Explain what it means for a system of linear equations to have no solutions, one solution, two solutions, and infinite solutions.

not a standard Television question, maybe try Facebook for a answer?

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.

The 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)

You can solve equations two ways: The EQN computational mode or the SOLVE application.
IN the EQN mode you can solve linear equations in one, two and three unknowns; the quadratic equation; and the cubic equation.
Enter the EQN mode by following the instructions on the following screen capture. Press 5:EQN

You have three types of equations.

Once you select the type of equations, a template is displayed where you enter the relevant coefficients.


Under each label (a, b, etc. ) enter a coefficient and press ENTER to move to the next. With a cubic equation, or a linear equation in 3 unknowns, you have to use the Right arrow to display the cells where you enter the value of d.
After you input all necessary coefficients press the = sign to get the solution.

To use the SOLVE feature, see the screen capture that follows. It comes from the Casio manual.






Good luck

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.

I am afraid you cannot use the TI8xPlus family of calculators to solve linear systems in matrix form. In this calculator, matrices must have real coefficients.
You can however separate (expand) the problem into a linear system of 4 equations in 4 unknowns and try to solve it with the calculator.

1. define X = a+i*b
2. Define Y=c+i*d
3. Rewrite the first equation substituting a+i*b for X and c+i*d for y.
4. Gather all real terms together, and all imaginary terms together on the left.
5. You will have (8a-15b-8c) +i(15a+8b-8d) = 10 +0i
6. This equation can be split into two equations by saying that

• the real part pf the left member is equal to the real part of the right member, this gives 8a-15b-8c=10, and
• the imaginary part of the left member is equal to the imaginary part of the right member, this gives 15a+8b-8d=0

Similarly, you rewrite the second complex equation substituting a+ib for X and c+id for Y, then expand the binomial products, gather the real parts on the left together, and the imaginary parts on the left together. After that you equate the real part on the left to the real part on the right, and do the same for the imaginary parts.

If I did not make mistakes during the expansions and the gathering of terms you should get the following equation
-(8a+8c+10d) +i*(-8b+10c-8d) =0+i*0 from which you extract an equation for the real parts, -(8a+8c+10d)=0 and another for the imaginagy parts i*(-8b+10c-8d) =i*0

If I did not make mistakes (you should be able to find them, if any) your system of two linear equations with complex coefficents has been converted to a system of 4 linear equations with real coefficients.

8a-15b-8c=10
15a+8b-8d=0
8a+8c+10d=0
-8b+10c-8d =0

Now, you can in theory solve this system with help of the calculator, to find a, b, c, and d. When these are found, you can reconstruct the X and Y solutions.

Now get to work: Ascertain that my extracted equations are correct, then solve for a, b,c, and d, and reconstruct X and Y.
I am no seer, but my hunch is that this system is degenarate. I will not explain what that means.

4x+7=11 Solve for "x".
subtract 7 from each side of the equation
4x +7(-7)=11(-7) = 4x=4
divide both sides of the equation by 4
4x/4=4/4
therefore
x=1
Correct?

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.

i tried to graph a simple function but it wont show the lines it will only show the graph

Consider the following system of 3 equations in 3 unknowns:
x + y = 2
2x + 3y + z = 4
x + 2y + 2z = 6Our goal is to transform this system into an equivalent system from which it is easy to find the solutions. We now do this step by step.

• Subtract 2*(Row1) from Row2 and place the result in the second row; subtract Row1 from Row2 and place in the third row. Leave Row1 as is.

x + y = 2
y + z = 0
y + 2z = 4

• Subtract Row2 from Row3, and place the result in row3. Leave Row1 and Row2 as they are.

x + y = 2
y + z = 0
z = 4 From the last form of the system we can deduce the following unique solution to the system:

z = 4, y = -4, and x = 2-(-4) = 6Equivalently, we say that the unique solution to this system is (x, y, z) = (6, -4, 4).