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:

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 = You should see the = sign appear after your expression (NO, do not press ENTER—the calculator will not simplify this expression for you!) Now, one by one, enter the first answer choice just as it appears. Double check to make sure you typed it in correctly. 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”.

That was a lot to grasp in one problem. Let’s try another one and practice with the calculator. (2003) 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 (2 nd Math) and select = Type in the first answer choice. Remember, “1” if it is equal; “0” if it is not. Try again with option B. You do NOT have to retype everything! Press 2 nd ENTER

That was a lot to grasp in one problem. Let’s try another one and practice with the calculator. (2003) Try again with option B. You do NOT have to retype everything! Press 2 nd ENTER Use the backspace to go to the equal sign. Either delete the first answer choice or type over it. That wasn’t the correct answer choice, either. 2 nd Enter, backspace to the =, and type in option C. Looks like the answer must be option D. Don’t assume! Check it out!

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.

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 Or we could use the graph and CALC features of the graphing calculator Enter the equation in y 1 and 104 in y 2. Adjust the window — You need y max to be higher than 104 Graph You need to be able to see in the window where the two lines intersect. That place looks way off to the right. Adjust the window again. Let’s try the x max at 50.

Alternate method Or we could use the graph and CALC features of the graphing calculator Enter the equation in y 1 and 104 in y 2. Adjust the window — You need y max to be higher than 104 Graph You need to be able to see in the window where the two lines intersect. That place looks way off to the right. Adjust the window again. Let’s try the x max at 50.

Alternate method Enter again for the second curve? And guess? Press 2 nd TRACE so that you get CALC. Now, select Intersect. Move the cursor to be close to the point of intersection.

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! 40

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 mess that up—especially with the subtraction sign! 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. Option F is not it. Neither is option G Option H is the one!

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! A is more than 12 And so is B Looks like C is the answer, but let’s check to be sure. Option D gives us 12. We want an answer that is less than 12. Sure enough, C is the answer.

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 – 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. 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.

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.

2006 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”. x Each leg is 5 more than double the base: 2x + 5 2x + 5 Perimeter is the sum of the lengths of the 3 sides: 5x x + 10 = 95 5x = 85 x = 17

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: = is the perimeter we want, so A is the answer.

2006 “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 x ≤ 5.11x ≤ 4.50 x ≤ … This part means $.11 times “x”

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) = 6. $6 is over Lisa’s limit. If minutes is 45, then (45) = 5.45 This value is also over Lisa’s limit If minutes is 40, then (40) = 4.9 $4.90 is under Lisa’s $5 limit, so option B is the correct one. X X

2003 “x” represents the number of plates Anna must sell Her costs: m = x 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. 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 = 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 is a NO 30 plates is also NO50 plates looks like 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… 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 X 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 = -6x x = 12 x = 2

2006 You are expected to know that the x-intercept is where the graph “touches” the x-axis. Option 2: graph the line and see where it crosses 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! y 1 = (-6x+12)/(2/3) x = 2 at this point

2003 Remember, the state graders are NOT looking at your graphs in your booklet. This grid is for you! 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!

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 = /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

2003 Eliminate A and D since the b-value, the y-intercept, is NOT -2. Make a point where the y- intercept is -2. Now, count the slope from (0,-2) to (5,-1). X X up 1right 5 The slope is 1/5.

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.

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. Draw the line between them. Count the slope: The y-intercept is at 3.5 7/2 = 3.5

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.

Now, let’s talk “linear systems” 2003 Neither of the given equations is in calculator form. We need to get them into that form first. That does NOT mean they have to be in slope-intercept form. 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 y 1 X X The line going through (0, 0) must go upward. The slope in F is too steep as compared to this graph. X

Now, let’s make sure… You were not asked, but let’s find the point of intersection for the two lines—the solution to the system. X X X Yes, this point is (-3, -2)

Count the slope, follow the pattern, and extend each line. The point of intersection, the solution, is (10,2).

Practice Problems