How can I solve this problem?

Question

2x+3y+7z=13
3x+2y-5z=-22
5x+7y-3z=-28

Answer 1

2x + 3y + 7z = 13 ---equation 1
3x + 2y - 5z = -22 ---equation 2
5x + 7y - 3z = -28 ---equation 3
---------------------------

Take ---equation 1 & ---equation 2

2x + 3y + 7z = 13 x5 ---equation 1
3x + 2y - 5z = -22 x7---equation 2
---------------------------------

10x + 15y + 35z = 65 ---equation 1
21x + 14y - 35z = -154 ---equation 2
-------------------------------
31x + 29y = -89 ---equation 4

Take ---equation 2 & ---equation 3

3x + 2y - 5z = -22 x3 ---equation 2
5x + 7y - 3z = -28 x-5 ---equation 3
------------------------
9x + 6y-15z = -66
-25x-35y+15z=140
-16x-29y = 74 ---equation 5


Now, Take ---equation 4 & ---equation 5

+31x + 29y = -89
-16x – 29y = 74
15x = -15
x=-15/15 = -1

So, x = -1

Take ---equation 4

31x + 29y = -89

31(-1) + 29y = -89

-31 + 29y = -89
29y = -89 +31 = -58

y = -58 / 29 = -2

Take ---equation 1

2x + 3y + 7z = 13

2(-1) + 3(-2) + 7z = 13

-2 – 6 +7z = 13
-8 + 7z = 13

7z = 13+8 = 21
z = 21 / 7 = 3

So, the results are x = -1 , y = -2 & z = 3

Answer 2

After following the entire process found at the source below, I was able to solve for:

x = -1
y = -2
z = 3

To check for correctness:

2x + 3y + 7z = 13
2(-1) + 3(-2) + 7(3) = 13
-2 - 6 + 21 = 13
13 = 13

3x + 2y - 5z = -22
3(-1) + 2(-2) - 5(3) = -22
-3 - 4 - 15 = -22
-22 = -22

5x + 7y - 3z = -28
5(-1) + 7(-2) - 3(3) = -28
-5 - 14 - 9 = -28
-28 = -28

Therefore, the solution is correct.

Answer 3

That is a system a liner equations. Most algebra textbooks will tell you to multiply the equations by a constant and then add and subtract them from one another to cancel the various terms until you're able to solve for the individul variable by removing the rest.

I prefer an appraoch using linear algebra. You can fit the coefficients of the seperate terms in each equation into a nxm matrix. In this case, a 3x4 matrix.

[2 3 7 13]
[3 2 -5 -22]
[5 7 -3 -28]

You can then put this matrix in reduced row echelon form (rref). Most graphing calculators have a button for it. The matrix will then be in such a form giving you the answers.

[1 0 0 value_of_x]
[0 1 0 value_of_y]
[0 0 1 value_of_z]

This method takes a lot of the work out of solving the system. However, be warned that your math teacher might not accept this approach on a test, quiz, or exam. Thus, you still probably need to learn how to do it the hard way.

Answer 4

2x+3y+7z=13
3x+2y-5z=-22
5x+7y-3z=-28
so
30x +45y +105z =195
30x +20y -50z = -220
30x + 42y -18z = -168
now you can eliminate x from second and third equations
and solve the simpler pair of y-z equations same way
substitute back into one of the original equations to get x

Answer 5

Multiply equation 1 by 5 and equation 2 by 7:-
10x + 15y + 35z = 65
21x + 14y - 35z = - 154------ADD
31x + 29y = - 89
Multiply equation 2 by -3 and equation 3 by 5
-9x - 6y + 15z = 66
25x + 35y - 15z = - 140-----ADD

16x + 29y = - 74
-31x - 29y = 89-----ADD
- 15x = 15
x = - 1

- 16 + 29y = - 74
29y = - 58
y = - 2

- 2 - 6 + 7z = 13
7z = 21
z = 3

Solution is x = -1, y = - 2, z = 3

Answer 6

What do you mean? what do you need to do?? It looks pretty solved for me!