Equations / Inequalities - PSAT Math

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."

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.

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.

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 .

First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)

_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).

Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).

This is a difference of squares. The difference of squares formula is a_2 – b_2 = (a + b)(a – b). In this problem, a = 6_x and b = 7_y.

So 36_x_2 – 49_y_2 = (6_x_ + 7_y_)(6_x_ – 7_y_).

Find two numbers that add to and multiply to

Factors of

You can use

Then make each factor equal 0.

and

and

Find all factors of 24

1, 2, 3,4, 6, 8, 12, 24

Now find two factors that add up to and multiply to ; and are the two factors.

By factoring, you can set the equation to be

If you FOIL it out, it gives you .

Set each part of the equation equal to 0, and solve for .

and

and

Factoring yields giving roots of and .

The numerator can be factored into .

Therefore, it can cancel with the denominator. So imples .

We can factor out a , leaving .

From there we can factor again to

.

First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)

_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).

Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).

Start by looking at your last term. Since this term is negative, you will need to have a positive group and a negative group:

Now, since the middle term is positive, you can guess that the positive group will contain the larger number. Likewise, since the coefficient is only , you can guess that the factors will be close. Two such factors of are and .

Therefore, your groups will be: