We’ve got free TI calculator programs to help you with nearly every formula you’ll need to use on the ACT or SAT. Make sure to practice using these tools so you feel completely comfortable with them on test day.

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Includes more programs and examples of how each program will help you answer questions on the test.

**Quick note:** we want to point out that to answer most questions on the ACT or SAT, you really do need to understand these formulas and the concepts behind them. Please don’t skimp on your algebra, plane geometry, and coordinate geometry review just because you have these calculator programs at your disposal. These programs will definitely save you time on the SAT or ACT, but any program is only as good as the numbers you input, and knowing what numbers to input requires knowledge of the basic concepts.

**Another quick note**: If you have a cable that connects your computer to your calculator, you can download all sort of helpful ACT and SAT tools. Check out the official TI website for loads of free programs. The programs below are simple programs designed to be quickly inputted by hand.

Pythagorean theorem when the hypotenuse is missing

Pythagorean theorem when one of the smaller legs is missing

Finding the equation of a line/slope of a line when given 2 points

Finding the point where two lines intersect

Solving a system of equations (also called simultaneous equations)

This program will compute the area of a circle if you plug in the radius. Please note that the program will calculate the numerical value of π.

Step-by-step directions | What your screen should look like when creating the new program

1. Hit PGRM and slide over to select New

2. Give the new program a name like ACIRC and hit enter

3. Hit PGRM, slide over to I/O, and select Prompt

4. Hit Alpha and the letter R. Hit Enter.

5. Put in the formula πR^{2}. Then hit the STO button (right above the ON button). After the little arrow, hit ALPHA and type in an A.

6. Go to PGRM, slide over to I/O and choose Disp. Then put an A.

7. Hit 2nd and then Quit to exit setup. Go to PGRM and EXEC and choose it from your program list to use it.

**What your screen should look like when programming**

PROGRAM: ACIRC

: Prompt R

: πR^{2} –> A

: DISP A

This program will compute the circumference of a circle if you plug in the radius. Please note that the program will calculate the numerical value of π.

Step-by-step directions | What your screen should look like when creating the new program

1. Hit PGRM and slide over to select New

2. Give the new program a name like CCIRC and hit enter

3. Hit PGRM, slide over to I/O, and select Prompt

4. Hit Alpha and the letter R. Hit Enter.

5. Put in the formula 2πR. Then hit the STO button (right above the ON button). After the little arrow, hit ALPHA and type in a C.

6. Go to PGRM, slide over to I/O and choose Disp. Then put a C.

7. Hit 2nd and then Quit to exit setup. Go to PGRM and EXEC and choose it from your program list to use it.

**What your screen should look like when programming**

PROGRAM: CCIRC

: Prompt R

: 2πR –> C

: DISP C

You’ve probably already noticed that the ACT and SAT folks love right triangles. This program will find the hypotenuse of a right triangle using the Pythagorean theorem. You will be prompted to input the other two sides of the triangle.

Step-by-step directions | What your screen should look like when creating the new program

1. Hit PGRM and slide over to select New

2. Give the new program a name like PYTHAG and hit enter

3. Hit PGRM, slide over to I/O, and select Prompt

4. Hit Alpha and A, B. Hit Enter.

5. Put in the formula √(A^{2 }+ B^{2} ) Then hit the STO button (right above the ON button). After the little arrow, hit ALPHA and type in a C.

6. Go to PGRM, slide over to I/O and choose Disp. Then put a C.

7. Hit 2nd and then Quit to exit setup. Go to PGRM and EXEC and choose it from your program list to use it.

**What your screen should look like when programming**

PROGRAM: PYTHAG

: Prompt A, B

: √(A^{2 }+ B^{2} ) –> C

: DISP C

You’ve probably already noticed that the ACT and SAT folks love right triangles. This program will find one of the shorter (non-hypotenuse) legs of a right triangle using the Pythagorean theorem. You will be prompted to input the other leg and the hypotenuse.

Step-by-step directions | What your screen should look like when creating the new program

1. Hit PGRM and slide over to select New

2. Give the new program a name like PYTHAGLEG and hit enter

3. Hit PGRM, slide over to I/O, and select Prompt

4. Hit Alpha and A, C. Hit Enter.

5. Put in the formula √(C^{2 –} A^{2} ) Then hit the STO button (right above the ON button). After the little arrow, hit ALPHA and type in a B.

6. Go to PGRM, slide over to I/O and choose Disp. Then put a B.

**What your screen should look like when programming**

PROGRAM: PYTHAGLEG

: Prompt A, C

: √(C^{2 –} A^{2} ) –> B

: DISP B

This isn’t really a program. It’s an easy 10 second calculator technique to find the equation (or just slope) when given two points. You’ve probably seen a question on the ACT or SAT testing this concept. Here’s how to find it without the sloppy work.

Say you had two points, (5,2) and (3,8), and you wanted to find the slope of line passing through those points. Press [Stat] > 1:Edit… and enter the x-coordinates in L1, and the y-coordinates in L2. So L1={5 3} and L2={2 8}.

Then, go to [Stat] > Calc > 4:LinReg(ax+b) then press Enter twice. You’ll be shown the results of the linear regression. For this particular problem, you’ll get results like this:

LinReg

y=ax+b

a=-3

b=17

I don’t know why the calculator chooses to use an ‘a’ instead of an ‘m’ when displaying the y=mx+b equation. But in any case, the line here would be y=-3x+17. Since the -3 is in the ‘m’ spot in the y=mx+b equation, the slope of the line is -3.

This isn’t really a program. It’s an easy 10 second calculator technique to find where two lines intersect. You’ve probably seen a question on the ACT or SAT testing this concept. Here’s how to find it without the sloppy work.

To graph your equations, press the “Y=” button at the top left hand corner of the calculator. Record each equation in the calculator and press “Graph.” To calculate the intersection of the lines, hit 2nd and CALC (located above the TRACE button) and scroll down to the option that says “Intersection.” Press the “Equals” button **three** times, and the points of intersection will appear in the lower left hand corner of your screen under the heading ‘intersection’.

This isn’t really a program. It’s an easy 15 second calculator technique to solve a system of equations. You’ve probably seen a question on the ACT or SAT testing this concept. Just to refresh your memory, here is an example system of equations:

2y = 4x + 2

4y = -2x + 24

Usually, you can solve such a system by substitution or combination. Depending on the complexity of the equations, using combination or substitution can take a lot of time and students often make careless mistakes. Here’s how to quickly solve it with your calculator.

First, move your equations around or simplify them to get them into Y = form. The two equations above would now be y = 2x+1 and y = -1/2 x + 6.

Now, just use the same calculator technique used to find the intersection of two lines. After all, these two lines will intersect at the point where both y’s and x’s are the same. And that’s what solving a system of equations is about–finding the y and x shared by both lines. Here’s what to do:

Graph your equations by pressing the “Y=” button at the top left hand corner of the calculator. Record each equation in the calculator and press “Graph.” To calculate the intersection of the lines, hit 2nd and CALC (located above the TRACE button) and scroll down to the option that says “Intersection.” Press the “Equals” button **three** times, and the points of intersection will appear in the lower left hand corner of your screen under the heading ‘intersection’. If your graphs do not intersect, there is no value that solves the system of the equations.

In the example above, plugging the equations into the calculator and calculating the intersection gives you x = 2, y = 5. Therefore, x =2 and y =5 are the solution to this system of equations.