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Archive for the ‘Algebra Tricks’ Category

How to quickly find the y-intercept (b-value) of a line


Of course there’s a standard way to find the y-intercept of any line, and there’s nothing wrong with using that approach.

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But the method I’ll present here is a bit faster and therefore easer. And hey, if we can save time when doing math, it’s worth it … right?

So first let’s recall that the y-intercept of any function is the y-value of the function when the x-value = 0. That’s because the y-intercept is the y-value where the function crosses or touches the old, vertical y-axis, and of course all along the y-axis the x-value is always 0 (zero).

So the standard slope-intercept formula is y = mx + b. In a problem asking for the y-intercept, you’ll be given one point that the line passes through (that point’s coordinates will provide you with an x-value and a y-value), and you will also be told the slope of the line (the line’s m-value).
So then, to get the b-value, which is the value of the y-intercept, you just grab your y = mx + b equation (dust it off if you haven’t used it in a while), and plug in the three value you’ve been given: those for x, y and m. Then you solve the equation for the one variable that’s left: b, the value of the y-intercept.

Let’s look at an example: a line with a slope of 2 passes through the point (3, 10). What is this line’s y-intercept.

Now, according to the problem, the m-value = 2, the x-value = 3, and the y=value = 10. We just take these values and plug them into the equation:
y = mx + b, like this:

10 = (2)(3) + b

After doing these plug-ins, you just solve the equation for b, finding that
b = 4. That means that the y-intercept of the line = 4.

Now let’s see how you can do the same problem, but a little bit faster.
To do so, we first need to play around with the y = mx + b equation by subtracting the mx-term from both sides, like this:

y = mx + b [Standard equation.]
– mx = – mx [Subtracting mx from both sides.]
y – mx = b [Result after subtracting.]
b = y – mx [Result after flipping left & right sides
of the equation above.]

Aha! Look at that final, beautiful equation. This equation has b isolated on the left-hand side. So now if we want to solve for b, all we do is plug in the x, y and m values into the right-hand side of the equation and simplify the value, and the value we get will be the b-value.

For the problem we just solved, with x = 3, y = 10, m = 2, watch how easy it is to solve:

b = y – mx
b = 10 – (2)(3)
b = 10 – 6
b = 4

So notice that this technique, just like the first technique, reveals that the
y-intercept of the line is 4, or (0, 4). The techniques agree, they just get to the same end in slightly different ways.

Notice that with the second, quicker technique, you don’t need to add or subtract any terms. And that’s a key reason that this technique is faster and easier to use than the standard method. So try it out and stick with it if you like it.



Factoring Trick: How to Flawlessly Factor any “Difference of Two Squares” Binomial


If you’re staring at two terms you need to factor, but feel like a deer looking at the headlights of an oncoming semi, here’s a way to leap to safety!

It’s called the “Difference of Two Squares” trick.

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It requires four simple steps.

  1. Figure out if each of the terms is a “perfect square.”
  2. If so, take the square root of each term.
  3. Put each square root in its proper place inside two (    ).
  4. Put a + sign inside the first (   ), and put a – sign inside the second (   ).

Let’s do an easy example. Suppose the terms you’re looking at are these:
x^2  – 9

Let’s go through the 4 steps together.

  1. Figure out if each term is a “perfect square.”

    So, what does it mean for a number or term to be a “perfect square”?  It means that you get the number or term by multiplying a number or term by itself. For example, 16 is a perfect square because you can get 16 by multiplying 4 by itself:  4 x 4 = 16.

    So when we look at our two terms, x^2 and 9, we notice that both
    are perfect squares.
    9 is just 3 times 3.
    And in the same way, x^2 is just x times x.

  2.  Take the square root of each term.
    The square root of x^2 is just x.
    And the square root of 9 is just 3.

  3. Put each square root in the proper place inside two sets of (    ).
    We put the square root of the term that was positive first, and the square root of the term that was negative second.Since the x^2 was the positive term, we put its square root, x, first inside each
    (   ).  So far, that gives us:  (x    ) (x     )

    Since the 9 was the negative term because it had the negative sign in front of it: – 9, we put its square root, 3, second inside each (   ). So our (   )s now look like this:  (x   3) (x   3)

  4. Finally, we just need to put in signs that connect the terms inside
    the (    )s.

    That’s easy. We put a + sign inside one (    ), and we put a – sign
    inside the other (    ).
    I prefer to put the + inside the first (   ), but it really doesn’t matter.The final factored form, then, looks like this:  (x + 3) (x – 3)
    That’s all there is to it.

Now try these problems for practice.

           a)  x^2 – 16
           b)  x^2 – 100
           c)   x^2 – 121
           d)   x^4 –  16x^2
           e)   49x^8 – 144y^12

Answers:

           a)   x^2 – 16   =  (x + 4) (x – 4)
           b)  x^2 – 100  = (x + 10) (x – 10)
           c)   x^2 – 121  = (x + 11) ( x – 11)
           d)   x^4 –  16x^2  = (x^2 + 4x) (x^2 – 4x)
           e)   49x^8 – 144y^12  = (7x^4 + 12y^6)(7x^4 – 12y^6)