Linear / Rational / Variable Equations - GRE Quantitative Reasoning

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.

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.

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.

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.

The quantity can be regrouped to be –3 = a – b. Thus, (a – b)2 = (–3)2 = 9.

The point of intersection for two lines is the same as the values of x and y that mutually solve each equation. Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y:

3x + 4y = 6

15x - 4y = 3

18x = 9; x = 0.5

You can now plug x into the first equation:

3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125

Therefore, our point of intersection is (0.5, 1.125)

From our data, we can come up with the following two equations:

0.50R + 0.75D = 50

2R = D

Replace the D value in the second equation into the first one:

0.5R + 0.75 * 2R = 50

0.5R + 1.5R = 50; 2R = 50; R = 25

However, note that the question asks for the number of diet cans, so this will have to be doubled to 50.

The cars have a distance from each other of 25 + 120t miles, where t is the number of hours, 25 is their initial distance and 120 is 50 + 70, or their combined speeds. Solve this equation for 400:

25 + 120t = 400; 120t = 375; t = 3.125

However, the question asked for minutes, so we must multiply this by 60:

3.125 * 60 = 187.5 minutes.

To solve, remove the radical by squaring both sides

(√3x) 2 = 92

3x = 81

x = 81/3 = 27

First, simplify the equation:

√(8y) + 18 = 4

√(8y) = -14

Then square both sides

(√8y) 2 = -142

8y = 196

y = 196/8 = 24.5

First let's solve for y using the second equation, x – y = 6.

x = 6 + y. Then plug this in to the other equation.

3 (6 + y) + 4_y_ = 5

18 + 3_y_ + 4_y_ = 5

18 + 7_y_ = 5

7_y_ = –13

y = –13/7. Now plug this value back into x = 6 + y.

x = 6 – 13/7 = 29/7. x is positive and y is negative, so clearly x is larger, so Quantity A is bigger.

We have to solve the equation 4(3_x_ – 2) = 12. First, we can distribute the left side.

4(3_x_) – 4(2) = 12

12_x_ – 8 = 12

Then we add 8 to both sides.

12_x_ = 20

Divide both sides by 12.

x = 20/12

Simplify 20/12 by dividing the numerator and denominator by 4.

x = 20/12 = 5/3

The answer is 5/3.

1. Add 7 to both sides

3(z + 4)3 – 7 + 7= 17 + 7

3(z + 4)3 = 24

2. Divide both sides by 3

(z + 4)3 = 8

3. Take the cube root of both sides

z + 4 = 2

4. Subtract 4 from both sides

z = –2