A linear equation is given below. Determine the steps to solve the equation for its inverse. y=6x+24

What do you mean by "solve matrix"? Do you want its determinant? Its inverse? The eigenvalues and eigenvector? Solve a system of linear equations?

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).

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.

Let your system of linear equations be [A][X]=[B] where [A] is a 3x3 matrix, [X] is column vector ( a 3x1 matrix) , and [B] is also a column vector or (3x1) matrix.

Assuming you are able to invert the [A] matrix to find [A^-1].
Multiplying on the keft by [A^-1] the system of equations you get
[A^-1].[A][X]=[A^-1][B].
But since the product [A^-1][A] is the (3x3) Identity matrix, the left side of the equation is jut [X], while the right side is [A^-1][B]. As you can see it is not [B][A^-1] .
To solve your problem, you should define your [A] matrix, define the [B] column vector (3x1) matrix, then perform the operation [A^-1][B] using the [X to -1] power key to calculate the inverse [A^-1].

What do you mean by "solve a matrix"? Find the inverse? Find the determinant? Find the solution to a system of linear equations? Within limits the fx-115ES can do any of these and more.

The procedures are described in the "Matrix Calculations" section of the manual. If you still have problems, please reply to this post specifying what you want to do.

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

The first step here is to draw a rough sketch of the two curves, and find the two points they meet. If you set x^3-6x^2+8x = x^3-4x, then you can rearrange this into the quadratic equation 6x^2-12x = 0. Solving this tells you x=0 or 2 : these are the two x values where the curves meet, and they give the bounds on your integral.

When you sketch the curves, you'll see that x^3-4x is below x^3-6x^2+8x, so the thing you need to integrate is (x^3-6x^2+8x)-(x^3-4x), that is, 12x-6x^2. If you integrate this from 0 to 2, you'll have the answer you need!

6x+6=4x+12
Since 4x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 4x from both sides.
6x+6-4x=12
Since 6x and -4x are like terms, add -4x to 6x to get 2x.
2x+6=12
Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.
2x=-6+12
Add 12 to -6 to get 6.
2x=6
Divide each term in the equation by 2.
(2x)/(2)=(6)/(2)
Simplify the left-hand side of the equation by canceling the common factors.
x=(6)/(2)
Simplify the right-hand side of the equation by simplifying each term.
x=3

Good Luck

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.

2x-3 = 3-3(x + 3)