So please help me to overcome this problem

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Here is a site dedicated to teaching simultaneous equations. Hopefully that'll help.

http://www.themathpage.com/alg/simultaneous-equations.htm

-John

Posted on Feb 13, 2008

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Posted on Jan 02, 2017

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There's a guide for that:

Canon 788SG User Instruction Simultaneous Linear Equations

Canon 788SG User Instruction Simultaneous Linear Equations

Jul 04, 2017 | Math

This calculator does not have a utility that solves equations of any type.

Nov 05, 2012 | Casio FX82MS Scientific Calculator

let c = cory's age now
let s = her sisters age now

cory's age now = sisters age + 6 years

=> c = s + 6

4 years ago, cory was 4 years younger ( i.e. c-4) and her sister was 4 years younger (s- 4)

4 years ago cory was 4 times older than her sister

=> c - 4 = 4*(s-4)

We can simplify this to c = 4s - 16 +4 c = 4s - 12

We can then use simultaneous equations to solve the two equations

(i) c = s + 6 (ii) c = 4s - 12

If we multiply both sides of equation (i) by 4 we get (iii) 4c = 4s + 24

We can then subtract equation (ii) from equation (iii) to eliminate the '4s' term => ( 4c = 4s + 24 ) - (c = 4s - 12) => 4c - c = 4s +24 - 4s +12 => 3c = 4s - 4s +36 => 3c = 36 => c = 36/3 => c = 12 => Cory's age now = 12

__For clarity:__
Her sister's age now is of course 6
4 years ago Cory would of been 8 and her sister would of been 2

I hope this helps and good luck! If you have more questions - ask away!

Please take the time to rate this answer

Many Thanks Don.

cory's age now = sisters age + 6 years

=> c = s + 6

4 years ago, cory was 4 years younger ( i.e. c-4) and her sister was 4 years younger (s- 4)

4 years ago cory was 4 times older than her sister

=> c - 4 = 4*(s-4)

We can simplify this to c = 4s - 16 +4 c = 4s - 12

We can then use simultaneous equations to solve the two equations

(i) c = s + 6 (ii) c = 4s - 12

If we multiply both sides of equation (i) by 4 we get (iii) 4c = 4s + 24

We can then subtract equation (ii) from equation (iii) to eliminate the '4s' term => ( 4c = 4s + 24 ) - (c = 4s - 12) => 4c - c = 4s +24 - 4s +12 => 3c = 4s - 4s +36 => 3c = 36 => c = 36/3 => c = 12 => Cory's age now = 12

I hope this helps and good luck! If you have more questions - ask away!

Please take the time to rate this answer

Many Thanks Don.

Sep 13, 2011 | Sunburst My Mathematical Life Single...

The absolutely-best, most-robust, software for solving systems of linear equations is the LINPACK software library.

It was originally written in FORTRAN.

But, see: http://www.greenecomputing.com/apps/linpack/

which indicates that a JAVA version exists.

It was originally written in FORTRAN.

But, see: http://www.greenecomputing.com/apps/linpack/

which indicates that a JAVA version exists.

Mar 02, 2011 | Computers & Internet

There are a few routines to solve simultaneous LINEAR equations that are embedded in the Operating system of the calculator.

If you need download and install instructions, you will find them here.

- solve(
- Simult(

If you need download and install instructions, you will find them here.

Aug 01, 2010 | Texas Instruments TI-89 Calculator

Press the [MODE] button 3 times then 1 to enter EQN mode. Then press the right pointing arrow. You can select the degree of the polynomial (2 or 3) or press again the right pointing arrow to select the number of unknowns for simultaneous linear equations( 2 or 3).

The form of the equations are as follows.

The form of the equations are as follows.

Jul 15, 2010 | Casio FX-115ES Scientific Calculator

Yes. Since there's far too much material for a single post, please refer to the chapter "Matrices" in the Guidebook. If you've misplaced your guidebook, you can download a new one from http://education.ti.com

Apr 28, 2010 | Texas Instruments TI-83 Plus 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

To view both graphs, the = signs on both y1 and y2 must be contained in black boxes. If you put your cursor over the = sign on the second equation and hit enter, it turns that graph off and you won't be able to see anything but the equation in y1 line. If there is not black box around the = sign of any equation in any of the lines you have typed an equation into, you will not see the graph.

Aug 12, 2009 | Texas Instruments TI-84 Plus Calculator

Hello,

I do not understand what you mean by solve complex equations, but if you want to manipulate complex numbers here are the tools you need to enter your complex numbers. Operations and function evaluations do not need special symbols.

Hope it helps.

I do not understand what you mean by solve complex equations, but if you want to manipulate complex numbers here are the tools you need to enter your complex numbers. Operations and function evaluations do not need special symbols.

Hope it helps.

Oct 31, 2008 | Casio FX-9860G Graphic Calculator

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