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.

We work from the inside out, so we start with the function f(x). f(4) = –1. Then we plug that value into g(x), so g(f(x)) = 3 * (–1) = –3.

f(–3) = (–3)2 + 5 = 9 + 5 = 14

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_).

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.

A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.

We have three variables and only two equations, so we will not be able to solve for each independent variable. We need to think of another solution.

Notice what happens if we line up the two equations and add them together.

(x + y) + (3_x –_ y + z) = 4x + z

and 5 + 3 = 8

Lets take this equation and multiply the whole thing by 3:

3(4_x_ + z = 8)

Thus, 12_x_ + 3_z_ = 24.

Step 1: 7_x_ + 3_y_

Step 2: 4 * (7_x_ + 3_y_) = 28_x_ + 12_y_

Step 3: 28_x_ + 12_y_ + x = 29_x_ + 12_y_

Step 4: 29_x_ + 12_y_ – (x – y) = 29_x_ + 12_y_ – x + y = 28_x_ + 13_y_

Plugging in numbers is not the best strategy here. Instead, let's see how we can equate the two expressions. Quantity A is actually a difference of squares. 256_x_2 = (16_x_)2 and 49_y_2 = (7_y_)2. These look like the expressions in Quantity B. The formula to remember here is the difference of squares formula, a very important one for this test! a_2 – b_2 = (a + b)(a – b). Thus, if a = 16_x and b = 7_y, 256_x_2 – 49_y_2 = (16_x_ – 7_y_)(16_x_ + 7_y_), and the quantities are equal.

(x + y)2 = x_2 + 2_xy + _y_2

Now, since there are no specifications on what x and y can equal, one or both of them could be 0, making the two columns equal. Any value other than 0 will make the columns unequal because of the additional 2xy term, so the answer cannot be determined.

Quantity A: (x – y)2 = x_2 – 2_xy + _y_2

Quantity B: (x + y)2 = x_2 + 2_xy + _y_2

Both have x_2 + y_2 so cancel those from both columns and just compare –2_xy in Quantity A to 2_xy in Quantity B. If x = 1 and y = 1, –2_xy_ = –2 and 2_xy_ = 2, so Quantity B is greater. But if x = 1 and y = –1, –2_xy_ = 2 and 2_xy_ = –2, so Quantity A is greater. The contradiction means the answer cannot be determined.

If the first integer is , then

This is the same as the median.

We need to work from the inside to the outside, so g(6) = 3(6) – 6 = 12.

Then f(g(6)) = 2(12) + 4 = 28.

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.