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# Things you should know in algebra before GRE

### Componendo and dividendo:

If x/y = z/t then; (x+y)/y=(z+t)/t.

__Componendo__

x/(x+y) =z/(z+t)

(x-y)/y = (z-t)/t

__Dividendo__

x/(x-y) =z/(z-y)

(x+y)/(x-y) = (z+t)/(z-t).

### Equations and Exponents

### Functions

f(x)= 9x + 5; where f is function of x (why? because it changes with x, it depends on x). value of f can be obtained by substituting x in '9x + 5', so f(2) =23 when x=2.

The**domain** of any function is set of all permissible inputs values of the variable (in above case it's all permissible values of x).

The**Range** is set of all values of y.

functions questions can also be represented with strange operators (like a # b) and the question might be (x # 2y) : just replace â€˜aâ€™ andâ€™ bâ€™, with â€˜xâ€™ and â€˜yâ€™. for example: a@ b =3a + b^{2}, then 5@2 = 3(5) +(2)^{2}
### INEQUALITIES:

The multiplication or division of an inequality by a negative number causes a reversal of direction of the inequality sign.E.g., 5 â€" 4x > 7 becomes -5 + 4x < 7 Note: Do not forget to charge the inequality sign (â€˜<â€™ or â€˜>â€™). "Reciprocal (or) inversion" also causes of direction of the inequality sign i.e., a/b > c/d becomes b/a < d /c.
x2>x You Cannot divided both sides by x and say x>1. x2-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

__-3x + 5 < 23__

for above inequality we need solutions for x.. above equation is equivalent to -3x < 18 or -x < 6 which means x > -6

hence x can be -5,-3.. 1,2... (infinite values)### Averages

Average of a set of n numbers is the sum of those numbers divided by n, Average = (sum of n numbers)/n or Average = sum/n.

**Arithmetic Sequence** If n numbers form an arithmetic sequence wherein the difference between any 2 consecutive terms is the same, then the average of the numbers is the middle term in the sequence, if n is odd and if n is even , the average of the numbers is the average of the two middle terms, or it also is the average of the first and the last numbers.

Distance, Speed, Time and RateDistance = rate x time
rate = distance /time
time = distance /rate.
Word problem that involves speed, Time and distance, we should always note whether the problem situation involves, Motion in the "same direction" Motion in the "opposite direction" or "A round trip".

If two objects with speeds x miles per hr and y miles per hr are moving in the same direction, then their relative speed is given by (x â€" y) miles/hr.

If they are moving in opposite direction, then their relative speed is given by (x + y) miles/hr.

The average speed is not the average of the two speeds in the problem.

x/(x+y) =z/(z+t)

(x-y)/y = (z-t)/t

x/(x-y) =z/(z-y)

(x+y)/(x-y) = (z+t)/(z-t).

- (x+y)
^{2}= x^{2}+ y^{2}+ 2xy - (x-y)
^{2}= x^{2}+ y^{2}- 2xy - (x+y)
^{3}= x^{3}+ y^{3}+ 3xy(x + y) - (x-y)
^{3}= x^{3}- y^{3}- 3xy(x - y) - x
^{2}- y^{2}= (x + y)(x - y) - x
^{-2}= 1/x^{2}, 2^{-4}= 1/16 = 1/2^{4} - (x
^{a})(x^{b}) = x^{a+b} - x
^{a}/x^{b}= x^{a-b}= 1/x^{b-a} - x
^{0}=1 - x
^{a}y^{a}= (xy)^{a}, 2^{2}3^{2}= 6^{2} - quadratic equation, for any given x if ax
^{2}+ bx + c =0 then x has 2 solutions

x=(-b+âˆš(b^{2}- 4ac)/2a, x=(-b-âˆš(b^{2}- 4ac)/2a - x
^{a}y^{b}is not equal to (xy)^{a+b}

The

The

functions questions can also be represented with strange operators (like a # b) and the question might be (x # 2y) : just replace â€˜aâ€™ andâ€™ bâ€™, with â€˜xâ€™ and â€˜yâ€™. for example: a@ b =3a + b

for above inequality we need solutions for x.. above equation is equivalent to -3x < 18 or -x < 6 which means x > -6

hence x can be -5,-3.. 1,2... (infinite values)

Distance, Speed, Time and Rate

If two objects with speeds x miles per hr and y miles per hr are moving in the same direction, then their relative speed is given by (x â€" y) miles/hr.

If they are moving in opposite direction, then their relative speed is given by (x + y) miles/hr.

The average speed is not the average of the two speeds in the problem.

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