Transcript TAKS_Tutorial__objectives_1-4_part_2 - Ayyadhury

TAKS Tutorial
Algebra Objectives 1 – 4
Part 2
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
Let’s start with expressions.
This problem was on the 2006 test.
This expression
certainly can be
simplified the
“traditional” way:
Distribute :
- 21x  12x  6x - 13  24x
Combine like terms :
2
3x  18 x  13
2
2
The only problem with the “traditional”
way is that some of you do not correctly
distribute the negative/subtraction signs!
So let’s look at an
alternate way to do
this problem with a
calculator.
It takes longer, but if
you are someone
who makes simple
mistakes…this is
for you!
The only variable in the expression is x. Store a
value into your calculator for x—any number except
1, 0, or -1.
Select and press your number,press STO, press
X,T,Θ,n , and ENTER
Now, enter your
expression, just as you
see it, into your
calculator…
-3x(7x-4)+6x-(13-24x^2)
Be sure to double check
that you typed the
expression correctly!
Now, press 2ND MATH to get TEST
We are looking for equal
expressions so select =
Now,
Youpress
should
ENTER.
see the
If this
= sign
answer
appear
choice
afterisyour
correct and
theexpression
two expressions are equal,
you will see a “1” as an
(NO, do not press
answer. If these two
Looks like option F is the
Now,
one
by
one,
enter
the
ENTER—the calculator
correct answer! We were
expressions are not equal in
firstlucky
answer
choice
just
as
it
will not simplify this
the first time. You
value, you will see “0” as an
might Double
check thecheck
other to
appears.
expression for you!)
answer.
options to see that you
make sureget
you
typed it in
“0”.
correctly.
That was a lot to grasp in one problem. Let’s try
another one and practice with the calculator.
(2003)
nd
Type
in
first
answer
Try
Type
Get
The
again
TEST
variable
in the
the
with
(2
expression
used
option
Math)
isB.
choice.
Remember,
You
exactly
and
“n”.do
select
Select
NOT
as
it=ahave
appears.
number
to “1”
ifDouble
is equal;
if itPress
is
retype
toit use
everything!
for
check
“n”“0”
and
store
2not.
correctness!
itndinENTER
the calculator.
That was a lot to grasp in one problem. Let’s try
another one and practice with the calculator.
(2003)
Looks
likebackspace
thethe
answer
That
Use
wasn’t
the
correct
toB.
go
Try
again
with
option
nd
must
beequal
optionsign.
D.
Don’t
answer
to
the
choice,
either.
Either
2
You do Check
NOT have
to
assume!
it
out!
Enter,
delete
the
first answer
toPress
the
retypebackspace
everything!
=,choice
and
type
or type
in option
over it.
C.
2nd
ENTER
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.
We will now move on to solving equations…
This problem was on the 2006 test
We can solve
this equation
the “traditional”
way—using the
“undo” process.
9
104  C  32
5
9
72  C
5
360  9C
40  C
Subtract 32
Multiply by 5
Divide by 9
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
Alternate method
You
need
to be
Or we
could
use
able
to see and
in the
the graph
window
where the
CALC features
of
two
thelines
graphing
intersect.
calculatorThat
place looks way
Enter
off tothe
theequation
right.
in y1 and 104 in y2.
Adjust the window
AdjustLet’s
the window
again.
try the
— You
need ymax to
xmax
at 50.
be higher than 104
Graph
Alternate method
You
need
to be
Or we
could
use
able
to see and
in the
the graph
window
where the
CALC features
of
two
thelines
graphing
intersect.
calculatorThat
place looks way
Enter
off tothe
theequation
right.
in y1 and 104 in y2.
Adjust the window
AdjustLet’s
the window
again.
try the
— You
need ymax to
xmax
at 50.
be higher than 104
Graph
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?
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
don’t want the
problem to be
scored as wrong
because you didn’t
bubble in the
answer properly!
4
0
This problem was on the 2003 test. Let’s
approach this one differently, since it is
multiple choice.
We could transform the equation so we can
use the calculator, but too many people
F is not it.
messOption
that up—especially
with the
Neither is
option G
subtraction
sign!
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.
Here’s an inequality problem from the 2003
test where they ask you to graph, but they
don’t look at the graph! We’ll solve this
problem the same as the last one.
In other words, don’t 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!
Let me again mention—the 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!
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 let’s check to be sure.
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.
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.
2006
Let’s reason. A total of
80 backpacks were
sold. That means, if 1
$35 backpack was sold,
79 of the $50 backpacks
were sold.
In each case, I subtracted the
number of $35 backpacks, x,
from the total of 80.
That is 80 – x .
If 2 $35 were sold, then
78 $50 were sold.
If 10 $35 were sold,
then 70 $50 were sold.
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).
2006
“Total” implies
addition. Adding b
and c should be 220.
b + c = 220
2006
As for the
money…Every
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.
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.
Perimeter is the sum of the
lengths of the 3 sides: 5x + 10
2006
2x + 5
2x + 5
5x + 10 = 95
x
5x
= 85
x = 17
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”.
Each leg is 5
more than
double the
base: 2x + 5
That is not the only way you can do this
problem. You can just use the answers and
figure it out.
Let’s 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.
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.
This part means $.11 times “x”
“Not more than”
means we have
an inequality—this
value is the upper
limit. Lisa wants to
spend $5 or less.
.50 + .11x ≤ 5
.11x ≤ 4.50
x ≤ 40.9090…
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.
Let’s start with option D
since it has the largest
upper limit.
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 Lisa’s limit.
X
X
If minutes is 45, then
.50 + .11(45) = 5.45 This
value is also over Lisa’s
limit
If minutes is 40, then
.50 + .11(40) = 4.9 $4.90 is
under Lisa’s $5 limit, so
option B is the correct one.
2003
Anna must get the
same amount of
money, m, for
selling the plates
as it takes to
make the plates
before she can
“x” represents the number of plates
make any profit.
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.
2003
750 + 10x = 25x
750
= 15x
50 = x
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
a NO
30
50plates
plates isis
looks
also
like
NO
a winner.
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. Nelson’s
website to complete the power point lesson!
About slope…
X
X
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.
Now, look for the
point (4, -3) on the
remaining 2 graphs.
The point must be
ON the line.
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.
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).
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
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
2/3—Not you! Tell the calculator to
do it. Just be sure to use
parentheses!
Option 2: graph the
line and see where
it crosses the xaxis.
y1= (-6x+12)/(2/3)
x = 2 at this point
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!
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 yintercept visually.
x = -2
-3y = 6 – x
y = (6 – x)/-3
here
2003
X
Eliminate A and D
since the b-value,
the y-intercept, is
NOT -2.
Make a point
where the yintercept is -2.
X
Now, count the
slope from (0,-2)
to (5,-1).
up 1
right 5
The slope is 1/5.
2006
Use your formula chart
and draw the line
between R and S.
You can see that the yintercept cannot be 4.5, so
eliminate those two choices.
Now, count the slope
between R and S.
up 12
3
  1.5
right 8 2
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 here—this is it!
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:
down 2  2  1


right 4
4
2
2004
Option 3
You could use the slope formula on
the formula chart to calculate the
slope. Then, you could use slopeintercept form or point-slope form,
which is also on the formula chart to
calculate the y-intercept.
y  mx  b
4-2
2
1


-1 - 3  4
2
1
(3)  b use the slope and (3,2) as(x, y)
2
4   3  2b get rid of the fraction by multiplyin g everything by 2
2
7  2b
7
b
2
add 3 to both sides
divide both sides by 2
Now, let’s talk “linear systems” 2003
X
X
X
The line going through (0, 0) must go
upward.
We
2xNeither
–just
3y need
= of
0 the
thegiven
y by
itself.
equations
Using parentheses
is in
-3y = 0 – 2x
willcalculator
allow us form.
the We
opportunity
need
y =to
(0get
–for
2x)/-3
them
the into
calculator
that formtofirst.
“calculate”
put this function in y1
instead of us.
That
does
The
slope
in FNOT
is too
steep
as they
compared
mean
havetoto
this graph.
be in slope-intercept
form.
Now, let’s make sure…
X
X
X
Yes, this point is
(-3, -2)
You were not asked,
but let’s find the point
of intersection for the
two lines—the solution
to the system.
Count the slope, follow the
pattern, and extend each
line.
The point of intersection,
the solution, is (10,2).
Practice Problems