the graph that f(x) = 4 when x = 5 or x = −1.

5( )f x x

Graphing gives approximate solutions Although making a graph is not the

fastest or most precise method of solving a rational equation, it provides visualization and is useful when problems are too difficult to solve algebraically

WhiteBoard WorkWhiteBoard Work

Problems From §6.6 Exercise Set• 34, 38, 62

Rational Expressions

All Done for TodayAll Done for Today

Remember:can NOTDivide by

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

AppendiAppendixx

srsrsr 22

Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls

-3 -2 -1 0 1 2 3 4 5

M55_§JBerland_Graphs_0806.xls -10

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls

bmayer@chabotcollege.edu mth55_lec-34_sec_6-6_rational_equations.ppt 1 bruce mayer, pe chabot...

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