Algebra - GRE Quantitative Reasoning

(2_x_ + 4)/(x + 2)

To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.

f(–3) = (–3)2 + 5 = 9 + 5 = 14

The inner function f(x) is like our y-value that we plug into g(y).

g(f(x)) = 2(7_x_2 – 3) + 9 = 14_x_2 – 6 + 9 = 14_x_2 + 3.

Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.

Solving for P will give the answer.

The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.

Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81

Solving for x gives Claire’s age, which can be used to find Nick’s age.

If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.

If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).

The y’s cancel leaving us with an answer of 5.

We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.

This is the difference of squares. You must know this formula for the GRE!

_a_2 – _b_2 = (a – b)(a + b)

Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).

Factor first. The numerators will not factor, but the first denominator factors to (x – 2)(x + 2) and the second denomintaor factors to x(x – 2). Multiplying fractions does not require common denominators, so now look for common factors to divide out. There is a factor of x and a factor of (x + 2) that both divide out, leaving 4 in the numerator and two factors of (x – 2) in the denominator.

6/8 X 20/3 first step is to reduce 6/8 -> 3/4 (Divide top and bottom by 2)

3/4 X 20/3 (cross-cancel the threes and the 20 reduces to 5 and the 4 reduces to 1)

1/1 X 5/1 = 5

If x = 0, (x – 5)2 = 25 and (x + 5)2 = 25, so the quantities are equal.

If x = 1, (x – 5)2 = (–4)2 = 16 and (x + 5)2 = 62 = 36, so B is greater.

This is a contradiction, so the answer cannot be determined.

Begin by expanding Quantity A:

Now in order to compare this to Quantity B:

A good method would be to subtract shared terms from each Quantity; in this case, both quantities have an and term. Removing them gives:

Quantity A' :

Quantity B' :

The question now is the sign of Quantity A'; if it's always positive, Quantity A is greater. If it's always negative, Quantity B is greater. If it is zero, the two are the same.

We only know that

If , then Quantity A' would be zero.

If , then Quantity A' would be positive.

Since values of x and y can be chosen to vary the relationship, th relationship cannot be determined.

In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes.

For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90).

Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90.

When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels.

We know that Quantity A = y / 3 = 15 /3 = 5.

If we plug in 15 for y, we can solve for x, for Quantity B.

y = x2 - 10

y = 15

15 = x2 - 10 (Add 10 to both sides.)

25 = x2

x = 5 or -5

Since 5 is equal to 5 but is greater than -5, we cannot determine the relationship between Quantities A and B.

To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = 4/3c.

Now we can plug into the last equation. We plug in 4/3c for a, 4c for b, and leave c as it is. We must find a common denominator (4/3c – 12/3c + 3/3c) and add the numerators to find that our equation equals –5/3c.

There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction).

If x2 + 5x – 24 = 0,

(x – 3)(x + 8) = 0 or x = –8 or +3.

y2 – 9y + 20 = 0, then

(y – 5)(y – 4) = 0, or y = +4 or +5.

y is always greater than x.

We can factor the equation x2 + kx - 12 = 0, knowing that we will have (x - 3) as one of the parentheses since the root is equal to 3.

x - 3 = 0

x = 3

We also know that the other root will be -4, because we multiply the 4 and -3 in (x + 4)(x - 3) to get our constant, -12.

This means that kx is equal to 4x - 3x = x. Therefore k = 1, and quantity A > quantity B.

Quantity A is greater.