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# Jennifer's Responses in WyzAnt Answers

#### how do I simplify?15x^2 + 24x +9/3x + 3

ok the question is how do i simplify the rational equation, If I can

15x2 + 24x + 9/ 3x + 3

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Asked by Jeremy from Milwaukee, WI
10

You take the same approach as with simplifying other rational expressions like... fractions!

Simplifying 15/6 is a no-brainer for most algebra students.
One method is to break up both the numerator and denominator into factors. --> (3*5)/(2*3)
Obviously, you would cancel out the common factor, 3, giving you 5/2 as the simplest form.

Same principle with ratios of polynomials! You want to create factors and look to cancel out common factors.

15x2 + 24x + 9 can be factored into (3x+3)(5x+3)
**if you need explanation on how to factor this polynomial, see below.

So your factored rational expression now looks like (3x+3)(5x+3)/(3x+3)(1).
Notice that the numerator and denominator both have 3x+3 as a factor, so that factor will cancel out, leaving you with (5x+3)/1 or just 5x+3.

**help on factoring the trinomial in the numerator begins here:
If you have a trinomial that looks like ax2+bx+c, it's called a quadratic trinomial.
I like to call the first and last term, ax2 and c, the "anchors."

To begin, start by thinking of factors for your anchors.
15x2 can be factored into 3x and 5x       or       x and 15x
9 can be factored into 3 and 3               or       1 and 9

So now, you have a few possibilities. A few (not all) of which are:
(3x ± 3)(5x ± 3) or (3x ± 9)(5x ± 1) or (x ± 3)(15x ± 3) or (x ± 1)(15x ± 9)

Now all the above possibilities are based on your "anchors." Here's where your middle term comes in, your bx, or in this case, 24x. When you "FOIL" the above factors, you want the two middle terms to add up to 24x.

As an arbitrary example, say we used (3x+9)(5x+1). If we FOIL, we get 15x2+3x+45x+9 = 15x2+48x+9
Notice, our anchors are "good," but our middle term is not 24x. It's 48x, and that's way off! Now, you could try (3x+9)(5x-1), but that would give you 15x2-3x+45x-9. Now you have the wrong middle term AND the wrong "anchor." (your anchor was a positive 9, not a negative 9, remember?) In order to guarantee a positive 9, you will see that your factors will have to be (a+b)(c+d) or (a-b)(c-d) because (a+b)(c-d) will not give you positive anchors.

It does require some trial-and-error, but with perseverance, you'll eventually find that the correct factors are (3x+3)(5x+3).

#### Need an Application of Linear Equation

An auto factory sends cars from plant 1 and 2 to dealership a and b. Plant 1 has 28 cars to send, Plant 2 has 8 to send. Dealer A needs 20 cars, Dealer B needs 16. Transportation costs are \$220 from 1 to A, \$300 from 1 to B, \$400 from 2 to A, and \$180 from 2 to B. Transportation costs are limited to \$10640. How many cars should be sent from each plant to each of the dealerships?

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Asked by Betty from Hillsboro, NH
00

Here's one way of solving this problem:

Plant 1 has 28 cars, x of which will be sent to dealership A, and the rest, 28-x, will be sent to dealership B.

Transportation cost for Plant 1 to Dealerships A and B will be 220(x) + 300(28-x)

Plant 2 has 8 cars, y of which will be sent to dealership A, and the rest, 8-y, will be sent to dealership B.

Transportation cost for Plant 2 to Dealerships A and B will be 400(y) + 180(8-y)

This would make the total cost the sum of the costs for Plant 1 and Plant 2:

[220(x) + 300(28-x)] + [400(y) + 180(8-y)] = 10640

-80x + 220y +9840 = 10640

Now, since we have 2 variables, we must have 2 unique equations to solve for both of the variables. We have one equation above. Recall that Dealer A needs 20 cars. Plant 1 is sending x many cars to Dealer A, and Plant 2 is sending y many cars to Dealer A. This means x + y = 20. This is your second equation.

Using elimination or substitution, you can now solve for x and y. Let's use substitution.

-80x + 220(20-x) +9840 = 10640

x = 12

y = 8

Plant 1 has 28 cars, 12 of which will be sent to dealership A, and the rest, 28-12, or 16, will be sent to dealership B.

Plant 2 has 8 cars, 8 of which will be sent to dealership A, and the rest, 8-8, or 0, will be sent to dealership B.

#### How much money will be available to use for college tuition __ years later?

When david was born, his uncle deposited \$8,000 into an account paying 6% interest, compounded quarterly. How much money will be available to use for college tuition 18 years later?

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Asked by Cool from Thermal, CA
00

If you have an account paying 6% interest compounded quarterly, that means after the first 3 months, you get 6%/4 = 1.5% interest on your current account balance. Then after the next 3 months, you get another 1.5% interest paid on the current balance. Then after the next 3 months, another 1.5%, and so on and so forth.

If you start out with \$8,000, then 3 months later, you will have \$8,000*(1 + .015).
3 months after that, you will have [\$8,000*(1 + .015)]*(1 + .015).
3 months after that, you will have [[\$8,000*(1 + .015)]*(1 + .015)]*(1 + .015).
3 months after that, you will have [[[\$8,000*(1 + .015)]*(1 + .015)]*(1 + .015)]*(1 + .015)

or \$8,000*(1 + .015)4

So, in a year, if interest is being compounded quarterly, it's being compounded four times in a year, which explains why the exponent is a 4.
6% = .06 and that divided by four gives us the .015 in the expression above.
This should help you understand why the general formula for problems of this sort is:

A = P*(1+r/n)nt where r is interest rate, n is the number of times it's compounded in a year, and t is the number of years. P is the principal, what you initially invest, and A is the amount you have in your account after t years.

Thus...

A = 8000*(1+.06/4)4*18

A = 8000*(1.015)72

A = 23,369.26

David will have \$23,369.26 available to use for college tuition 18 years later. Hopefully.

#### solve for each equation for the variable indicated 1. s+4t= r is s2. 3m-7m= p for m

it sf or algebra 1

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Asked by Alisssa from Mountain Home A F B, ID
00

1. s + 4t = r, solve for s

Your goal is to get s = something, so isolate the s! Right now, your s is followed by "+ 4t."
To get s by itself, you want to get rid of 4t from the left-hand-side of your equation.
To that end, subtract 4t from each side of the equation. (thus keeping your equation "balanced")

(s + 4t) - 4t = r - 4t

s + 4t - 4t = r - 4t

s + 0 = r - 4t

s = r - 4t

2. 3m - 7m = p, solve for m

Your goal is to get m = something, but you have more than just one m term.
In fact, you have 3 m's from which you are subtracting 7 m's!
That means you have "negative 4" m's. (algebraically speaking, since you can't have a negative number of things in real life)

-4m = p

Now at this point, many students make the mistake of adding 4 to each side. DON'T DO THAT! Why?
Well, what's going on between -4 and m? Are you adding them? Subtracting? Multiplying? Dividing?
Right, the operation that's not written, but implied, is multiplication. -4m is (-4)*(m).
Hence, if you want to get rid of the -4 that's being multiplied to the m, you'll want to use multiplication or division to get rid of the -4, not addition or subtraction.

This gives you two options:

1) divide each side by -4
2) multiply each side by -1/4 (because dividing by a number is the same as multiplying by the reciprocal of that number)

-4m = p

(-4*m)/(-4) = p/(-4)

m/1 = p/(-4)

m = - p/4

*note: - p/4 = -p/4 = p/-4

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