Word Problems For Rational Expressions
The rational equation is the one of the most important topic (concept) in algebra chapter
. The word rational arises from term ratio. We know that a ratio like 5:3 can also be written as 5/3. Here 5 and 3 are natural numbers. Similarly, the ratio of two integers x and y (y '!=' 0), which is x:y can be written in the form x/y. this is the form which rational numbers are expressed. Not only in mathematics, as well as physics and mechanics subjects also used for rational equations and word problems in many places.
Algebra Rational Word Problems:
This equation has used for many applications in more fields in mathematics subject. In algebra we have to use this rational word problem in cosine law, tangent law and lines law. These laws are usually expressed as rational equations. In the algebraic chapter majority of the application problems involves in rational equations, and most of the mechanical problems involves in skill manipulations of rational equations in the word problems in algebra.
Example Word Problem for Algebra Rational Expression:
A carpenter worked on a job for 10 days and is then joined by an assistant. both of them can finish the job in 6 more days. The assistant could have done the job alone in 30 days. How long would it have taken the carpenter to do the job alone?
Solution:
First we have considered the amount of work done per day.
The carpenter can do the job in k days.
In one day, he can do 1 / k of the work.
In x days, he can do x / k of the work.
The assistant can do the work in 30 days.
In one day he can do 1 / 30 of the work.
In x days, he can do x / 30 of the work.
The carpenter worked alone for 10 days.
So he already did 10 / k of the work part of work.
Then he and the assistant worked together for 6 days.
They did: (6 / k + 6 / 30) of the workthe rest of work.
So we have: 10/ k + (6/k + 1/5) = 1 the full (whole) work, this is our equation, so we solve for k
Therefore, 10 / k + 6 / k = 16 / k
16 / k + 1 / 5 = 1
16 / k = 1 1 / 5
16 / k = 4 / 5
(Cross multiply the above values) and we get,
4k = 80
So, k = 20.
Answer is: K = 20.
So the carpenter is worked alone, he will be do it alone in 20 days.
Introduction to word problems for rational expressions:
The rational expressions are the algebraic expressions where they are represented in the form 'P/Q' , where P and Q are usually the polynomial expressions and Q value is not equal to zero from the given expressions. In this, the problems are given in the sentence formats where it has to be converted to equations and to be solved. Now we see about the word problems rational expressions.
About Word Problems for Rational Expressions:
It is known that the rational expressions in algebra are undefined as the number is divided by zero. The rational expression which does not includes all the values that make the denominator to be 0.
The word problems are represented in the sentence formation where it has to be solved by forming the equations and the equations has been simplified. In the equations, the expressions will be in the form of the fractional. The equations will be in the form of the fractions. The word problems must be understand correctly or it will gives the wrong solutions.
Word Problems for Rational Expressions:
Example 1:
A farmer has some butter, which is 24% fat and some butter that is 18% fat. How many quarts of each must he use to produce 90 quarts of butter, which is 22% fat?
Solution:
Let us take x for the quarts of butter with 24% fat, and y denotes the quarts of butter with 18% fat. We know that:
x + y = 90
Now we solve for x as follows,
x = 90 - y
We know that the equation is written as
0.24 x + 0.18 y = 0.22 * 90
Now substitute in x from above and we have to solve for y
0.24 (90 - y) + 0.18 y = 19.8
21.6 - 0.24y + 0.18 y = 19.8
-0.06y = 19.8 - 21.6
-0.06y = -1.8
y = 30
Now we have to find for x
x = 90 - y
x = 90 - 30
x = 60
Example 2:
Peter climbed up a hill at 6 km/h and then climbed down at 10 km/h. Determine that how many kilometers did he travel in all if his total climbing time was 1 and '1/3' hours?
Solution:
Let us consider x as the time climbing up hill and y the time climbing going down the hill. We know
x + y = 1 '1/3'
x + y = '4/3'
Now we solve for x value,
x = '4/3' - y
We know that the distance travelled up and down are same, so we have to find as follows,
6x = 10y
6x - 10y = 0
Now substitute value in x and solve for y value as follows,
6(4/3 - y) - 10y = 0
8 - 6y - 10y = 0
-16y = -8
y = '8/16'
y = '1/2'
pre algebra practice testWe don't need to find the value of x, but we have to do it as follows,
x = '4/3' - y
x = '4/3' - '1/2'
x = '(4*2 - 1*3) / 6'
x = '(8 - 3) / 6'
x = '5/6'
Total distance is the sum of distance going up and down
Distance = 6x + 10y
= 6'(5/6)' + 10'(1/2)'
= 5 + 5
= 10
by: johnharmer
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