How to find the solution to an equation - SAT Math

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.

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

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

This question gives two equalities and one inequality. The inequality (sv ≠ 0) simply says that neither s nor v is 0. The two equalities tell us that 4_xs and ks are both equal to v, which means that 4_xs_ and ks must be equal to each other--that is, 4_xs_ = ks. Dividing both sides by s gives 4_x_ = k, which is our solution.

1. Substitute y in the equation for 4.
2. You now have 6 * 4 = 10z + 4
3. Simplify the equation: 24 = 10z + 4
4. Subtract 4 from both sides: 24 – 4 = 10z + 4 – 4
5. You now have 20 = 10z
6. Divde both sides by 10 to solve for z.
7. z = 2.

Create 3 separate equations to solve for each variable separately.

1. 2_x_ = 48

2. 3_y_ = 48

3. 6_z_ = 48

x = 24

y = 16

z = 8

x * y * z = 3072

This goes up at a constant number between values, making it an arthmetic sequence. The first number is 2, with a difference of 3. Plugging this into the arithmetic equation you get An = 2 + 3 (n – 1). Plugging in 20 for n, you get a value of 59.

Looking at the sequence you can see that it doubles each term, making it a geometric sequence. Since it doubles r = 2 and the first term is 5. Plugging this into the geometric equation you get An = 5(2)n–1. Setting n = 6, you get 160 as the 6th term.

The zeroes of the equation are where f(x) = 0 (aka x-intercepts). Setting the equation equal to zero you get x2 = 9. Since a square makes a negative number positive, x can be equal to 3 or –3.

In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.

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.

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.