Simplifying Algebraic Expressions - GMAT Quantitative

Let's first look at the numerator and denominator separately.

$x^{2}$+6x+5: We need two numbers that multiply to 5 and add to 6. The numbers 1 and 5 work. So, $x^{2}$+6x+5 = (x+5)(x+1)

$x^{2}$+10x + 25: We need two numbers that multiply to 25 and add to 10. The numbers 5 and 5 work. So, $x^{2}$+10x + 25 = (x+5)(x+5)

Putting this together, $$\frac{x^{2}$$$+6x+5}{x^{2}$+10x+25} = $\frac{(x+5)(x+1)}{(x+5)(x+5)}$ = $\frac{x+1}{x+5}$

Let's combine like terms.

, so the equation has no solution.

From (1), we can easily get the result .

Then from (2), we need to multiply by . This gives us .

The last step is to subtract from :

You have to isolate by moving around the separate components in the problem. The steps should go as follows:

To simplify algebraically, we combine like terms. First, we should get the expression in one long string, by removing the parentheses. So remembering the communitive property, the first group in parentheses will have no changes when we remove the parentheses. So simplifies to

However, note the second group in parentheses is being subtracted. So we must invert all the signs in the group to simplify properly. So the previous expression simplifies to

Finally we reorder and combine like terms to get

By the Binomial Theorem, if you expand , writing the result in standard form, the term (with the terms being numbered from 0 to ) is

Set , , , and (again, the terms are numbered 0 through , so the third term is numbered 2) to get

Simplify.

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Therefore:

Let the value of the first integer be . This means that the consecutive integers will be , , and . The sum must be 12 which means that

Since the consecutive integers are 3, 4, and 5. The middle integer is 4.

The best way to illustrate the answer to this question is to do these operations to the number 1.

First, divide by 4:

Now move the decimal point right three spaces:

This has the effect of multiplying the number by 250.

By the Binomial Theorem, the term of is:

The coefficient of is therefore:

You can use the pattern for cubing a binomial sum, setting :

Rewrite, distribute, and simplify where possible:

The simplest way to solve this problem is to start by picking the smallest positive integer that can be divided by 24. That would be 24, since . Then, since we need a number that when divided by 24, leaves a remainder of 6, simply add 6 to 24. That gives us 30.

30 is the smallest positive integer that leaves a remainder of 6 when divided by 24.

Finally, divide 30 by 10.

The remainder is 0.

Let's set up the problem using algebraic symbols.

,

where Q is the quotient of the answer, and r is the remainder.

,

which means that .

Hence, the remainder MUST be a multiple of 7. The only multiple of 7 in the answer choices is 21, so that is our answer.

We want to find the smallest number that fits initial conditions, so we need to write two equations to represent the two conditions. If X leaves a remainder of 3 when divided by 5, that is equivalent to:

, where is some integer greater than or equal to 0.

Likewise, the second condition can be expressed by:

, where is some integer greater than or equal to 0.

For the first equation, X could be 3, 8, 13, 18, 23, 28, 33, 38...

For the second equation, X could be 4, 12, 20, 28, 36, 42...

The first number that fits our conditions is 28.

So, finally, what's the remainder when 28 is divided by 9?

The answer is 1.