TAKS Tutorial

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TAKS Tutorial

• Algebra Objectives 1 4
• Part 2

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Remember

• This tutorial and all the ones presented
  previously can be found on Mrs. Nelsons website.
• There is a link to her site on the Spring High
  website.
• If you missed Objectives 1 4 part 1, you
  definitely want to go to the website to go
  through the lesson.

3
Today, we have a great deal to cover! Topics
include

• Simplifying expressions
• Setting up solving linear equations
• Setting up solving linear systems
• Setting up solving linear inequalities
• Writing linear equations from information
• Slope and intercepts

4
Lets start with expressions.This problem was on
the 2006 test.
This expression certainly can be simplified the
traditional way
5
The only problem with the traditional way is
that some of you do not correctly distribute the
negative/subtraction signs!
So lets look at an alternate way to do this
problem with a calculator. It takes longer, but
if you are someone who makes simple mistakesthis
is for you!
6
The only variable in the expression is x. Store a
value into your calculator for xany number
except 1, 0, or -1.

• Select and press your number, press STO, press
  X,T,T,n , and ENTER

Now, enter your expression, just as you see it,
into your calculator -3x(7x-4)6x-(13-24x2) Be
sure to double check that you typed the
expression correctly!
7
Now, press 2ND MATH to get TEST
We are looking for equal expressions so select
You should see the sign appear after your
expression (NO, do not press ENTERthe calculator
will not simplify this expression for you!)
Now, press ENTER. If this answer choice is
correct and the two expressions are equal, you
will see a 1 as an answer. If these two
expressions are not equal in value, you will see
0 as an answer.
Looks like option F is the correct answer! We
were lucky the first time. You might check the
other options to see that you get 0.
Now, one by one, enter the first answer choice
just as it appears. Double check to make sure you
typed it in correctly.
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That was a lot to grasp in one problem. Lets try
another one and practice with the calculator.
(2003)
Type in the first answer choice. Remember, 1 if
it is equal 0 if it is not.
The variable used is n. Select a number to use
for n and store it in the calculator.
Type in the expression exactly as it appears.
Double check correctness!
Get TEST (2nd Math) and select
Try again with option B. You do NOT have to
retype everything! Press 2nd ENTER
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That was a lot to grasp in one problem. Lets try
another one and practice with the calculator.
(2003)
Looks like the answer must be option D. Dont
assume! Check it out!
Use the backspace to go to the equal sign. Either
delete the first answer choice or type over it.
That wasnt the correct answer choice, either.
2nd Enter, backspace to the , and type in option
C.
Try again with option B. You do NOT have to
retype everything! Press 2nd ENTER
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We will now move on to solving equations This
problem was on the 2006 test
We are looking for C when they have given us F.
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We will now move on to solving equations This
problem was on the 2006 test
We can solve this equation the traditional
wayusing the undo process.
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Alternate method
We can solve this equation by using the table
feature of the graphing calculator.
Enter the equation.
Go to the table.
Scroll down the table until you find 104 in the
y-column
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Alternate method
You need to be able to see in the window where
the two lines intersect. That place looks way off
to the right.
Or we could use the graph and CALC features of
the graphing calculator
Enter the equation in y1 and 104 in y2.
Adjust the window again. Lets try the xmax at 50.
Adjust the window You need ymax to be higher
than 104
Graph
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Alternate method
Press 2nd TRACE so that you get CALC. Now, select
Intersect.
Move the cursor to be close to the point of
intersection.
Enter again for the second curve? And guess?
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This problem was NOT multiple choice. You have to
bubble in your answer correctly!

• Be careful!!! After going through all that work
  to get the correct answer, you dont want the
  problem to be scored as wrong because you didnt
  bubble in the answer properly!

4
0
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This problem was on the 2003 test. Lets approach
this one differently, since it is multiple choice.

• We could transform the equation so we can use the
  calculator, but too many people mess that
  upespecially with the subtraction sign!

Option F is not it.
Neither is option G
Option H is the one!
Since we were given answer choices for y, we will
use the calculator to substitute and simplify
until we find the y-value that gives us 18 for an
answer. We already know x is 3.
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Heres an inequality problem from the 2003 test
where they ask you to graph, but they dont look
at the graph! Well solve this problem the same
as the last one.

• In other words, dont be fooled by how
  complicated a problem looks!
• Study the problem. See what you know that can be
  used to make the problem easy to do!

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Let me again mentionthe state graders are not
going to look in the test booklets for your
graph! We are NOT going to use the grid to do
this problem!
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We are going to substitute these coordinates into
the expression. We want an answer that is less
than 12!
Option D gives us 12. We want an answer that is
less than 12. Sure enough, C is the answer.
A is more than 12
And so is B
Looks like C is the answer, but lets check to be
sure.
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TAKS will not always give you an expression,
equation or inequality.

• You may just have to read a word problem and come
  up with one of your own.
• The next several problems fit in this category.

21
2006
First, you want to find the profit on ONE tool
set. If it sells for 19.95 and is made for
4.37, the profit is the difference in the two
prices. 19.95 4.37 Eliminate B C
Now, the profit is made on every tool set sold,
which is s. We would multiply s to the profit
(19.95 4.37) on each tool set.
22
2006
Lets reason. A total of 80 backpacks were sold.
That means, if 1 35 backpack was sold, 79 of the
50 backpacks were sold. If 2 35 were sold, then
78 50 were sold. If 10 35 were sold, then 70
50 were sold.
In each case, I subtracted the number of 35
backpacks, x, from the total of 80. That is 80
x .
23
2006
Since x is the number of 35 backpacks, I would
need to multiply 35 and x together to get the
amount of money made by those purchases.
Since (80 x) is the number of 50 backpacks
sold, I would need to multiply 50 and (80 x).
24
2006
Total implies addition. Adding b and c should
be 220. b c 220
25
2006
As for the moneyEvery brownie sold for .75 so
.75 needs to be multiplied to b, the number of
brownies.
Please note that .50, the cost of each cookie is
multiplied to c.
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In some cases, you will be expected to use the
information given in a problem situation to set
up the procedure and to solve for the requested
information.

• How you do the problem is up to you and will not
  be checked. That you come up with the correct
  answer is the important piece this time.

27
2006
Perimeter is the sum of the lengths of the 3
sides 5x 10
Anytime a figure is used, DRAW LABEL it in the
test booklet!
We have no idea how long the base is so that is
x.
5x 10 95 5x 85 x 17
x
Each leg is 5 more than double the base 2x 5
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That is not the only way you can do this problem.
You can just use the answers and figure it out.
Lets look at option A. If the base is 17, then
based on what the problem says, the legs are each
2(17) 5 39
The perimeter means to add the two legs and the
base 17 39 39 95
95 is the perimeter we want, so A is the answer.
29
2006
This part means .11 times x
x is the number of minutes Lisa can talk.
Notice all graphs have a start at zero. That is
because Lisa cannot talk for less than 0 minutes,
which is no talking at all.
The connection fee is charged, no matter what.
30
2006
This part means .11 times x
Not more than means we have an inequalitythis
value is the upper limit. Lisa wants to spend 5
or less.
.50 .11x 5 .11x 4.50
x 40.9090
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Again, on this problem we could use the answer
choices to work toward the solution.
All of the graphs have 0 as a lower limit of
minutes. That makes the 0 a non-issue.
We need to look at the upper limit of the graphs
and see which one keeps us at 5 or less.
Lets start with option D since it has the
largest upper limit.
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Again, on this problem we could use the answer
choices to work toward the solution.
If minutes is 50, then .50 .11(50) 6. 6 is
over Lisas limit.
If minutes is 45, then .50 .11(45) 5.45 This
value is also over Lisas limit
X
If minutes is 40, then .50 .11(40) 4.9
4.90 is under Lisas 5 limit, so option B is
the correct one.
X
33
2003
Anna must get the same amount of money, m, for
selling the plates as it takes to make the plates
before she can make any profit.
x represents the number of plates Anna must sell
Her costs m 750 10x
Her revenue m 25x
As we discussed earlier, you can solve by
graphing the two equations and finding the
intersection. Or you can use the table and find
where the two y values are the same. Or you can
use substitution and solve algebraically.
34
2003
750 10x 25x
750 15x 50 x
35
2003
And, of course, you always have the option of
working with the answer choices to see which one
gives you equal values for her costs and how much
she makes selling the plates.
20 plates is a NO
30 plates is also NO
50 plates looks like a winner.
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Now, it is time to move on to the mechanics of
lines.

• We need to talk about
• Slope
• Y-intercepts
• X-intercepts
• Equations of lines through points
• If we do not get everything covered before the
  end of the session, please visit Mrs. Nelsons
  website to complete the power point lesson!

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About slope
The first thing you need to note is that the
slope is negative. That means the line should be
going downward. Eliminate the ones going up.
X
Now, look for the point (4, -3) on the remaining
2 graphs. The point must be ON the line.
X
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Rate of change is slope
Graph the line and see what it looks like.
The line is horizontal. It is running but it is
NOT rising. It has 0 rate of change because the
y-coordinates of every point on the line are
exactly the same.
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You want to remember

• Horizontal lines have 0 slope
• Vertical lines have undefined slope
• Diagonal lines have some fraction for a slope
• If the diagonal line goes up as it goes right,
  the slope is positive
• If the diagonal line goes down as it goes right,
  the slope is negative
• The steeper the line (the closer it is to the
  y-axis), the bigger the slope
• The flatter the line (the closer it is to the
  x-axis), the closer the slope is to zero (which
  would be horizontal).

40
2006
You are expected to know that the x-intercept is
where the graph touches the x-axis. You are
also expected to know that the y-coordinate is 0
for an x-intercept.
Option 1 put 0 in for y and find x.
2/3(0) -6x 12 0 -6x 12 6x
12 x 2
41
2006
You are expected to know that the x-intercept is
where the graph touches the x-axis.
This equation is not in the correct form for our
calculator. So, divide by 2/3Not you! Tell the
calculator to do it. Just be sure to use
parentheses!
Option 2 graph the line and see where it crosses
the x-axis.
y1 (-6x12)/(2/3)
x 2 at this point
42
2003
Find the point (5, -1) on the grid.
You need to use the equation x 3y 6 to
find the y-intercept.
There are a few ways to do that!
Remember, the state graders are NOT looking at
your graphs in your booklet. This grid is for you!
43
2003
The x-coordinate of the y-intercept is always
zero. Substitute 0 in for x and find y. 0 3y
6 -3y 6 y -2
You could transform the equation x 3y 6 into
slope-intercept form to find the y-intercept. -3y
6 x y -2 1/3 x
You could put the equation into a form that the
calculator will graph and find the y-intercept
visually.
-3y 6 x y (6 x)/-3
x -2 here
44
2003
Eliminate A and D since the b-value, the
y-intercept, is NOT -2.
X
Make a point where the y-intercept is -2.
Now, count the slope from (0,-2) to (5,-1).
up 1
right 5
X
The slope is 1/5.
45
2006
Use your formula chart and draw the line between
R and S.
You can see that the y-intercept cannot be 4.5,
so eliminate those two choices.
Now, count the slope between R and S.
46
2004
There are numerous ways to do this problem.
You could type each answer choice in your
calculator, go to the table, and look for BOTH
points.
BOTH points are herethis is it!
47
2004
Option 2
You are given graph paper at the end of the math
section. Make yourself a set of axes and graph
the two points.
The y-intercept is at 3.5 7/2 3.5
Draw the line between them.
Count the slope
48
2004
Option 3
You could use the slope formula on the formula
chart to calculate the slope. Then, you could use
slope-intercept form or point-slope form, which
is also on the formula chart to calculate the
y-intercept.
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Now, lets talk linear systems 2003
X
X
Neither of the given equations is in calculator
form. We need to get them into that form first.
We just need the y by itself. Using parentheses
will allow us the opportunity for the calculator
to calculate instead of us.
2x 3y 0 -3y 0 2x y (0
2x)/-3 put this function in y1
X
That does NOT mean they have to be in
slope-intercept form.
The slope in F is too steep as compared to this
graph.
The line going through (0, 0) must go upward.
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Now, lets make sure
X
X
x 2y -7 2y -7 x y (-7
x)/2 put this function in y2
X
Sure enough, we have a match.
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Now, lets make sure
X
X
You were not asked, but lets find the point of
intersection for the two linesthe solution to
the system.
X
Yes, this point is (-3, -2)
52
2006
Count the slope, follow the pattern, and extend
each line.
The point of intersection, the solution, is
(10,2).
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