SAT Math - Solution Sets

Solve _x_2 – 48 = 0.

x = 4 or x = –4

x = 4√3

x = 4√3 or x = –4√3

x = –√48

x = 0

Explanation

No common terms cancel out, and this isn't a difference of squares.

Let's move the 48 to the other side: _x_2 = 48

Now take the square root of both sides: x = √48 or x = –√48. Don't forget the second (negative) solution!

Now √48 = √(3*16) = √(3*42) = 4√3, so the answer is x = 4√3 or x = –4√3.

Solve _x_2 – 48 = 0.

x = 4 or x = –4

x = 4√3

x = 4√3 or x = –4√3

x = –√48

x = 0

Explanation

No common terms cancel out, and this isn't a difference of squares.

Let's move the 48 to the other side: _x_2 = 48

Now take the square root of both sides: x = √48 or x = –√48. Don't forget the second (negative) solution!

Now √48 = √(3*16) = √(3*42) = 4√3, so the answer is x = 4√3 or x = –4√3.

Solve and describe your answer in both inequality notation and interval notation:

$10< -3a+10\leq 34$

$-8 \leq a< 0$

![\left -8,0\left \right )

$\left ( 8,0 \right )$

$-a \leq -8$

![\left ( 0, -8]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/2818/gif.latex)

$-8\leq a> 0$

Explanation

This is a question with double inequality.

First solve the left side which will be $10< -3a+10$ which will give you $a<0$ and then solve the right side which is $-3a+10\leq 34$ and solution is $-a\leq 8$ which is really equal to $a\geq -8$

What is the sum of all solutions to the equation

$\dpi{100} \small |x+3| = 10$ ?

$\dpi{100} \small -6$

$\dpi{100} \small 7$

$\dpi{100} \small 14$

$\dpi{100} \small 13$

$\dpi{100} \small -13$

Explanation

If $\dpi{100} \small |x+3| = 10$ , then either

$\dpi{100} \small x+3 = 10$ or $\dpi{100} \small x+3 = -10$ .

These two equations yield $\dpi{100} \small 7$ and $\dpi{100} \small -13$ as answers.

$\dpi{100} \small 7+(-13)=-6$

Solve and describe your answer in both inequality notation and interval notation:

$10< -3a+10\leq 34$

$-8 \leq a< 0$

![\left -8,0\left \right )

$\left ( 8,0 \right )$

$-a \leq -8$

![\left ( 0, -8]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/2818/gif.latex)

$-8\leq a> 0$

Explanation

This is a question with double inequality.

First solve the left side which will be $10< -3a+10$ which will give you $a<0$ and then solve the right side which is $-3a+10\leq 34$ and solution is $-a\leq 8$ which is really equal to $a\geq -8$

What is the sum of all solutions to the equation

$\dpi{100} \small |x+3| = 10$ ?

$\dpi{100} \small -6$

$\dpi{100} \small 7$

$\dpi{100} \small 14$

$\dpi{100} \small 13$

$\dpi{100} \small -13$

Explanation

If $\dpi{100} \small |x+3| = 10$ , then either

$\dpi{100} \small x+3 = 10$ or $\dpi{100} \small x+3 = -10$ .

These two equations yield $\dpi{100} \small 7$ and $\dpi{100} \small -13$ as answers.

$\dpi{100} \small 7+(-13)=-6$

The set contains all multiples of . Which of the following sets are contained within ?

I. The set of all multiples of .

II. The set of all multiples of .

III. The set of all multiples of .

III only

I only

II only

I and II

I, II, and III

Explanation

Think of the multiples of 10: 10, 20, 30, 40, 50, 60, 70, . . .

I. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, . . .

Some of these already are not contained in S.

II. Multiples of 5: 5, 10, 15, 20, 25, . . .

Some of these already are not contained in S.

III. Multiples of 20: 20, 40, 60, 80, 100, . . .

All of these are also multiples of 10. Thus, our answer must be III only.

$0.1(x-5)+0.03(2x+5)=0.5(x)$

$-\frac{35}{34}$

$35$

$34$

$-34$

$-35$

Explanation

First multiply each decimal number in each term by 100 to remove the decimals (to get a whole number you have to multiply 0.03 by 100 to get 3). You need to do this for terms on both sides of the equal sign.

The second method would be to look for the number of digits to the right of the decimal point (e.g., 0.03 has two digits). So in this method shift the decimal point to the right two places.

Now the equation looks as follows:

$10x-50+6x+15=50x$

Now solve for $x$ and $x$ will be equal to $-\frac{35}{34}$ .

Using the ordered pairs listed below, which of the following equations is true?

(0, –4)

(2, 0)

(4, 12)

(8, 60)

$y=x^{2}-4$

$y=x^{2}+4$

$y=2x^{2}-8$

$y=2x^{2}+8$

Explanation

You can solve this problem using the guess and check method by substituting the first number in the ordered pair for "x" and the second number for "y". Therfore the correct answer is $y=x^{2}-4$

–4 = 0 – 4

0 = 4 – 4

12 = 16 – 4

60 = 64 – 4

The set contains all multiples of . Which of the following sets are contained within ?

I. The set of all multiples of .

II. The set of all multiples of .

III. The set of all multiples of .

III only

I only

II only

I and II

I, II, and III

Explanation

Think of the multiples of 10: 10, 20, 30, 40, 50, 60, 70, . . .

I. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, . . .

Some of these already are not contained in S.

II. Multiples of 5: 5, 10, 15, 20, 25, . . .

Some of these already are not contained in S.

III. Multiples of 20: 20, 40, 60, 80, 100, . . .

All of these are also multiples of 10. Thus, our answer must be III only.

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