Kathryn's Responses in WyzAnt Answers

When you're solving an equation for a variable (in this case x) you are trying to get x on one side of the equal sign all by itself. To do that, you do the opposite of what is being done to it (so that you can get the number in front of the x to equal 1). So when you have:

17 x = -85 

x is multiplied by 17 -- to get rid of that 17, we have to divide by 17 on both sides of the equation. You have to do it on both sides of the equation to keep the equation balanced.  

17 x ÷ 17 = -85 ÷ 17

On the left side of the equation the 17s get canceled out (specifically because 17/17 = 1), so what you're left with is:

x = -85 ÷ 17

That you can punch into a calculator to get 

x = -5

The way I see it is like this:

He ended with $42.15 -- the last thing that he did before that was buy the umbrella
$42.15 + $23.75 (umbrella) = $65.90 

The next thing backwards in order is that he spent "half of his remaining money on a painting", so what he has left is one half what he had before he bought the painting. 
$65.90 + $65.90 (cost of the painting) = $131.80

Before that, he spent $25 on dinner and $20 on a cab. So we add those all up:
$131.80 + $25 (dinner) + $20 (cab) = $176.80 (how much money he had at the start of the weekend)

To write it as a single equation:

$42.15 = -$23.75 + (-$25.00 + -$20.00)/2

end of the weekend = cost of umbrella + (cost of dinner + cost of cab)/cost of painting

Because the things to the right of the equal sign are costs, they are losses of money and therefore negative numbers. You don't have a cost for how much the painting is until you know how much he had before he bought the umbrella because it said that he spent "half of his remaining money on a painting", so he must have had twice the amount of money as how much he had before he bought the umbrella.

:) 

The way that I would solve this problem is by graphing both the purse snatcher (thief) and the police officer, and the point where they intersect is where the officer catches up to the thief. The y-axis can represent how many meters were run, and the x-axis can represent how many seconds it has taken. 

Step 1. Figure out the slope for both the thief and the officer.

Thief: Can run 20 meters in 4 seconds. If we write that as a slope, it would be 20 meters/4 seconds. We can simplify that down to 5 meters/second. Therefore our slope will be 5/1 or 5. 
Officer: Can run 32 meters in 4 seconds. If we write that as a slope, it would be 32 meters/4 seconds. We can simplify that down to 8 meters/second. Therefore our slope will be 8/1 or 8.

Step 2. Figure out the y-intercepts for both the thief and the officer. 

Thief: Started 9 meters ahead of the officer. So the y-intercept would be (0,9). 
Officer: Started 9 meters behind the theif. So the y-intercept would be (0,0). 

Step 3. Write the equations for the line for the thief and the officer.

Thief: y = 5x + 9
Officer: y= 8x + 0

Step 4. Set the lines equal to each other and solve for x. This is because the point at which they intersect will be the same point for both lines, so if you find x you can put that value in either equation to find the value for y. 

y = 5x + 9 and y = 8x + 0
5x + 9 = 8x + 0
    -9       -9

5x = 8x -9
-8x  -8x

-3x = -9
/-3   /-3

x = -9/-3
x = 3

Then plug in the x value into either of the equations:

y = 5x + 9     or     y = 8x + 0
y = 5(3)+9     or     y = 8(3)+0
y = 15 + 9     or     y = 24 + 0
y = 24         or     y = 24

Step 5. Analyze your results.

The point that you end up with is x = 3, y = 24 or (3,24). Since the y-axis is how many meters were traveled, and the x-axis is how many seconds. That means that in 3 seconds, the officer would have caught up to the thief. As well, we can tell that the officer caught the thief 24 meters away from the starting point. 

http://www.mathsisfun.com/definitions/monomial.html

General Information:

For x² ± # x ± # you have a ( x ± # )( x ± # ) The two numbers at the end have to add together to get to # x  and have to multiply together to get to the ending # 

When you have a number in front of the x² that number must be split up so that when you multiply ( x ± # )( x ± # ) your "x"s will get you back to that number. So if you have a problem that starts with 6x2 you can have either ( 1x ± # )( 6x ± # ) OR ( 2x ± # )( 3x ± # ) because 1*6=6 and 2*3=6.

Your first problem:

9x² - 16

The first thing that we have to worry about it the 9. You can have either (9x ± #)(1x ± #) OR (3x ± #)(3x ± #). What I would usually do from here is figure out what multiplies together to get sixteen: 1*16, 2*8, 4*4, 8*2, and 16*1 (order matters!). Then (becasue we have to take into account the numbers in front of x), I would plug them into each of the equations until I found one where the singular "x"s canceled out. That would give us our 9x² +0x -16

Because I've been at this a while, I know that 16 is a perfect square (4*4=16), So, I would choose the second of the two potential answer bases (3x ± #)(3x ± #) to use. Because 4 is the same either way, and I know that they need to cancel out in the middle, one would be positive and the other would be negative to look like this: (3x + 4)(3x - 4)  then, like any good problem, you should multiply it back out and check that yes (3x + 4)(3x - 4) = 9x²-16