Variables - PSAT Math

With this problem, you need to take the trinomials out of parentheses and combine like terms. Since the two trinomials are being added together, you can remove the parentheses without needing to change any signs:

The next step is to combine like terms, based on the variables. You have two terms with , two terms with , and two terms with no variable. Make sure to pay attention to plus and minus signs with each term when combining like terms:

With this problem, you need to distribute the two fractions across each of the trinomials. To do this, you multiply each term inside the parentheses by the fraction outside of it:

The next step is to combine like terms, based on the variables. You have two terms with , two terms with , and two terms with no variable. Make sure to pay attention to plus and minus signs with each term when combining like terms. Since you have a positive and negative , those two terms will cancel out:

To add these two trinomials, you will first begin by combining like terms. You have two terms with , two terms with , and two terms with no variable. For the two fractions with , you can immediately add because they have common denominators:

To add trinomials, identify and group together the like-terms: . Next, factor out what is common between the like-terms:. Finally, add what is left inside the parentheses to obtain the final answer of .

All operations are addition, so we can first remove the parentheses:

Now rearrange the terms so that like terms are next to each other:

Combine like terms:

Find the sum:

Write the answer in standard form.

Combine like terms:

Write the answer in standard form (terms with the highest degree first):

Set up an equation for the sentence: 15 – 3 = 2x and solve for x. X equals 6. If you plug in 6 for x in the expression 24/x, you get24/6 = 4. 4 is only choice greater than a.

Notice that - (a-b) = b-a, so statement I & III are true after substituting the expression. Substitute the expression for statement II gives ((a+b)/(a-b))((a+b)/(b-a))=((a+b)(b+a))/((-1)(a-b)(a-b))=-1 〖(a+b)〗2/〖(a-b)〗2 =-((a+b)/(a-b))2 = -(a♦b)2 ≠ (a♦b)2

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.

Cancel by subtracting the exponents of like terms:

By the remainder theorem, if a polynomial is divided by the linear binomial , the remainder is - that is, the polynomial evaluated at . The remainder of dividing by is the dividend evaluated at , which is

Find the quotient:

Step one: Factor the numerator

Step two: Simplify

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.