Algebra - PSAT Math

We need to rewrite this problem in mathematic terms.

8 less than 32 can be written as 32 – 8

so we will get the equation

18 – w = 32 – 8.

Now, we need to solve for w.

–w = 32 – 8 – 18

–w = 6

w = –6

Find the value of the given expression, , by plugging in –6 for w.

For F(1/2), x2=1/4, which is less than 1, so we use the bottom equation to solve. This gives F(1/2)= 4(1/2)2 – 2(1/2) + 2 = 1 – 1 + 2 = 2

Subtract 5 from both sides of the inequality:

Multiply both sides by 5:

Therefore only I must be true.

Add the two equations together to yield 5x + 5y = 9a + b, then factor out 5 to get 5(x + y) = 9a + b; divide both sides by 5 to get x + y = (9a + b)/5; subtract the two equations to get x - y = -a - 3b. So, x2 – y2 = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–\[(9a)\]2 – 28ab – \[(3b)\]2)/5

The best way to solve this problem is to plug in an appropriate value for a. For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4.

Then 3_a + 5_, where a = 11, gives us 38. Then 38 divided by 3 gives a remainder of 2.

The algebra method is as follows:

a divided by 7 gives us some positive integer b, with a remainder of 4.

Thus,

a / 7 = b 4/7

a / 7 = (7_b +_ 4) / 7

a = (7_b_ + 4)

then 3_a + 5 =_ 3 (7_b_ + 4) + 5

(3_a_+5)/3 = \[3(7_b_ + 4) + 5\] / 3

= (7_b_ + 4) + 5/3

The first half of this expression (7_b_ + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

Divide the coefficients and subtract the exponents.

You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative

(x –__)(x –__).

You should realize that 6 fits into both blanks.

You must now set each set of parenthesis equal to 0.

x – 6 = 0; x – 6 = 0

Solve both equations: x = 6

We first expand the right hand side as x2+2x+tx+2t and factor out the x terms to get x2+(2+t)x+2t. Next we set this equal to the original left hand side to get x2+rx +6=x2+(2+t)x+2t, and then we subtract x2 from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.

We are told that x3 - y3 = 56. We can factor the left side of the equation using the formula for difference of cubes.

x3 - y3 = (x - y)(x2 + xy + y2) = 56

Since x - y = 2, we can substitute this value in for the factor x - y.

2(x2 + xy + y2) = 56

Divide both sides by 2.

x2 + xy + y2 = 28

Because we are trying to find x2 + y2, if we can get rid of xy, then we would have our answer.

We are told that 3xy = 24. If we divide both sides by 3, we see that xy = 8.

We can then substitute this value into the equation x2 + xy + y2 = 28.

x2 + 8 + y2 = 28

Subtract both sides by eight.

x2 + y2 = 20.

The answer is 20.

ALTERNATE SOLUTION:

We are told that x - y = 2 and 3xy = 24. This is a system of equations.

If we solve the first equation in terms of x, we can then substitute this into the second equation.

x - y = 2

Add y to both sides.

x = y + 2

Now we will substitute this value for x into the second equation.

3(y+2)(y) = 24

Now we can divide both sides by three.

(y+2)(y) = 8

Then we distribute.

y2 + 2y = 8

Subtract 8 from both sides.

y2 + 2y - 8 = 0

We need to factor this by thinking of two numbers that multiply to give -8 but add to give 2. These numbers are 4 and -2.

(y + 4)(y - 2) = 0

This means either y - 4 = 0, or y + 2 = 0

y = -4, or y = 2

Because x = y + 2, if y = -4, then x must be -2. Similarly, if y = 2, then x must be 4.

Let's see which combination of x and y will satisfy the final equation that we haven't used, x3 - y3 = 56.

If x = -2 and y = -4, then

(-2)3 - (-4)3 = -8 - (-64) = 56. So that means that x= -2 and y = -4 is a valid solution.

If x = 4 and y = 2, then

(4)3 - 23 = 64 - 8 = 56. So that means x = 4 and y = 2 is also a valid solution.

The final value we are asked to find is x2 + y2.

If x= -2 and y = -4, then x2 + y2 = (-2)2 + (-4)2 = 4 + 16 = 20.

If x = 4 and y = 2, then x2 + y2 = (4)2 + 22 = 16 + 4 = 20.

Thus, no matter which solution we use for x and y, x2 + y2 = 20.

The answer is 20.

First, add 4 to both sides:

Divide both sides by 2:

First, subtract 3 from both sides in order to obtain an equation that equals 0:

The left side can be factored. We need factors of that add up to . and work:

Set both factors equal to 0 and solve:

To solve the left equation, add 1 to both sides. To solve the right equation, subtract 3 from both sides. This yields two solutions:

Only one of these solutions is negative, so the answer is 1.

can be seen to fit the pattern

:

where

can be factored as , so

, as the sum of squares, is a prime polynomial, so the complete factorization is

,

making the only prime factor, and "one" the correct choice.

Cancel by subtracting the exponents of like terms:

To begin, factor out any like terms from the expression. In this case, the term can be pulled out:

Next, recall the difference of squares:

Here, and .

Thus, our answer is

.

You need to find two numbers that multiply to give 72 and add up to give 18

easiest way: write the multiples of 72:

1, 72

2, 36

3, 24

4, 18

6, 12: these add up to 18

(x + 6)(x + 12)

Let's try to factor x2 – y2 – z2 + 2yz.

Notice that the last three terms are very close to y2 + z2 – 2yz, which, if we rearranged them, would become y2 – 2yz+ z2. We could factor y2 – 2yz+ z2 as (y – z)2, using the general rule that p2 – 2pq + q2 = (p – q)2 .

So we want to rearrange the last three terms. Let's group them together first.

x2 + (–y2 – z2 + 2yz)

If we were to factor out a –1 from the last three terms, we would have the following:

x2 – (y2 + z2 – 2yz)

Now we can replace y2 + z2 – 2yz with (y – z)2.

x2 – (y – z)2

This expression is actually a differences of squares. In general, we can factor p2 – q2 as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.

x2 – (y – z)2 = (x – (y – z))(x + (y – z))

Now, let's distribute the negative one in the trinomial x – (y – z)

(x – (y – z))(x + (y – z))

(x – y + z)(x + y – z)

The problem said that factoring x2 – y2 – z2 + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.

(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.

The answer is 2.

Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.

So 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4

Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us

9_x_2 + 12_x_ + 4

= 9_x_2 + 6_x_ + 6_x_ + 4

= 3_x_(3_x_ + 2) + 2(3_x_ + 2)

= (3_x_ + 2)(3_x_ + 2)

This is as far as we can factor.

is a difference of squares.

The difference of squares formula is .

Therefore, = .