SAT Math - How to find a solution set

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.

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

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

$-\frac{35}{34}$

$35$

$34$

$-34$

$-35$

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 $-\frac{35}{34}$ .

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.

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

Give the solution set of the inequality

$(-\infty, -4.5) \cup (23.5, \infty)$

$(-\infty, -23.5) \cup ( 4.5, \infty)$

The inequality has no solution.

Explanation

In an absolute value inequality, the absolute value expression must be isolated on one side first. We can di this by first subtracting 42 from both sides:

Divide by , reversing the direction of the inequality symbol since we are dividing by a negative number:

$\frac{- |-2x+19| }{-1}> \frac{-28}{-1}$

This inequality can be rewritten as the compound inequality

Solve each simple inequality separately.

Subtract 19 from both sides:

Divide by , remembering to reverse the symbol:

$\frac{-2x}{-2} > \frac{-47}{-2}$

In interval notation, this is $(23.5, \infty)$ .

Carry out the same steps on the other simple inequality:

$\frac{-2x}{-2} <\frac{9}{-2}$

In interval notation, this is $(-\infty, -4.5)$ .

Since the two simple inequalities are connected by an "or", their individual solution sets are connected by a union; the solution set is

$(-\infty, -4.5) \cup (23.5, \infty)$ .

Give the solution set of the inequality

$\left ( -7 \frac{1}{6}, 11 \frac{5}{6} \right )$

$\left ( - 11 \frac{5}{6}, 7 \frac{1}{6} \right )$

$\left (- \infty , -7 \frac{1}{6} \right ) \cup \left ( 11 \frac{5}{6}, \infty \right )$

$\left (- \infty , - 11 \frac{5}{6} \right ) \cup \left ( 7 \frac{1}{6} , \infty \right )$

$\left (- \infty , \infty \right )$

Explanation

In an absolute value inequality, the absolute value expression must be isolated on one side first. We can do this by subtracting 21 from both sides:

This can be rewritten as the three-part inequality

Subtract 14 from all three expressions:

Divide all three expressions by , reversing the inequality symbols since you are dividing by a negative number:

$-71\div (-6) >-6x \div (-6)> 43 \div (-6)$

$11 \frac{5}{6} >x > -7 \frac{1}{6}$

In interval notation, this is $\left ( -7 \frac{1}{6}, 11 \frac{5}{6} \right )$ .

Solve for x: $(x-8)^{2}=36$

x = 14 or 2

x = 14 or –2

x = 2 or –14

x = –2 or –14

x = 14

Explanation

First, take the square root of both sides:

$x-8=\pm 6$

Therefore, $x-8=6$ or $x-8 = -6$

Add 8 to both sides of the equation; therefore, $x=14$ or $x=2$

Different colored marbles are placed in a bag. There are red marbles, black marbles, and green marbles in the bag. What is the probability that a green marble will be chosen?

$\frac{4}{11}$

$\frac{4}{7}$

$\frac{5}{11}$

$\frac{2}{11}$

$\frac{4}{14}$

Explanation

When doing probability problems, we are looking for the number of successes over number of possible outcomes. There are 4 chances to successfully choose a green marble. The number of possible outcomes are 11, one for each of the 11 marbles in the bag. When we write the fraction, we get our answer.

In mathematical words we get the following:

$P=\frac{G}{T}=\frac{4}{5+2+4}=\frac{4}{11}$

How to find a solution set

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