I am trying to find the value of x from this equation f(x)= 1/75 *e^ -(x/75)

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X is a function, therefore it has infinite solutions. you need the function to equal a number to solve for a specific x value

ex: f(x)= 1/75*e^(-x/75)

20= 1/75*e^(-x/75), then under catolog choose solve.

solve((1/75)*e^(-x/75)=20,x) the x is what you are solving for at the end.

long hand would be multiply by 75 to get rid of 1/75..so now you have

1500=e^(-x/75)

now Ln gets rid of e so Ln(1500) = -x/75, so x = -75*ln(1500)

hope this all helps

Posted on Apr 27, 2010

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Posted on Jan 02, 2017

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If you have the equation of the line (of best fit), enter a value of x in the equation to get the value of y. Similarly, if you have a target value of y, enter it in the equation and solve for the value of x. That's about it.

Jun 06, 2014 | Casio FX-9750GII Graphing Calculator

In equation mode, you have system of linear equations (3 unknown) you have polynomial (quadratic and cubic), and solver. Use the solver foe any type of equation (nonlinear, polynomial of order higher than 4, trigonometric, exponential, logs).

May 21, 2012 | Casio Scientific Calculator Fx-570 Fx570...

Thhe Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Mar 17, 2012 | Casio FX-9860G Graphic Calculator

To find the solution, first find the value of y for each equation.

Then substitue one equation into the other so that you only the x variable left.

Then just solve for x.

Once you have a value for x, then you can easily solve for y.

So for the first equation:

3y - 6x = -3

3y = 6x - 3

**y = 2x - 1**

Now for the second equation:

2y + 8x = 10

2y = -8x + 10

**y = -4x + 5**

Since both equations equal y, they also equal each other, therefore:

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1

6x = 6

**x=1**

Now substitute x=1 into either original equation:

y = 2x - 1

y = 2 (1) - 1

y = 2 - 1

**y = 1**

Therefore the solution is x=1 and y=1

Good luck, I hope that helps.

Joe.

Then substitue one equation into the other so that you only the x variable left.

Then just solve for x.

Once you have a value for x, then you can easily solve for y.

So for the first equation:

3y - 6x = -3

3y = 6x - 3

Now for the second equation:

2y + 8x = 10

2y = -8x + 10

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1

6x = 6

y = 2x - 1

y = 2 (1) - 1

y = 2 - 1

Good luck, I hope that helps.

Joe.

Nov 09, 2011 | Texas Instruments TI-84 Plus Silver...

Apply what you learned, especially that this system is quite simple.

Elimination

1. entails eliminating one variable to find (in this case) a single equation involving the other variable.

2. Solve that new equation, meaning isolate the variable that was not eliminated.

3. Substitute the value found in the last step and replace it in one of the original equation to obtain the other variable.

x+y=10

x-y=-6

Add the left sides to get x+y+x-y=2x

Add the right sides of the system to obtain 10+(-6)=4

Write Sum of left sides =Sum of right sides or 2x=4.

Since this new equation involves only x, you can solve it for x, getting x=4/2=2

Now you know the value of x (=2)

Take one of the original equations (for example x+y=10) and put 2 in place of x.

The equation becomes 2+y=10

Solve it to obtain y=10-2=8

Thus your solution is x=2, y=8

Check: use one equation in which you substitute 2 for x and 8 for y: x-y=-6 becomes 2-8=-6

Verified.

In the general case you will have to multiply by certain values to obtain opposite coefficients for the same variable. Here that was not necessary because the coefficient of y is 1 in the first equation and -1 in the second.

Elimination

1. entails eliminating one variable to find (in this case) a single equation involving the other variable.

2. Solve that new equation, meaning isolate the variable that was not eliminated.

3. Substitute the value found in the last step and replace it in one of the original equation to obtain the other variable.

x+y=10

x-y=-6

Add the left sides to get x+y+x-y=2x

Add the right sides of the system to obtain 10+(-6)=4

Write Sum of left sides =Sum of right sides or 2x=4.

Since this new equation involves only x, you can solve it for x, getting x=4/2=2

Now you know the value of x (=2)

Take one of the original equations (for example x+y=10) and put 2 in place of x.

The equation becomes 2+y=10

Solve it to obtain y=10-2=8

Thus your solution is x=2, y=8

Check: use one equation in which you substitute 2 for x and 8 for y: x-y=-6 becomes 2-8=-6

Verified.

In the general case you will have to multiply by certain values to obtain opposite coefficients for the same variable. Here that was not necessary because the coefficient of y is 1 in the first equation and -1 in the second.

Aug 16, 2011 | Casio Office Equipment & Supplies

You are looking for a line (y=m*x+b) and have two points. From this information you can generate two equations with two unknowns (m and b are unknown).

First plug in the first point (1,5) to the general form:

5(the y value of the point) = m*1(the x-value at this point)+b

Do the same for the second point you're given.

From here solve the first equation for m in terms of b.

Plug this value of 'm' into the second equation so you will end up with something like:

3=(something in terms of b)*(-2)+b

This final equation can be solved for b (try factoring)

You now have a value for the y-intercept. Plug that into y=m*x+b

Choose either of the two points, plug into the equation on the last line with the value of b known

You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

First plug in the first point (1,5) to the general form:

5(the y value of the point) = m*1(the x-value at this point)+b

Do the same for the second point you're given.

From here solve the first equation for m in terms of b.

Plug this value of 'm' into the second equation so you will end up with something like:

3=(something in terms of b)*(-2)+b

This final equation can be solved for b (try factoring)

You now have a value for the y-intercept. Plug that into y=m*x+b

Choose either of the two points, plug into the equation on the last line with the value of b known

You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

Apr 23, 2011 | Texas Instruments TI-83 Plus Calculator

This expression 5x^2+15x+3 is not an equation, therefore it can have any value depending on the value of x. You have to make it an equation before looking for the particular value of x that satisfy your equation **5x^2+15x+3 =0**

This calculator cannot solve equations because it does not have an equation solver.

This calculator cannot solve equations because it does not have an equation solver.

Dec 15, 2010 | Sharp ELW535 Calculator

I assume that if you want to input the equation it is with the intent to solve it. If that be the case then what follows will help you input and solve any equation (linear, quadratic, trig, etc.)

Make sure the calculator angle unit is the same as your problem requires (degree, radian, grad). Here are some screen capture to help you. Change the equation to yours.

To enter the equation in screen above, use the right arrow.

You have to enter an initial guess on line X=, a default value may be supplied by calculator: It depends on last value stored in X. Once a value is entered press the F6 key to SOLV

Press F1: Rept if you want to solve another equation.

Make sure the calculator angle unit is the same as your problem requires (degree, radian, grad). Here are some screen capture to help you. Change the equation to yours.

To enter the equation in screen above, use the right arrow.

You have to enter an initial guess on line X=, a default value may be supplied by calculator: It depends on last value stored in X. Once a value is entered press the F6 key to SOLV

Press F1: Rept if you want to solve another equation.

Nov 13, 2010 | Casio FX-9860G Graphic Calculator

Hello,

Let us assume you have two simultaneous linear equations :

**a_1*x+ b_1*y+c_1=0**

a_2*x +b_2*y+c_2=0

where a_1, a_2, b_1, b_2, c_1,c_2 are coefficients (numerical or algebraic).

The problem is to obtain the particular values of the unknowns x and y for which the two equations are both satisfied: If you substitute the particular values of x and y you find in any of the two equations you discover that both equalities are true.

A small system of equations like the one above can be solved by some very simple algorithms (elimination, substitution, combination) which can be carried out by hand.

The solution of large systems of linear equations can be sought by making use of the concepts of matrices (plural of matrix), determinants, and certain rules called Cramer's rules.

Due to its repetitive nature, the algorithm ( a well defined, limited sequence of steps) is suitable for a calculating machine (computer or calculator).

Certain calculators have, embedded in their ROM, a program that solves linear systems of simultaneous equations. Usually you are asked to enter the values of the coefficients a_1, etc. in a set order, then you press ENTER or EXE (Casio) . If a solution exits (not all linear systems have solutions) the calculator displays it.

Hope that satisfies your curiosity.

Let us assume you have two simultaneous linear equations :

a_2*x +b_2*y+c_2=0

where a_1, a_2, b_1, b_2, c_1,c_2 are coefficients (numerical or algebraic).

The problem is to obtain the particular values of the unknowns x and y for which the two equations are both satisfied: If you substitute the particular values of x and y you find in any of the two equations you discover that both equalities are true.

A small system of equations like the one above can be solved by some very simple algorithms (elimination, substitution, combination) which can be carried out by hand.

The solution of large systems of linear equations can be sought by making use of the concepts of matrices (plural of matrix), determinants, and certain rules called Cramer's rules.

Due to its repetitive nature, the algorithm ( a well defined, limited sequence of steps) is suitable for a calculating machine (computer or calculator).

Certain calculators have, embedded in their ROM, a program that solves linear systems of simultaneous equations. Usually you are asked to enter the values of the coefficients a_1, etc. in a set order, then you press ENTER or EXE (Casio) . If a solution exits (not all linear systems have solutions) the calculator displays it.

Hope that satisfies your curiosity.

Aug 12, 2009 | Sharp EL-531VB Calculator

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Apr 27, 2009 | Casio CFX-9850G Plus Calculator

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