Equations / Inequalities - PSAT Math

The equation simplifies to x > -1.41. -1 is the answer.

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

This inequality could be rewritten as:

4x + 14 > 30 OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

For this problem, we must take into account the absolute value.

First, we solve for 2_x_ – 2 > 20. But we must also solve for 2_x_ – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).

First step:

2_x_ – 2 > 20

2_x_ > 22

x > 11

Second step:

2_x_ – 2 < –20

2_x_ < –18

x < –9

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

Subtract from both sides:

Subtract 7 from both sides:

Divide both sides by :

Remember to switch the inequality when dividing by a negative number:

Since is not an answer, we must find an answer that, at the very least, does not contradict the fact that is less than (approximately) 4.67. Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that is less than 5.

Subtract 5 from both sides of the inequality:

Multiply both sides by 5:

Therefore only I must be true.

The correct answer is (x + 4)(x – 4)

We neen to factor x2 – 16 to solve. We know that each parenthesis will contain an x to make the x2. We know that the root of 16 is 4 and since it is negative and no value of x is present we can tell that one 4 must be positive and the other negative. If we work it from the multiple choice answers we will see that when multiplying it out we get x2 + 4x – 4x – 16. 4x – 4x cancels out and we are left with our answer.

First, let's factor x3 – y3 using the formula for difference of cubes.

x3 – y3 = (x – y)(x2 + xy + y2)

We are told that x2 + xy + y2 = 6. Thus, we can substitute 6 into the above equation and solve for x – y.

(x - y)(6) = 30.

Divide both sides by 6.

x – y = 5.

The original questions asks us to find x2 – 2xy + y2. Notice that if we factor x2 – 2xy + y2 using the formula for perfect squares, we obtain the following:

x2 – 2xy + y2 = (x – y)2.

Since we know that (x – y) = 5, (x – y)2 must equal 52, or 25.

Thus, x2 – 2xy + y2 = 25.

The answer is 25.

This can be solved by substitution and factoring.

x2 – y = 34 can be written as y = x2 – 34 and substituted into the other equation: x – y = 4 which leads to x – x2 + 34 = 4 which can be written as x2 – x – 30 = 0.

x2 – x – 30 = 0 can be factored to (x – 6)(x + 5) = 0 so x = 6 and –5 and because only 6 is a possible answer, it is the correct choice.

When a = 9, then x_2 + 2_ax + 81 = 0 becomes

x_2 + 18_x + 81 = 0.

This equation can be factored as (x + 9)2 = 0.

Therefore when a = 9, x = –9.

Multiply both sides by LCD :

or

There are two solutions of unlike sign.

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.

In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):

(x – (–1)) = x + 1

(x – 0) = x

and (x – 2).

This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).

We can immediately eliminate the function f(x) = _x_2 + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = _x_2 – x – 2.

Let's look at the function f(x) = _x_3 – x_2 + 2_x. When we factor this, we are left with x(_x_2 – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.

Next, let's examine f(x) = _x_4 + _x_3 – 2_x_2 .

We can factor out _x_2.

_x_2 (_x_2 + x – 2)

When we factor _x_2 + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1.

The only function with the right factors is f(x) = _x_3 – x_2 – 2_x.

When we factor out an x, we get (_x_2 – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.

The answer is f(x) = _x_3 – x_2 – 2_x.

First, translate the problem into three equations. The statement, "Joule is three years older than twice Newton's age" is mathematically translated as

where represents Joule's age and is Newton's age.

The statement, "Newton is Toby's age younger than eleven years" is translated as

where is Toby's age.

The third statement, "Toby is one year younger than Joule" is

.

So these are our three equations. To figure out the age of these dogs, first I will plug the third equation into the second equation. We get

Plug this equation into the first equation to get

Solve for . Add to both sides

Divide both sides by 3

So Joules is 9 years old. Plug this value into the third equation to find Toby's age

Toby is 8 years old. Use this value to find Newton's age using the second equation

Now, we have the age of the following dogs:

Joule: 9 years

Newton: 3 years

Toby: 8 years

Since this is a long word problem, it might be easy to confuse the two behaviors and come up with the wrong answer. Let's avoid this problem by turning each behavior into a variable. If we call "sitting quietly" and "completing assignments" , then we can easily construct a simple system of equations,

and

.

We can multiply the first equation by to yield .

This allows us to cancel the terms when we add the two equations together. We get , or .

A quick substitution tells us that . So, sitting quietly is worth 3 tokens and completing an assignment on time is worth 7.

6,035 total tickets were sold, and the total number of tickets is the sum of the adult and child tickets, .

Therefore, we can say .

The amount of money raised from adult tickets is \$11 per ticket mutiplied by tickets, or dollars; similarly, dollars are raised from child tickets. Add these together to get the total amount of money raised:

These two equations form our system of equations.

Let's call "" the cost of one jersey (this is the value we want to find)

Let's call the amount of money Jack starts with ""

Let's call the amount of money Aaron starts with ""

We know Jack buys a glove for and bats for each, and then has left over after. Thus:

simplifying, so Jack started with

We know Aaron buys hats for each and jerseys (unknown cost "") and spends all his money.

The last important piece of information from the problem is Aaron starts with dollars more than Jack. So:

From before we know:

Plugging in:

so Aaron started with

Finally we plug into our original equation for A and solve for x:

Thus one jersey costs

Since we know that stands for weeks, the answer has to have something to do with the weeks. This eliminates "the number of candles she has remaining for the month." Also, we can eliminate "the number of weeks that she has sold candles this month" because that would be our value for , not what we'd multiply by. The correct answer is, "the number of candles that she sells each week."

Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.

The numbers by which _x_6 – _y_6 is divisible will be all of its factors. In other words, we need to find all of the factors of _x_6 – _y_6 , which essentially means we must factor _x_6 – _y_6 as much as we can.

First, we will want to apply the difference of squares rule, which states that, in general, _a_2 – _b_2 = (a – b)(a + b). Notice that a and b are the square roots of the values of _a_2 and _b_2, because √_a_2 = a, and √_b_2 = b (assuming a and b are positive). In other words, we can apply the difference of squares formula to _x_6 – _y_6 if we simply find the square roots of _x_6 and _y_6.

Remember that taking the square root of a quantity is the same as raising it to the one-half power. Remember also that, in general, (ab)c = abc.

√_x_6 = (_x_6)(1/2) = x(6(1/2)) = _x_3

Similarly, √_y_6 = _y_3.

Let's now apply the difference of squares factoring rule.

_x_6 – _y_6 = (_x_3 – _y_3)(_x_3 + _y_3)

Because we can express _x_6 – _y_6 as the product of (_x_3 – _y_3) and (_x_3 + _y_3), both (_x_3 – _y_3) and (_x_3 + _y_3) are factors of _x_6 – _y_6 . Thus, we can eliminate _x_3 – _y_3 from the answer choices.

Let's continue to factor (_x_3 – _y_3)(_x_3 + _y_3). We must now apply the sum of cubes and differences of cubes formulas, which are given below:

In general, _a_3 + _b_3 = (a + b)(_a_2 – ab + _b_2). Also, _a_3 – _b_3 = (a – b)(_a_2 + ab + _b_2)

Thus, we have the following:

(_x_3 – _y_3)(_x_3 + _y_3) = (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2)

This means that x – y and x + y are both factors of _x_6 – _y_6 , so we can eliminate both of those answer choices.

We can rearrange the factorization (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2) as follows:

(x – y)(x + y)(_x_2 + xy + _y_2)(_x_2 – xy + _y_2)

Notice that (x – y)(x + y) is merely the factorization of difference of squares. Therefore, (x – y)(x + y) = _x_2 – _y_2.

(x – y)(x + y)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2) = (_x_2 – _y_2)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2)

This means that _x_2 – _y_2 is also a factor of _x_6 – _y_6.

By process of elimination, _x_2 + _y_2 is not necessarily a factor of _x_6 – _y_6 .

The answer is _x_2 + _y_2 .