Week of February 13, 2017

Algebra1

Systems of Linear Equations:
  Solving by Addition / Elimination

The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations.
If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.

x + 6 = 11
    –6    –6
x       =   5

You'll do something similar with the addition method.

• Solve the following system using addition.
2x + y = 9
3x – y = 16
Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:

2x + y = 9
3x – y = 165x      = 25
Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:

2(5) + y = 9
  10 + y = 9
          y = –1
Then the solution is (x, y) = (5, –1).

It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:

3(5) – y = 16
  15 – y = 16
        –y = 1
          y = –1

...which is the same result as before.

• Solve the following system using addition.
x – 2y = –9x + 3y = 16
Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!)
I'll multiply the second equation.

The "–1R₂" notation over the arrow indicates that I multiplied row 2 by –1. Now I can solve the equation "–5y = –25" to get y = 5. Back-solving in the first equation, I get:

x – 2(5) = –9
x – 10 = –9
x = 1
Then the solution is (x, y) = (1, 5).

A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y-value first and then the x-value second, and of course in points the x-value comes first. So just be careful to write the coordinates for your solutions correctly. 


Algebra2

You can also evaluate compositions symbolically. It is simpler to evaluate a composition at a point because you can simplify as you go, since you'll always just be plugging in numbers and simplifying. Evaluating a symbolic compositon, where you're first plugging x into some function and then plugging that function into some other function, can be much messier. But the process works just as the at-a-number composition does, and using parentheses to be carefully explicit at each step will be even more helpful.

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find ( f o g)(x).
In this case, I am not trying to find a certain numerical value. Instead, I am trying to find the formula that results from plugging the formula for g(x) into the formula for f(x). I will write the formulas at each step, using parentheses to indicate where the inputs should go:

( f o g)(x) = f (g(x))     = f (–x² + 5)     = 2(             ) + 3     ... setting up to insert the input formula     = 2(–x² + 5) + 3     = –2x² + 10 + 3     = –2x² + 13

If you plug in "1" for the x in the above, you will get ( f o g)(1) = –2(1)² + 13 = –2 + 13 = 11, which is the same answer we got before. Previously, we'd plugged a number into g(x), found a new value, plugged that value into f(x), and simplified the result. This time, we plugged a formula into f(x), simplified the formula, plugged the same number in as before, and simplified the result. The final numerical answers were the same. If you've done the symbolic composition (the composition with the formulas) correctly, you'll get the same values either way, regardless of the value you pick for x. This can be a handy way of checking your work.
Here's another symbolic example:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find (g o f )(x).
(g o f )(x) = g( f(x))     = g(2x + 3)     = –(           )² + 5    ... setting up to insert the input     = –(2x + 3)² + 5     = –(4x² + 12x + 9) + 5     = –4x² – 12x – 9 + 5     = –4x² – 12x – 4

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There is something you should note from these two symbolic examples. Look at the results I got:

( f o g)(x) = –2x² + 13 (g o f )(x) = –4x² – 12x – 4

That is, ( f o g)(x) is not the same as (g o f )(x). This is true in general; you should assume that the compositions ( f o g)(x) and (g o f )(x) are going to be different. In particular, composition is not the same thing as multiplication. The open dot "o" is not the same as a multiplication dot "•", nor does it mean the same thing. While the following is true:

f(x) • g(x) = g(x) • f(x)            [always true for multiplication]

...you cannot say that:

( f o g)(x) = (g o f )(x)           [generally false for composition]

That is, you cannot reverse the order in composition and expect to end up with the correct result. Composition is not flexible like multiplication, and is an entirely different process. Do not try to multiply functions when you are supposed to be plugging them into each other.

Usually composition is used to combine two functions. But sometimes you are asked to go backwards. That is, they will give you a function, and they'll ask you to come up with the two original functions that they composed. For example: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Given h(x) = (x + 1)² + 2(x + 1) – 3, determine two functions f (x) and g(x) which, when composed, generate h(x).
This is asking you to notice patterns and to figure out what is "inside" something else. In this case, this looks similar to the quadratic x² + 2x – 3, except that, instead of squaring x, they're squaring x + 1. In other words, this is a quadratic into which they've plugged x + 1. So let's make g(x) = x + 1, and then plug this function into f (x) = x² + 2x – 3:

( f o g)(x) = f (g(x))     = f (x + 1)     = (         )² + 2(         ) – 3     = (x + 1)² + 2(x + 1) – 3
Then h(x) may be stated as the composition of  f (x) = x² + 2x – 3 and g(x) = x + 1.

• Given h(x) = sqrt(4x + 1), determine two functions f (x) and g(x) which, when composed, generate h(x).
Since the square root is "on" (or "around") the "4x + 1", then the 4x + 1 is put inside the square root. I need to take x, do "4x + 1" to it, and then take the square root of the result:

g(x) = 4x + 1,  f(x) = sqrt(x), and h(x) = ( f o g)(x).

© Elizabeth Stapel 2003-2011 All Rights Reserved