# Algebra

GMAT Study Guide - a prep wikibook

## Basic Rules for Inequalities

(in the example: $a>b>0$, $c>d>0$)

You need to flip signs when both side are multiplied by a negative number:

$-a<-b$, $-c<-d$

You need to flip signs when 1 is divided by both side: $\frac{1}{a}<\frac{1}{b}$, $\frac{1}{c}<\frac{1}{d}$

You can only add or multiply them when their signs are in the same direction:

• $a+c>b+d$
• $ac>bd$

You can only apply substractions and divisions when their signs are in the opposite directions:

• $a>b$, $d
• $a-d>b-c$
• $\frac{a}{d}>\frac{b}{c}$
• You can't say $\frac{a}{c}>\frac{b}{c}$. It is WRONG.

Deal with negative numbers:

$-a<-b<0$, $-c<-d<0$

Then

• $-a-c<-b-d<0$
• $-a-(-d)<-b-(-c)$

However the sign needs to be flipped one more time if you are doing multiplication or division (because you are multiplying/dividing a negative number):

• $(-a)*(-c)>(-b)*(-d)$
• $\frac{-a}{-d}>\frac{-b}{-c}$

For example:

If $x<-4$, $y<-2$, we know that $xy>8$, but we don't know how $\frac{x}{y}$ compare to $\frac{-4}{-2}=2$ since you can only do division when their signs are in different directions.

If $x>-4$ and $y<-2$ then $\frac{x}{y}<2$ but we don't know how xy is compared to 8 since we can only do multiplication when their signs are the same direction.

It is easier to do the derivation, though, if you first change them to postive. For example:

• If $x<-4$, $y<-2$, then $-x>4$, $-y>2$, $xy>8$
• If $x<-4$, $y<2$, then $-x>4$, $y<2$, $-\frac{x}{y}>2$, $\frac{x}{y}<-2$

## Cancelling out "Common Terms"

You need to be very careful when you do algebra derivations. One of the common mistakes is to divide both side by "a common term". Remember you can only do this safely if the "common term" is a constant. However you CAN't do it if it contains a variable.

Example:

$x(x-2)=x$

You can't cancel out the $x$ on both side and say $x=3$ is the solution. You must move the $x$ from the right side to the left.

$x(x-2)-x=0$

$x(x-2-1)=0$

The solutions are: $x=0$ and $x=3$.

The reason why you can't divided both sides by $x$ is that when $x$ is zero, you can't divide anything by zero.

Equally important if not more, is that you CAN'T multiply or divide a "common term" that includes a variable from both side of an inequality. Not only it could be zero, but it could also be negative in which case you would need to flip the sign.

Example:

$x^2>x$

You CAN'T divided both sides by $x$ and say $x>1$. What you have to do is to move the right side to the left:

$x^2-x>0$

$x(x-1)>0$

Solution would be either both $x$ and $x-1$ are greater than zero, or both $x$ and $x-1$ are smaller than zero. So your solution is: $x>1$ or $x<0$.

Example:

$x>\frac{1}{x}$

Again you CAN'T multiply both sides by $x$ because you don't know if $x$ is positive or negative. What you have to do is to move the right side to the left:

$x-\frac{1}{x}>0$

$\frac{x^2-1}{x}>0$

• If $x>0$, then $x^2-1>0 \right x>1$
• If $x<0$, then $x^2-1<0 \right x>-1$

Therefore, your solution is $x>1$ or $0>x>-1$.

You could also break the original question to two branches from the beginning:

$x>\frac{1}{x}$

• If $x>0$, then $x^2>1 \right x>1$
• If $x<0$. then $x^2<1 \right x>-1$

Therefore, your solution is $x>1$ or $0>x>-1$.

## Absolute values

The absolute value of a number is its numeric value no matter what the sign is. The absolute value is sometimes referred to as "magnitude". The expression $|x|$ reads "the absolute value of $x$."

### Equations

Here is an example of an absolute value equation.

$| x + 2 | = 4$

To solve this equation, we will need to open up the absolute value signs and solve two equations:

$x + 2 = 4$ and $x + 2 = -4$; we will have two roots: 2 and -6.

### Inequalities

The way to solve this kind of questions is to break the equation (inequality) into two parts, one is when the value is non negative, the other is when the value is negative.

For example:

$|x-4| < 9$

You break it into two parts:

• If $x-4 \ge 0$, then $x-4 < 9$, solve for both you get $x \ge 4$, $x < 13$. So your solution is $4 \le x < 13$
• If $x-4 < 0$, then $-(x-4) < 9$, i.e. $x-4 > -9$. Solving for both, you get $x < 4$, $x > -5$. So your solution for this part is $-5 < x < 4$.

Combining the two solutions, you get $-5 < x < 13$ as your final solution.

Another example:

$|x+4| > 4$

• If $x+4 \ge 0$, then $x+4 > 4$. Solving for both, you get $x \ge -4$, $x > 0$. So the solution is $x>0$.
• If $x+4 < 0$, then $-(x+4) > 4$, i.e. $x+4 < -4$. Solving for both, you get $x < -4$, $x < -8$. So, your solution is $x < -8$.

The final solution is $x > 0$ or $x < -8$.

One more example:

$|y| > |y+1|$

• if $y \ge 0$, $y+1 \ge 0 \right y > y+1$, no solution
• if $y < 0$, $y+1 < 0 \right -y > -(y+1)$, solution is $y < -1$
• if $y \ge 0$, $y+1 < 0 \right y > -(y+1)$, no solution
• if $y < 0$, $y+1 \ge 0 \right -y > y+1$, solution is $-1 \le y < -\frac{1}{2}$.

So your final solution is $y < -\frac{1}{2}$

You could also solve this question by going the square root:

$y^2 > (y+1)^2$

$y^2 > y^2+2y+1$

$2y+1 < 0$

$y < -\frac{1}{2}$

{{#x:box|

• If $d$ is POSITIVE and $|x| < d$, then $-d < x < d$
• If $d$ is NEGATIVE and $|x| < d$, then there is no solution
• If $d$ is POSITIVE and $|x| > d$, then $x < -d$ OR $x > d$
• If $d$ is NEGATIVE and $|x| > d$, then $x$ is all real numbers

}}

The same strategy can apply to square inequalities. For example:

$(x+4)^2 > 4$

You could solve it this way:

$x^2+8x+12 > 0$

$(x+2)(x+6) > 0$

$x > -2$ or $x < -6$

Or you can solve it this way:

• If $x+4 \ge 0$, then $x+4 > 2$. Solving for them you get $x > -2$
• If $x+4 < 0$, then $x+4 < -2$. Solving for them you get $x < -6$.

## Short multiplication

### Squares

• $(a\pm b)^2=a^2\pm 2ab+b^2$
• $(a+b)(a-b)=a^2-b^2$
• $(a+b-c)^2=a^2+b^2+c^2+2ab-2ac-2bc$

### Cubes

• $(a\pm b)^3=a^3\pm 3a^2 b+3ab^2\pm b^3$
• $(a\pm b)(a^2\mp ab+b^2) = a^3\pm b^3$

### Other

$a^4-b^4=(a-b)(a+b)(a^2+b^2)$

## Exercises

In order to practice you can try to prove the following statements:

1. If $a>x$ , $b>y$ , $c>z$ then $a+b+c>x+y+z$ and $abc>xyz$.
2. $(n!)^2 > n^n$ when $n \gt 2$
3. $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} > 4$
4. $a^2*b +b^2*c + c^2*a \ge 3abc$
5. For any positive integer $n$, $2 \le (1 + \frac{1}{n})^n \le 3$
6. If $a$,$b$,$c$ are positive and not equal then
1. $(a+b+c)(ab+bc+ca) > 9abc$
2. $(b+c)(c+a)(a+b) > 8abc$
7. If $x > y$, then $x - y$ divides $x^n - y^n$.
1. $a^n - b^n$ is divisible by $a + b$ if $n$ is even.
2. $a^n + b^n$ is divisible by $a+b$ if $n$ is odd, and not divisible by $a+b$ if $n$ is even.
3. $a^n-b^n$ is divisible by $a-b$ whether $n$ is odd or even.