# Module in Mathematics Ii

## Module in Mathematics Ii

• Date Submitted: 06/15/2013 5:45 PM
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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,...