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

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.

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 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)