Solving matrix -3x+4y+5z = 7 4x+3y+2z = 9 -5x+5y+3z = -10

This is best written as two separate equations:

8x+3y = -23 and 34x+ 5y = -23
Solving the first one for x:
8x = -23-3y
x = -23/8 - 3/8y
Substituting this value for x into the second equation:
34(-23/8 - 3/8y) + 5y = -23
-97.75 - (34)(.375)y + 5y = -23
-97.75 - 12.75y + 5y = -23
-97.75 -7.75y = -23
-7.75y = 97.75-23=74.75
y = -74.75/7.75 = -9.645161
Substitution back into the equation for x:
x = -23/8 - 3/8(-9.645161)
x = -2.875 + 3.616935
x =.741935

First, I graphed the lines and the point using Desmos.com.

I noticed that the two lines are perpendicular to each other and the point (1,-1) appears to be on the right side of the circle, on a line parallel to 3x -4y-10=0. The equation of this line is y= 3/4x - 1.75. The y-intercept is -1.75. Now we have two points on the opposite sides of the circle, (1, -1) and (0,-1.75). The midpoint formula will give you the centre of the circle and the distance formula will provide the radius.

Let me know if you have any questions.

Good luck.

Paul
Desmos Beautiful Free Math

This is no linear system. You cannot solve it like that using the matrix techniques. Haven't you made a mistake in writing the equations?
If that is tryly the system you want to solve, I suggest that you make a change of variables as follows:
X=1/x , Y= 1/y, Z=1/z (it being understood that x, y, z cannot be equal to 0). You will have to exclude the values x=0, y=0, z=0
Not I am not being sloppy, X and x are different entities, same with Y and y, Z and z.
Your system becomes
2X+3Y-1Z=26
1X+3Y-2Z=36
2X+4Y-5Z=52
Now that is a linear system. Solve it using matrices or Cramer's rule, When you obtain X, Y, and Z, get x=1/X, y=1/y, z=1/Z
The actual implementation of the solution method will depend on the exact model of calculator you are using. Not knowing that, I cannot advise you how to do it.

If I have not made any mistakes, the results are X=-58/9,Y=106/9, Z=-32/9. And x, y, z are just the reciprocals of their namesake.

Write the equality in the form y=(5X+3)/(4X-5). Insert parentheses to ensure a correct result.

1. Multiply both sides of the equality by (4X-5). This gives (4X-5)y=(5X+3).
2. Open the parentheses as 4Xy-5y=5X+3
3. Subtract 5X from both sides 4Xy-5y-5X=5X-5X+3
4. Add 5y to both sides 4Xy-5X-5y+5y=5y+3 or 4Xy-5X=5y+3
5. Factor the X on the left side X(4y-5)=5y+3
6. If 4y-5 does not vanish, you can isolate X by dividing both members of the equality by (4y-5).
7. You get X=(5y+3)/(4y-5)=f(y)

Compare the initial function f(x) and the function just found: They have the same form: X and y have switched places.

You have to put the equations into matrix form first. To do this, each variable has one column in the first matrix and you fill in the co-efficients for the variables. The second matrix has one column and contains all the numbers.

{ 2 -1 1 -1} = Matrix A
{ 1 3 -2 0}
{ 3 -2 0 4}
{-1 -3 -3 -1}

{-1} = Matrix B
{-5}
{ 1}
{-6}

{x=-2} = A*(B^-1)
{y=-.2}
{z=3}
{w=1/6}

I used excel for all my calculation and a helpful tutorial can be found here. I hope this helps and have a nice day!

Type solve(exp1 and exp2 and exp3, {x,y,z})
See cap images

You did not provide a general (n) term. However, I figured it to be something involving n^2. Hence the sum to n terms can be of the form: xn^3 + yn^2 + zn. Using the partial sums: 1, 6 (1+5), and 18 (1+5+12) we can build a linear equations system:
1= x + y +z (n=1)
6 = 8x + 4y +2z (n=2, n^2=4, n^3 = 8)
18 = 27x + 9y + 3z (n=3, n^=9, n^=27)
Solving for x,y,z we get x = 1/2, y= 1/2, z =0
Hence 1+5+12+22+..+ f(n^2)= 1/2(n^3+n^2)

You can solve it with following method.

5x+3y=6 2x-4y=5
So 5x=6-3y so 2[(6-3y)/5]-4y=5
So x=(6-3y)/5 so 12-6y-20y=25
so -26y=25-12
so -26y=13
so y= -(1/2)
2x-4y=5
so 2x=5+4y
so 2x=5+4(-1/2)
so 2x=(10-4)/2
so 2x=6/4
so x =3/2

The value of x=3/2 and value of y= -1/2

Let me know if you need further assistance.
Thanks for using FixYa.

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).

(13)3/8-(6)15/16