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HP 35s Calculator - Solving Equations

Description
The HP Solve feature can be used to find the value of any variable in any type of operation.
For example, consider the following equation:
x2 - 3y = 10
If the value of y is known in this equation, then Solve can calculate the value for x. If the value of x is known, then Solve can find the value for y. This concept works for other types of problems with unknown variables as well:
Markup x Cost = Price
If any two variables are known, then Solve can calculate the value of the third.
To solve an equation for an unknown variable
  1. Press and display the desired equation.
  2. Press then press the key for the unknown variable. For example, press , X to solve for x. The equation then prompts for a value for every other variable in the equation.
  3. For each prompt, enter the desired value:
    • If the displayed value is the one desired, press .
    • If a different value is desired, type or calculate the value and press .
Pressing or can halt a running calculation.
When the root is found, it is stored in the relation variable, and the variable value is viewed in the display. In addition, the X–register contains the root, the Y–register contains the previous estimate value, or Zero, and the Z–register contains the value of the root D-value (which should be zero).
For some complicated mathematical conditions, a definitive solution cannot be found — and the calculator displays NO ROOT FOUND.
For certain equations, it helps to provide one or two initial guesses for the unknown variable before solving the equation. This can speed up the calculation, direct the answer toward a realistic solution, and find more than one solution, if appropriate.
Example of solving the equation of linear motion
The equation of motion for a free falling object is: d = v0t + ½ gt2.
Where d is the distance, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity.
Type in the equation:
Key
Display
Description
(3ALL)
(Y)
Clears memory.
3*3 lin. solve
EQN LIST TOP
Selects Equation mode.
Starts the equation.
D=VxT+_
=VxT+0.5xGxT^2
D=VxT+0.5xGxT^2
Terminates the equation and displays the left end.
CK=FB3
CLN=15
Checksum and length.
g (acceleration due to gravity) is included as a variable so it can be changed for different units (9.8 m/s2 or 32.2 ft/s2).
Example of calculating how many meters an object falls
Calculate how many meters an object falls in 5 seconds, starting from rest. Since Equation mode is turned on and the desired equation is already in the display, start solving for D:
Key
Display
Description
SOLVE_
Prompts for unknown variable.
V? value
Selects D; prompts for V.
T? value
Stores 0 in V; prompts for T
G? value
Stores 5 in T; prompts for G.
SOLVING
D
=122.5000
Stores 9.8 in G; solves for D.
Example of calculating the time it takes for an object to fall
Using the same equation, how long does it take an object to fall 500 meters from rest?
Key
Display
Description
D=VxT+0.5xGxT^2
Displays the equation.
D? 122.5
Solves for T; prompts for D.
V? 0
Stores 500 in D; prompts for V.
G? 9.8
Retains 0 in V; prompts for G.
SOLVING T=10.1015
Retains 9.8 in G; solves for T.
  note:
For this example, using a body at rest, g is considered a positive value.
What if the body is in motion? For an example using a body that has an initial displacement and an initial velocity greater than zero, enter an equation with a variable, R, to represent the displacement in the equation:
D=D0 + V0T - ½gt2
In this instance, D is the final displacement, D0 is the initial displacement, and V0 is the value of the initial velocity; it is positive to indicate the body is in motion in an upward direction, as a ball travels when thrown in the air. However, because the ball will be acted upon by gravity at some time, T, in the opposite direction, acceleration due to gravity, g, must have a negative value in this case.
When entering this equation into the 35s, consider using variables for the displacement that are easy to remember, as only one variable can be assigned to the letter 'D'.
Example of solving the Ideal Gas Law equation
The Ideal Gas Law describes the relationship between pressure, volume, temperature, and the amount (moles) of an ideal gas:
P × V = N × R × T
Where P is pressure (in atmospheres or N/m2), V is volume (in liters), N is the number of moles of gas, R is the universal gas constant (0.0821 liter–atm/mole–K or 8.314 J/mole–K), and T is temperature (Kelvins: K=°C + 273.1).
Enter the equation:
Key
Display
Description
Px_
Selects Equation mode and starts the equation.
PxV=NxRxT_
PxV=NxRxT
Terminates and displays the equation.
CK=EDC
LN=9
Checksum and length.
Example of calculating an Ideal Gas' pressure
A 2-liter bottle contains 0.005 moles of carbon dioxide gas at 24 degrees Celsius. Assuming that the gas behaves as an ideal gas, calculate its pressure. Since the Equation mode is turned on and the desired equation is already in the display, start by solving for P. The example is calculated in RPN mode.
Key
Display
Description
V? value
Solves for P; prompts for V.
N? value
Stores 2 in V; prompts for N.
R? value
Stores .005 in N; prompts for R.
T? value
Stores .0821 in R; prompts for T.
N? 297.1000
Calculates T (Kelvins).
SOLVING P=0.0610
Stores 297.1 in T; solves for P in atmospheres.
Example of calculating the density of a gas
A 5–liter flask contains nitrogen gas. The pressure is 0.05 atmospheres when the temperature is 18°C. Calculate the density of the gas (N × 28/V, where 28 is the molecular weight of nitrogen).
Key
Display
Description
PxV=NxRxt
Displays the equation.
P? 0.0610
Solves for N; prompts for P.
V? 2.0000
Stores .05 in P; prompts for V.
R? 0.0821
Stores 5 in V; prompts for R.
T? 297.1000
Retains previous R; prompts for T.
T? 291.1000
Calculates T (Kelvins).
SOLVING N=0.0105
Stores 291.1 in T; solves for N.
0.2929
Calculates mass in grams, N × 28.
0.0105
0.0586
Calculates density in grams per liter.

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