# This will not have any fractions in

This lesson will explain how to solve algebra problems that include fractions by modifying the problems to make them easier. It will explain two methods that are closely related to one another.

## Overview: Changing Fractions to Integers

Fractions have always been a difficult obstacle for math students. Even for those who know how to handle fractions, integers (numbers that are not fractions) are still easier to use. Well, what if there was a way to change a problem that has fractions into one that doesn’t? This is your lucky day because that is exactly what we’re going to cover. Now, please realize this doesn’t mean fractions cannot be part of the answer, but it will eliminate them for much of the calculations.There are two concepts that we need to review.1) The top number of a fraction is called the numerator, and the bottom number is called the denominator. This denominator is the value you are dividing by.

For example, 3/4 means you are dividing 3 by 4.2) When you are solving an equation, the objective is to get the variable (letter) on one side of the equal sign by itself with a coefficient (number part) of a positive one. Remember, the positive one is understood and does not need to be written next to the variable. Continue to cancel out operations until this result is achieved.3) The least common multiple, LCM, of denominators is the smallest value that all of the denominators will divide into without leaving a remainder.

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## Example 1: Multiply by Each Denominator

Let’s suppose the following equation is presented to you. You see several fractions, and you may be tempted to run and hide.

Don’t be too hasty! There is a rather quick and easy modification that you can make to ease your concerns.(3/4)n + 2 – (4/3) = (1/6)Multiply each term by all of the denominators: 4, 3, and 6. This will cause each original denominator to be canceled leaving only multiplication. The result of the multiplication will not have any fractions in it.(3/4)n (4)(3)(6) + (2) (4)(3)(6) – (4/3) (4)(3)(6) = (1/6) (4)(3)(6)(3)n(3)(6) + (2)(4)(3)(6) – (4)(4)(6) = (1)(4)(3)54n + 144 – 96 = 1254n + 48 = 1254n = -36n = (-36/54) = -2/3Notice that the answer can still be a fraction, but the equation changed into one without any fractions. The only down side to this method is that the resulting equation has the potential for rather large numbers.

Even so, it should still be easier than working with the fractions along the way.

## Example 2: Multiply by Each Denominator

Here is another example.(-11/4)x – (5/3)x = (53/9)Each term will be multiplied by 4, 3, and 9.(-11/4)x (4)(3)(9) – (5/3)x (4)(3)(9) = (53/9) (4)(3)(9)(-11)x(3)(9) – (5)x(4)(9) = (53)(4)(3)-297x – 180x = 636-477x = 636x = (636/-477) = -4/3Let me be clear on this. The numbers can get quite large. If you would prefer not to deal with numbers so large, there is another method closely related to this one.

## Example 3: Multiplying by the LCM

Instead of multiplying each term by each of the denominators, multiply each term by the least common multiple of the denominators.

In the first example, this LCM would be 12. The result would look like this.(3/4)n + 2 – (4/3) = (1/6)(3/4)n (12) + (2) (12) – (4/3) (12) = (1/6) (12)The factor of 12 will simplify with each denominator.

(3)n(3) + (2)(12) – (4)(4) = (1)(2)9n + 24 – 16 = 29n + 8 = 29n = -6n = (-6/9) = -2/3The answer is the same as we expected, but this time the values were smaller and therefore easier to work with.

## Lesson Summary

We cannot always avoid problems that contain fractions, but we can manipulate them so that fractions aren’t a stumbling block. Multiply each term by either all of the denominators or by the least common multiple of the denominators. Reduce each term. Solve the remaining equation by getting the variable isolated on one side of the equal sign with a coefficient of a positive one.

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