Algebra

E

Identifying Least Common Multiples

Finding least common multiples is useful in combining algebraic fractions. The least common multiple (LCM) of a set of numbers is the smallest number into which each number in the set will divide evenly. The LCM of 2, 3, 4, and 6, for example, is 12.

The LCM can be calculated by factoring numbers into their prime components. The LCM is the product of the highest power of each prime factor of the given numbers. For example, to find the LCM for the three numbers 27, 63, and 75, each number is first factored: 27 = 3³, 63 = 3²·7, and 75 = 3·5². The prime factors of these three numbers are 3, 5, and 7, and the highest powers of those three factors are 3³, 5², and 7. The LCM, therefore, is 3³·7·5² = 4,725; 4,725 is the smallest number into which 27, 63, and 75 will all divide evenly.

Given several algebraic expressions, the least common multiple is the expression of lowest degree and least coefficient that can be divided evenly by each of the expressions. To find a common multiple of the terms 2x²y, 15x²y², and 6ay³, all three expressions could simply be multiplied together: (2x²y)(15x²y²)(6ay³) = 180ax⁴y⁶. However, 180ax⁴y⁶ is not the least common multiple. To determine which is the least, each of the terms is reduced to its prime factors. For the numerical coefficients 2, 15, and 6, the prime factors are 2, 3·5, and 2·3, respectively; the least common multiple for the numerical coefficients is therefore 2·3·5, or 30. Similarly, because the constant a appears only once, it too must be a factor. Of the variables, x² and y³ are required because they are the highest powers of the two variables that appear in any of the expressions. The LCM of the three terms, therefore, is 30ax²y³. Each term will evenly divide this expression:

IV

Solving Equations

The means of manipulating equations outlined in the previous section can be employed to solve equations. Given an equation, algebra supplies solutions based on the general idea of the identity a = a. As long as the same arithmetic or algebraic procedure is applied simultaneously to both sides of the equation, the equality remains unaffected. The basic strategy is to isolate the unknown term on one side of the equation and the solution on the other.

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Suppose eight people each have an equal but unknown number of pennies. The people have a scale with 12 pennies on one side and 6 on the other. They start putting their pennies on this unbalanced scale, trying to make it balance. The scale balances once five of the people put all of their pennies on the side that already has six and the other three people put all of their pennies on the side that already has 12. The number of pennies each person originally had can be determined by turning this situation into a linear equation with one unknown:

5x + 6 = 3x + 12

To solve this equation, the variable terms are isolated on one side and the constant terms on the other. The term 3x can be removed from the right side by subtracting; 3x must then be subtracted from the left side as well:

The number 6 is then subtracted from both sides:

To isolate x on the left side, both sides of the equation are divided by 2:

The solution then follows directly: x = 3. Each person, then, originally had three pennies. This can easily be verified by substituting the solution value x = 3 back into the original equation:

5x + 6 = 3x + 12

5(3) + 6 = 3(3) + 12

15 + 6 = 9 + 12

21 = 21

A

Factoring Quadratic Equations

It is not so easy to isolate the variable on one side of an equation in which more than one degree of the variable appears. Quadratic equations are the simplest such equations. Given any quadratic equation of the general form

ax² + bx + c = 0

a number of approaches are possible depending on the specific nature of the equation in question. If the equation can be factored, then the solution is straightforward. For instance, if a = 1, b = -3, and c = -10, then

x² – 3x – 10 = 0

can be factored as follows:

(x – 5)(x + 2) = 0

The only way to get 0 when multiplying numbers together is when one of the numbers is 0. Therefore this equation can only be true when one or the other of the individual factors is equal to zero—that is, when x - 5 = 0 or x + 2 = 0. Thus the equation has two solutions: x = 5 and x = -2. That these are the solutions to the equation may again be verified by substituting them back into the original equation:

5² – 3(5) – 10 = 25 - 15 - 10 = 0

and (-2)² – 3(-2) - 10 = 4 + 6 - 10 = 0.

B

Completing the Square

If, on inspection, no obvious means of factoring the equation directly can be found, an alternative might exist. For example, in the equation

4x² + 12x = 7

the expression 4x² + 12x could be factored as a perfect square if it were 4x² + 12x + 9, which equals (2x + 3)². This can easily be achieved by adding 9 to both sides of the equation, completing the square:

4x² + 12x + 9 = 7 + 9

Factoring and simplifying yields:

(2x + 3)² = 16

By taking the square root of both sides, this can be reduced to

(2x + 3) = Â

which is the same as

2x + 3 = 4 and 2x + 3 = -4

because  has two values, positive 4 and negative 4. The first equation leads to the solution x = y because 2x + 3 = 4 becomes 2x = 1 after subtracting 3 from both sides, and 2x = 1 becomes x = y after dividing both sides by 2. The second equation leads to the solution x = -7/2, or x = -3y. Both solutions can be verified by substituting the two solutions in question back into the original equation, 4x² + 12x + 9 = 7 + 9:

4(y)² + 12(y) + 9 = 7 + 9

4(‚) + 6 + 9 = 16

1 + 6 + 9 = 16

16 = 16

4(-3y)² + 12(-3y) + 9 = 7 + 9

4(12‚) - 42 + 9 = 16

49 - 42 + 9 = 16

16 = 16

C

The Quadratic Formula

Sometimes it is not possible or readily apparent how to factor an equation or complete its square. However, all quadratic equations that can be put into the form

ax² + bx + c = 0

can be solved using the quadratic formula:

For example, to find the roots of

x² – 4x = -3

the equation is first put into the standard form

x² – 4x + 3 = 0

In this equation a = 1, b = -4, and c = 3. These terms are then substituted into the quadratic formula:

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