How do I sole system of equations - Office Equipment & Supplies

Plugging the coefficients 2, -2, and -30 into the quadratic equation produces -3 and 5.

If this is homework, be sure to show your work.

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).

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)

Definition

A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income ÷ Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

1. Linear Equations y= mx + b (standard form of linear equation)
2. Quadratic Equations y= ax^2+bx+c
3. Exponential Equations y= ab^x
4. Cubic Equations y=ax^3+ bx^2+cx+d
5. Quartic Equations y= ax^4+ bx^3+ cx^2+ dx+ e
6. Equation of a circle (x-h)^2+(y-k)^2= r^2
7. Constant equation y= 9 (basically y has to equal a number for it to be a constant equation).
8. Proportional equations y=kx; y= k/x, etc.

And your question is? If you just want to know how to solve this system, you only need to substitute one equation into the other and solve for the remaining unknown.

y=3x (equation 2), so plug it into equation 1

7x-2y=1
7x-2(3x)=1
7x-6x=1
x=1

from equation 2 then, you know y=3x=3

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.

There are no "free" product keys, giving away keys is illegal.

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.

Select a blank cell in workbook. Click the function icon located informula bar("fx"). In the popup Insertwindow Select category as ALL.Select the function you want to get help and click on "help on this function " located in the left down page. You can get all the formula help bu using this method.