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McGraw Hill Glencoe Algebra 2, 2012 View details

5. Law of Cosines

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Exercise 41 Page 828

Multiply both sides of the equation by the least common denominator (LCD).

4, 23/15

Practice makes perfect

We want to solve the given rational equation. 1/x - 1 + 5/8 = 23/6x Let's begin by highlighting all of the different factors in the denominators. This will help us find the least common denominator (LCD). 1/x - 1 + 5/8 = 23/6 x We will solve the equation by multiplying each side by the LCD to clear denominators.

1/x - 1 + 5/8 = 23/6x

MultEqn

LHS * (x-1) (x)=RHS* (x-1) (x)

(1/x - 1 + 5/8)* (x - 1) (x) = 23/6x* (x - 1) x

▼

Multiply

Distr

Distribute (x-1)(x)

1/x - 1(x - 1)(x) + 5/8(x - 1)(x) = 23/6x(x - 1)(x)

MoveRightFacToNum

a/c* b = a* b/c

(x - 1)(x)/x - 1 + 5/8(x - 1)(x) = 23(x - 1)(x)/6x

CancelCommonFac

Cancel out common factors

(x - 1)(x)/x - 1 + 5/8(x - 1)(x) = 23(x - 1)(x)/6x

SimpQuot

Simplify quotient

x + 5/8(x - 1)(x) = 23(x - 1)/6

MultEqn

LHS * 24=RHS* 24

24x + 120/8(x - 1)(x) = 552(x - 1)/6

CalcQuot

Calculate quotient

24x + 15(x - 1)(x) = 92(x - 1)

▼

Simplify

Distr

Distribute x

24x + 15(x^2 - x) = 92(x - 1)

Distr

Distribute 15

24x + 15x^2 - 15x = 92(x - 1)

Distr

Distribute 92

24x + 15x^2 - 15x = 92x - 92

SubTerm

Subtract term

15x^2 + 9x = 92x - 92

SubEqn

LHS-92x=RHS-92x

15x^2 - 83x = - 92

AddEqn

LHS+92=RHS+92

15x^2 - 83x + 92 = 0

Note that we have a quadratic equation now. Let's identify the values of a, b, and c. 15x^2 - 83x + 92 = 0 ⇕ 15x^2 +( - 83)x+ 92=0 We have that a= 15, b= - 83, and c= 92. Let's substitute these values into the Quadratic Formula and solve for x.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 83)±sqrt(( - 83)^2-4( 15)( 92))/2( 15)

▼

Simplify right-hand side

NegNeg

- (- a)=a

x=83±sqrt((- 83)^2-4(15)(92))/2(15)

CalcPowProd

Calculate power and product

x=83±sqrt(6889-5520)/30

SubTerm

Subtract term

x=83±sqrt(1369)/30

CalcRoot

Calculate root

x = 83 ± 37/30

Let's calculate both solutions by using the positive and negative signs.

x = 83 ± 37/30
x=83 + 37/30	x=83 - 37/30
x=120/30	x=46/30
x=4	x=23/15

Finally, we will check if either x=4 or x= 2315 is an extraneous solution. To do this, we need to substitute 4 and 2315 for x in the original equation. Let's start with x = 4.

1/x - 1 + 5/8 = 23/6x

Substitute

x= 4

1/4 - 1 + 5/8 ? = 23/6( 4)

▼

Simplify

Multiply

1/4 - 1 + 5/8 ? = 23/24

SubTerm

Subtract term

1/3 + 5/8 ? = 23/24

ExpandFrac

a/b=a * 8/b * 8

8/24 + 5/8 ? = 23/24

ExpandFrac

a/b=a * 3/b * 3

8/24 + 15/24 ? = 23/24

AddFrac

Add fractions

23/24 = 23/24 ✓

Now we will check our second solution, x= 2315.

1/x - 1 + 5/8 = 23/6x

Substitute

x= 2315

1/2315 - 1 + 5/8 ? = 23/6( 2315)

▼

Simplify

OneToFrac

Rewrite 1 as 15/15

1/2315 - 1515 + 5/8 ? = 23/6( 2315)

SubFrac

Subtract fractions

1/815 + 5/8 ? = 23/6( 2315)

ReduceFrac

a/b=.a /23./.b /23.

1/815 + 5/8 ? = 1/6( 115)

MoveLeftFacToNumOne

a* 1/b= a/b

1/815 + 5/8 ? = 1/615

ReduceFrac

a/b=.a /3./.b /3.

1/815 + 5/8 ? = 1/25

Recall that dividing by a fraction is the same as multiplying by its reciprocal. 1/815 + 5/8 ? = 1/25 ⇕ 1( 15/8 ) + 5/8 ? = 1(5/2 ) Let's simplify to finally see whether the equation is satisfied when x= 2315.

1( 15/8 ) + 5/8 ? = 1(5/2 )

▼

Simplify

MultByOne

\a * 1=a

15/8 + 5/8 ? = 5/2

AddFrac

Add fractions

20/8 ? = 5/2

ReduceFrac

a/b=.a /4./.b /4.

5/2 = 5/2 ✓