Yeti Math Bloggie 34: Alina’s Cookies :P

Yo peeps, Yeti here! As a lover of cookies 😛 I shall put a math related cookie problem in this post, because I merely feel like it. We start with the problem itself:

When Alina was hungry she ate c cookies, which was 30% of all the cookies she had. How many cookies did she have in the beginning?

You sees, there are 3 numbers the problem:

  • The total number of cookies: x
  • The number of cookies she ate: c
  • % of the cookies she ate: 30%

There is a word in this problem that can help us with solving it: of                       This word means to multiply the #’s in front on behind the word. More can be found out in post #17 X of Y, of this series. “… 30% of all the cookies she had. …” basically means, we get that (assuming x is the total number of cookies) c is 30% of x.

Translating that into an algebraic equation gets you:

(1)   \begin{equation*} x \times \dfrac{3}{10} = c \end{equation*}

Solving for x gets you

(2)   \begin{equation*} x = \dfrac{10c}{3} \end{equation*}

Yeeto Yeets!

Yeti Math Bloggie 32: Solving Inequalities

Yo peeps! Yeti here, explaining the steps of inequality solving. We shall start by examining the following problem:

(1)   \begin{equation*} x-3 > 5 \end{equation*}

Our first step would be to make the expression x-3 equal to 5. We can then get the following equation:

(2)   \begin{equation*} x-3 = 5 \end{equation*}

After solving, we can get that the x=8. But if you remember, the original problem doesn’t have x-3 equal to 5. x-3 is actually greater than 5. We can draw this table to help:

x 7 8 9
x-3 4 5 6

See how we put x in the top row and the expression x-3 in the bottom row. For x, we put a value less than what we solved, then our solved value (8), and finally a value greater than 8. Now we can plug these values to x in the expression x-3 and get the values 4, 5, and 6.

Since the problem asks us to make x-3 larger than 5, we can look at our table and find all the values of x-3 that make our value larger than 5. We can see then when x is 9, our value is 6, which is bigger than 5. In fact, any # bigger than 8 will make our value larger than 5 (for example for x=8.1, x-3=5.1 which is larger than 5). Therefore, we have solved our problem and thus, our answer:

(3)   \begin{equation*} x>8 \end{equation*}

Now, let’s solve this more complicated inequality, with absolute value.

(4)   \begin{equation*} |2x-3| > 7 \end{equation*}

We can start off by finding those x for which |2x-3| is 7.

(5)   \begin{equation*} |2x-3| = 7 \end{equation*}

Since absolute value equations (usually 😛 ) have 2 answers, our table will look a bit different. First, we need to find the 2 values, which in this case are 5 and -2. Now that we have our values, we can commence the table:

x -2 5
|2x-3|

We first place the 2 values for x in ascending order. Next, we can put more values, (1 less than, 1 in between, and 1 more than) to keep the ascending order.

x -4 -2 2 5 7
|2x-3|

We want values that are close to the 2 original values, but still easy to compute. Finally, we plug in all the values for x to the expression |2x-3| to get:

x -4 -2 2 5 7
|2x-3| 11 7 1 7 11

Notice how we already knew the values of -2 and 5 because we already solved for those values, using the equation |2x-3|=7 (Maybe this is also a good time to verify your answers… 😛 )

Its pretty easy to now (using the fully filled in table) find the inequalities, noticing to make |2x-3| larger than 7,  x has to be smaller than -2, or x to be bigger than 5. We than get the answers:

(6)   \begin{equation*} x<-2 \text{ or } x>5 \end{equation*}

Hope you found this post helpful, Till next time peeps, Bai! 😀

Yeti Math Bloggie 31: Distance, Speed and Time

Yo peeps, Yeti here, explaining the topic of speed, distance and time. Now, Yeti here is on a very tight schedule here and so this post shall be very short and basic. So, lets start with relationships.

(1)   \begin{equation*} \text{Time} \times \text{Speed} = \text{Distance} \end{equation*}

(2)   \begin{equation*} \text{Time} = \frac{\text{Distance}}{\text{Speed}} \end{equation*}

(3)   \begin{equation*} \text{Speed} = \frac{\text{Distance}}{\text{Time}} \end{equation*}

Now, we shall present you with a problem and guide you through a step by step explanation on how to solve it. Thus, the problem:

One hour after a hiker left a camp, a cyclist set out to catch up. If the hiker traveled 3 miles per hour and the cyclist traveled 6 miles per hour faster, how long will it take the cyclist to catch up with the hiker?

To organize the information, you can make a table:

Person Speed Time Distance
Hiker 3 mph x+1 3x+3
Biker 9mph x 9x

We can get the speed from the problem and we can assign x to the biker’s time, or duration. Since the hiker left one hour earlier, you can put x+1 for his duration. Using the above formula, we can calculate the distance covered the the hiker and the biker.

Once the biker caught up, the distances would be equal so we can put:

(4)   \begin{equation*} 3x+3=9x \end{equation*}

After doing immensely hard and deep calculations, we find the x= \dfrac{1}{2}, so the biker will catch up to the hiker in 30 minutes, or \dfrac{1}{2} of an hour.

Yeti Math Bloggie 28: The Wonders of Absolute Value Explained

So first, Yeti must say zat absolute value is pretty easy to learn! I, Yeti shall also explain how they can be used in equations and other helpful stuffs. Lets start by reviewing the definition shall we?

Absolute value: How far away is a # on a number line from zero

Now, if you do some very deep and logical thinking, you can find that the absolute value is ALWAYS positive (with the exception of 0 of course! 😛 ). Tis just a fact, you must believe it. I must also mention the the number itself can be negative, just not the absolute value of the number it self. Now, lets jump into it! 😀

First, we shall display some very simple + basic examples:

(1)   \begin{equation*} |3| = 3 \end{equation*}

(2)   \begin{equation*} |-10| = 10 \end{equation*}

(3)   \begin{equation*} |0| = |-0| = 0 \end{equation*}

But since, zee great great world of algebra is developing 😀 , we shall display zat branch of math by expressing things with letters! Here are some examples (Plz remember a could stand for any number 😀 ):

(4)   \begin{equation*} \begin{split} &a > 0    \quad   |a| = a \\ &a = 0   \quad  |a| = a = 0 \\ &a < 0  \quad  |a| =  -a \end{split} \end{equation*}

But as you probably already know, algebra can get real fancy! Now, lemme give a concrete example by using 4x -y:

(5)   \begin{equation*} \begin{split} &4x-y > 0    \quad   |4x-y| = 4x-y \\ &4x-y = 0   \quad  |4x-y| = 4x-y = 0 \\ &4x-y < 0  \quad  |4x-y| =  -(4x-y) \end{split} \end{equation*}

Diddja notice the subtle/ extremely mini and tiny difference of the last equation? If you happen to have extremely good eyesight, or simply a prodigy on math, or you happen to be very lucky, you would have noticed zat in the last equation, there was a – sign applied to the expression which is enclosed by parenthesis. Why should we do that? Look back at 4, it says when a<0, |a| = -a, here a stands for any number, or any expression, so when we remove the absolute value operator, the negative sign should be applied to a as a whole, in the above case, the expression in the absolute value operator is 4x-y, so the negative sign should be applied to the expression as a whole as -(4x-y) 😀 .

For some simpler expression, we can figure out more about its relationship with 0, for example x-3. You see, it all depends on what x in the expression stands for. If say x stands for a whopping 9876 (or any number larger than 3 for that matter) the value would be positive. If say x equaled exactly 3, we would then know that the value would be zero. Lastly, if x equals a mere 0.093739 (or any number smaller than 3 for that matter) the value of the expression would be negative.

(6)   \begin{equation*} \begin{split} &x > 3    \quad   |x-3| = x-3 \\ &x = 3   \quad  |x-3| = x-3 = 0 \\ &x < 3 \quad  |x-3| =  -(x-3) \end{split} \end{equation*}

For a slightly complicated expression like 2x-8, when x = 4, we get 8-8 which is 0. This means that the expression equals to zero.

This also means that when x is like say 8, than we get 16-8=8. You could also put any other number for x, as long as it satisfies this condition: x>4. This expression gets an answer larger then zero.

Similarly, when x = 1, we get 2-8 which equals -6. -6 is less than zero. As you probably figured out, for any value less than 4, (a.k.a. x < 4), the value of the expression would then be less than zero.

Exercises

Let’s do a few exercises to check your understanding now!

  • Rewrite the expression |3x - 6| without absolute operators.
  • Simplify expression |z-9| - |z-7| when z<7.
  • |x+1|=2, what are the possible values of x?

Yeti Math Bloggie (27): Definition Of Slope and the Equation of a Straight Line

As the following picture shows, the slope of a straight line can be found by using the coordinates of two points P1(x_1, y_1) and P2(x_2, y_2) on the line as:

(1)   \begin{equation*} \text{slope} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \end{equation*}

A straight line consists of many points, and every point has an abscissa  x and an ordinate y. For a straight line there is an equation which describes the relationship between  the abscissa x and the ordinate y as below.

(2)   \begin{equation*} y=kx+b \end{equation*}

What does this equation mean? For an arbitrary point P(x, y), the equation decides if the point P is on the straight line or not. After you put the abscissa and ordinate of P into the lines equation. If the two sides equal to each other, then the P is on the line, otherwise P is not on the line.

The equation can also help you to calculate abscissa or ordinate of P if you know the point is on the line.

  1. If you know that P is on the line, and you know the abscissa of P, you can calculate the Ordinate of P.
  2. If you know that P is on the line, and you know the ordinate of P, you can calculate abscissa of P.

For example, if a straight line with equation y = x + 1, is the point (3, 4) on that line? Zee point is on zee line because 3+1=4. On the other hand, point (1, 0) is not on the line, because the left side has y=0, but on the right side we have x+1 = 1+1 = 2, the two sides do not equal.

There is a key fact that, if a line’s equation is y=kx+b, then the slope of the line is k.     Here is zee proof zat zee piece of information is tru. It’s kinda a fact but an anonymous peep wishes a certain author to prove it so here we go! 😀

Suppose there are two points P1(x_1, y_1) and P2(x_2, y_2) on the line, since the equation of the line is y = kx + b, apparently we have:

(3)   \begin{equation*} y_1 = k x_1 + b \end{equation*}

(4)   \begin{equation*} y_2 = k x_2 + b \end{equation*}

Now from the definition of slope, we have:

(5)   \begin{equation*} \begin{split} \text{slope} &= \frac{y_2 - y_1}{x_2 - x_1} \\ &=\frac{(kx_2+b)-(kx_1+b)}{x_2-x_1} \\ &=\frac{kx_2+b-kx_1-b}{ x_2-x_1} \\ &=\frac {kx_2-kx_1}{x_2-x_1} \\ &=\frac {k(x_2-x_1)}{x_2-x_1} \\ &=k \end{split} \end{equation*}

And so,this is the proof that for a line with equation y=kx +b, its slope is k. Here you have it kids, proof for lots of math calculations and stuffs, ’till next time kids, ADIOS!

Exercises

  • Find the slope of the line passing through each of the following pairs of points: (4, 0) and (-2, 8).
  • If the equation of a line is 7x - 5y = 12, what is the slope of the line?