Equations / Inequalities - SAT Math

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

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

There are two values where _x_2 = x, namely x = 0 and x = 1. All values between 0 and 1 get smaller after squaring. All values greater than 1 get larger after squaring.

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.

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.

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.

Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.

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

First, distribute, making sure to watch for negatives.

Combine like terms.

Subtract 7x from both sides.

Add 18 on both sides and be careful adding integers.