7.3 solving equations with rational expressions7.3 solving equations with rational...
TRANSCRIPT
7.3 Solving Equations with Rational Expressions
Solving equations with rational expressions involves first and foremost, finding the least common denominator of the whole equation. Once you find the LCD, then you multiply each term in the numerator by it. To solve equations with rational expressions, it is a 4 step process:
STEP 1: Find the LCD of the whole equation
STEP 2: Multiply each term in the equation by the LCD
STEP 3: Simplify each fraction. If you multiply by the correct LCD, there will be no fractions left in the equation.
STEP 4: Determine what type of equation you have, and solve the equation appropriately.
Ex. Solve the following equations:
o. !!− 2+ !
!= −5 STEP 1: The LCD of 4 and 3 is 12
12 !!− 2+ !
!= −5 STEP 2: Multiply each term by 12
!"!!− 24+ !"!
!= −60
3m – 24 + 4m = -‐60 STEP 3: Simplify each fraction
7m – 24 = -‐60 STEP 4: This is a linear equation, +24 +24 so solve accordingly.
7m = -‐36 7 7
m = − !"!
p. !!!= !
!− !
!"STEP 1: The LCD of 2x, x and 12 is
12x
12! !!!= !
!− !
!"STEP 2: Multiply each term by 12x
!"!!!
= !"!!− !"!
!"
30 = 24 – x STEP 3: Simplify each fraction
30 = 24 – x STEP 4: This is a linear equation, -‐24 -‐24 solve accordingly.
6 = -‐x -‐1 -‐1
-‐6 = x
q. 10− !!!= − !
! STEP 1: The LCD of x2 and x is x2
!! 10− !!!= − !
!STEP 2: Multiply each term by x2
10!! − !!!
!!= − !!
!
10x2 -‐ 3 = -‐x STEP 3: Simplify each fraction
10x2 – 3 = -‐x STEP 4: This is a quadratic equation,
10x2 + x – 3 = 0 so solve accordingly
(5x + 3)(2x -‐1) = 0
5x + 3 = 0 2x – 1 = 0 -‐3 -‐3 +1 +1 5x = -‐3 2x = 1 5 5 2 2
x = − !! !"# !
!
q. !!!!
= !!!!
STEP 1: The LCD of a+1 and a+2
is (a+1)(a+2)
! + 1 (! + 2) !!!!
= !!!!
STEP 2: Multiply each term by
(a+1)(a+2)
! !!! (!!!)!!!
= ! !!! (!!!)!!!
5(a+2) = 4(a+1) STEP 3: Simplify each fraction
5(a+2) = 4(a+1) 5a + 10 = 4a + 4 STEP 4: This is a linear equation -‐4a -‐4a so solve accordingly
a + 10 = 4 -‐10 -‐10 a = -‐6
r. !!!!
+ !!!!
= !!!!!!!!!"
STEP 1: Note first, that the
denominator of the fraction on the right hand side can be factored. If this is possible, it is important to factor first because that will help us find the LCD.
!!!!
+ !!!!
= !!!!! (!!!)
Now we can find the LCD. The LCD
of (x+5), (x+4), and (x+5)(x+4) is (x + 5)(x + 4)
! + 5 (! + 4) !!!!
+ !!!!
= !!!!! (!!!)
STEP 2: Multiply each term by
(x+5)(x+4)
! !!! (!!!)!!!
+ ! !!! (!!!)!!!
= !! !!! (!!!)!!! (!!!)
2(x+4) + 3(x+5) = 2x STEP 3: Simplify each fraction
2(x+4) + 3(x+5) = 2x 2x + 8 + 3x + 15 = 2x STEP 4: This is a linear equation, so solve accordingly.
2x + 8 + 3x + 15 = 2x
5x + 23 = 2x -‐5x -‐5x 23 = -‐3x -‐3 -‐3
− !"!= !
s. !!!!!!!!"
+ !!!!!!!!
= 1 STEP 1: We begin by factoring the
denominators so we can find the LCD.
!!!!!(!!!)
+ !!!!!(!!!)
= 1 The LCD of 5(x+2) and 4(x+2) is
20(x+2).
20(! + 2) !!!!!(!!!)
+ !!!!!(!!!)
= 1 STEP 2: Multiply each term by
20(x+2)
!" !!! (!!!!)!(!!!)
+ !" !!! (!!!!)!(!!!)
= 1×20(! + 2)
4(2x – 3) + 5(3x – 2) = 20(x + 2) STEP 3: Simplify each fraction
8x -‐ 12 + 15x -‐ 10 = 20x + 40 STEP 4: This is a linear equation, 23x -‐ 22 = 20x + 40 so solve accordingly. -‐20x -‐20x
3x -‐ 22 = 40 +22 +22
3x = 62 3 3
x = !"!
4.7: Applications of Rational Equations
In this section, we discuss numerous applications of rational equations. Now that we know how to solve rational equations, we need to discuss how to set up word problems using rational equations, and then review how to solve them. Consider the examples below:
t. One number is four times another. The sum of the reciprocals of the numbers is !!.
Find the numbers.
If we let the first number equal m, then the second number (which is four times the first number is 4m).
One number = m
Second number = 4m
The information given states the sum of the reciprocals. The reciprocal of m is !! and
the reciprocal of 4m is !!!
. Sum implies addition. The sum of the reciprocals is 5/8 translates
into the following equation:
!!+ !
!!= !
!
Now that we have the equation, we must solve it. The LCD of m, 4m and 8 is 8m, so we multiply each term in the equation by 8m:
8! !!+ !
!!= !
! !!
!+ !!
!!= !"!
!
Simplifying this give us:
8 + 2 = 5m
10 = 5m 5 5
2 = m
If one of the numbers is 2, and the second number is four times the first number, so the second number is 8. Therefore, the two numbers are 2 and 8.
u. The sum of a number and three times its reciprocal is !"!. Find the numbers.
If we let the first number equal x, then its reciprocal is 1/x, and three times its reciprocal
is 3 !!= !
!. Then, the sum of a number an three times its reciprocal is !"
! translates to:
! + !!= !"
!
To solve this equation, we first have to find the LCD. The LCD of x and 4 is 4x. We then multiply each term in the equation by 4x.
4! ! + !!= !"
! 4!! + !"!
!= !"!
!
Simplifying this gives us,
4!! + 12 = 19!
This is a quadratic equation, so we rewrite in standard form for solving a quadratic equation, and we get:
4x2 – 19x + 12 = 0
Factoring, we get:
4x2 -‐ 3x – 16x + 12 = 0
x(4x -‐3) – 4(4x – 3) = 0
(x – 4)(4x – 3) = 0
X – 4 = 0 and 4x – 3 = 0
X = 4 and x = ¾
The answer is either 4 ( 4 plus three times its reciprocal which is ¾ equals !"! )
Or ¾ ( ¾ plus three times its reciprocal which is 4 equals !"! ).
Distance = Rate times Time problems
"Distance" word problems, often also called "uniform rate" problems, involve something travelling
at some fixed and steady ("uniform") pace ("rate" or "speed"), or else moving at some average speed.
Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of
the distance equation, d = rt, where d stands for distance, r stands for the (constant or average) rate of
speed, and t stands for time.
Warning: Make sure that the units for time and distance agree with the units for the rate. For
instance, if they give you a rate of feet per second, then your time must be in seconds and your distance
must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and
convert to the correct units.
For the next two examples, we will use the Distance = rate × time equations. We begin with a set up of a table:
We can use the above chart to help us fill in what we don’t know, and to determine what we need to know. We will be given two out of the three pieces of information in distance, rate and time, and each problem will be divided into 2 general parts. Take, for example, the following problems:
Distance Rate Time Part 1 Part 2
v. A train travels 30 mph faster than a car. If the train covers 120 miles in the sameamount of time the car travels 80 miles, what is the speed of each of them?
We know first of all, that the two “parts” we are breaking this problem up into is car and train. Let’s start with information we can determine from the train. First, we know that the speed (or rate) of the train is 30 miles per hour faster than the car. If the car has a rate of x, then the rate of the train is 30 + x or x + 30
We also know that the train covers 120 miles in the same amount of time that the car travels 80 miles, so that the distance covered by the train is 120, and the distance covered by the car is 80
How does time relate in this problem? Well, the important part is that the train covers the distance in the SAME AMOUNT OF TIME that the car travels. This means that the time the train travels and the time the car travels are EQUAL!
So, we need to rewrite each of these equations in terms of time. If D = R x T, then
!!= !. So, we can express the time as a ratio of distance divided by rate.
Distance Rate Time
Train Car
Distance Rate Time
Train x + 30 Car x
Distance Rate Time
Train 120 x + 30 Car 80 x
Knowing that the times are the same, set them equal to each other and solve:
120! + 30 =
80!
Multiply both sides of the equation by the LCD, the LCD of x + 30 and x is x(x + 30)
!(! + 30) !"#!!!"
= !"!
! !!!" !"#!!!"
= ! !!!" !"!
120x = 80(x+30)
120x = 80x + 2400 -‐80x -‐80x
40x = 2400 40 40
X = 60
Recall that x is the speed of the car, so that the car travels 60 miles per hour, and that the train travels x + 30 or 60 + 30 or 90 miles per hour.
w. A boat, which moves at 18 miles per hour in still water, travels 14 miles downstreamin the same amount of time it takes to travel 10 miles upstream. Find the speed of the current.
For this example, we are breaking the two parts we are breaking the problem up into are upstream and downstream. Let’s review other information.
Distance Rate Time
Train 120 x + 30 120! + 30
Car 80 x 80!
Distance Rate Time Upstream Downstream
First, the problem asks us for the speed of the current. Let’s call the speed of the current x.
The current will either add to the speed (make it go quicker) if the boat is going downstream, or subtract from the speed (make it slower) if it is going upstream. If the boat moves at 18 miles per hour in still water (with no current), this means that the speed of the boat going upstream will be 18 – x ( or 18 mph minus the current’s speed) and the speed of the boat going downstream will be 18 + x ( or 18 mph plus the current’s speed). Recall that speed is the RATE.
We also know that the boat travels 14 miles downstream in the same amount of time it takes to travel 10 miles upstream. So, we have distances:
What about the time? Well, we know that the boat travels 14 miles downstream In THE SAME AMOUNT OF TIME it takes to travel 10 miles upstream. So, the Time downstream = Time upstream. This means we have to write the equations in terms of time. From the above
example, we know that !!= !. So, we can express the time as a ratio of distance divided by
rate.
Distance Rate Time Upstream 18-x Downstream 18+x
Distance Rate Time Upstream 10 18-x Downstream 14 18+x
Distance Rate Time Upstream 10 18-x 10
18− ! Downstream 14 18+x 14
18+ !
Setting the times equal to each other, we get: 10
18− ! = 14
18+ !
Multiplying both sides by the LCD, which is (18-‐x)(18 + x), we get:
18− ! (18+ !) !"!"!!
= !"!"!!
!"!! (!"!!)!"!"!!
= !" !"!! (!"!!)!"!!
10(18 + x) = 14(18 – x)
10(18 + x) = 14(18 – x)
180 + 10x = 252 – 14x -‐10x -‐10x
180 = 252 – 24x -‐252 -‐252
-‐72 = -‐24x -‐24 -‐24
X = 3
This means that the speed of the current is 3 mph.