How to find the solution to an equation - GRE Quantitative Reasoning

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

To solve for x:

bx + c = e – ax

bx + ax = e – c

x(b+a) = e-c

x = (e-c) / (b+a)

Start by setting up an equation using Quantity A and Quantity B. In other words, you can solve an inequality where Quantity A > Quantity B. You would have one of four outcomes:

1. Quantity A = Quantity B: the two quantities are equal.
2. The inequality is always satisfied: Quantity A is always larger.
3. The inequality is never satisfied (but the two are unequal): Quantity B is always larger.
4. The inequality is not always correct or incorrect: the relationship cannot be determined.

So solve:

–5x + 4 > 8 – 2x (Quantity A > Quantity B)

+2x +2x

–3x + 4 > 8

–4 –4

–3x > 4 or x < –4/3

*remember to switch the direction of the inequality when you divide by a negative number

As the inequality \[x < –4/3\] is always false for \[x>0\], Quantity B is always larger.

First we need the area or the rectangle. 24 * 8 = 192. So now we know that 10 gallons will cover 75 ft2 and x gallons will cover 192 ft2. We set up a simple ratio and cross multiply to find that 75_x_ = 1920.

x = 25.6

First, solve for x:

11 + 3_x_ = 29

29 – 11 = 3_x_

18 = 3_x_

x = 6

Then, solve for 2_x_:

2_x_ = 2 * 6 = 12

For this type of problem, we use the formula:

When Karen catches up with Jen, their distances are equivalent. Thus,

We then make a variable for Jen's time, . Thus we know that Karen's time is (since we are working in hours).

Thus,

There's a logical shortcut you can use on "catching up" distance/rate problems. The difference between the faster (Karen at 60mph) and slower (Jen at 45mph) drivers is 15mph. Which means that every one hour, the faster driver, Karen, gains 15 miles on Jen. We know that Jen gets a 1/2 hour head start, which at 45mph means that she's 22.5 miles ahead when Karen gets started. So we can calculate the number of hours (H) of the 15mph of Karen's "catchup speed" (the difference between their speeds) it will take to make up the 22.5 mile gap:

15H = 22.5

So H = 1.5.

Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.

Start by combining like terms.

4_x_ + 5 = 13_x_ + 4 – x – 9

4_x_ + 5 = 12_x_ – 5

–8_x_ = –10

x = 5/4

First we solve for x.

Subtracting 3 from both sides gives us –3_x_ < 17.

Dividing by –3 gives us x > –17/3.

–6 is less than –17/3.

In order to solve for y, place x in terms of y in the first equation and then substitute that for x in the second equation.

The first equation would yield: .

Substituting into the second equation, we get: .

Simplify:

You can solve this problem by plugging in random values or by simply solving for k. To solve for k, put the s values on one side and the k values on the other side of the equation. First, subtract 4s from both sides. This gives 4s – 6k = –2k. Next, add 6k to both sides. This leaves you with 4s = 4k, which simplifies to s=k. The answer is therefore s.

This is a simple plug-in and PEMDAS problem. First, plug in x = 3 and y = –3 into the x and y. You should follow the orders of operation and compute what is within the parentheses first and then find the product. This gives 8 * 13 = 104. The answer is 104.

Start by plugging in x = 4 to solve for y: y = 3 * 4 + 5 = 17. Then 2 * 17 – 1 = 33

The best way to solve this problem is to turn the two statements into equations calling Sarah’s age S and Ron’s age R. So, S = 3(R – 2) and S = 14 + R. Now substitute the value for S in the second equation for the value of S in the first equation to get 14 + R = 3(R – 2) and solve for R. So R equals 10 so S equals 24 and the sum of 10 and 24 is 34.

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.