- Submitted By: beadatay
- Date Submitted: 06/15/2013 5:45 PM
- Category: Science
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MODULE IN INTERMEDIATE ALGEBRA

3RD QUARTER

PREPARED BY: Ma. Beatriz Ane G. Datay

I. SOLVING WORD PROBLEMS INVOLVING RATIONAL ALGEBRAIC EQUATIONS

A. Solving Number Problems Involving Rational Algebraic Expressions

1. The denominator of a fraction is 2 more than the numerator. If 1 is subtracted from both numerator and denominator, the resulting fraction has a value of12. Find the original fraction.

Let us now write the representations,

Solution:

Let x = numerator

Let x + 2 = denominator, because it is said in the problem that the denominator of a fraction is 2 more than the numerator, the representation for the numerator is x.

xx+2 = the original number or the original fraction.

Translate the worded problems to equations,

Equations:

=

x-1x+2-1

12

the resulting fraction is

If 1 is subtracted from both the

numerator and denominator

Thus, x-1x+1=12

Multiply both sides by the LCD, 2(x+1).

2x+1x-1x+1=2x+112, so it will become,

2x-1=x+1,

Applying distributive property,

2x-2=x+1,

x = 3

Substituting x by its value which is 3 the original fraction is xx+2= 35

2. If two-fifths of a certain number is subtracted from twice a number, the result is 4. Find the number.

Let us now write the representations,

Solution:

Let a = representation of the number

2a-25a = the number

Translate the worded problems to equations,

=

2a-25a

4

the result is

Equations:

If two-fifths of a certain number is

subtracted from twice a number

2a-25a=4

Checking:

a=52

2a-25a=4

252-2552=4

5-1=4

4=4

the next thing we are going to do is to multiply both sides by 5 to eliminate the denominator which is also 5.

52a-25a=(4)(5)

Using the distributive property it will become,

10a-2a=20, thus,

8a=20,

Dividing both sides by 8, we will get,

8a8=208=a=208

a=208 or 52

3. Find two consecutive integers such that the sum of one-fourth of the first and one-fifth of the second is 11.

Let us now write the representations,...