systems of Linear Equations in three variables?

Question

please solve this and tell how
2X+Y-Z=5
-5X-3Y+2Z=7
X+4Y-3Z=0

please solve this and tell how
2X+Y-Z=5
-5X-3Y+2Z=7
X+4Y-3Z=0

please tell me using substution or adding no hard stuff im only in 8th grade answers 1 and 2 are wrong and i dont get no3

Answer 1

2x + y - z = 5
-5x - 3y + 2z = 7
x + 4y - 3z = 0

The best way to solve this is to solve for one variable in terms of the others in one equation, and then plug this value into the other two equations.

Let's pick the third equation. x + 4y - 3z = 0 implies
x = -4y + 3z

Plugging this into the other two equations, 2x + y - z = 5 and
-5x - 3y + 2z = 7, we have

2(-4y + 3z) + y - z = 5
-5(-4y + 3z) - 3y + 2z = 7

Simplifying both equations, we have

-8y + 6z + y - z = 5
20y - 15z - 3y + 2z = 7

And then grouping like terms,

-7y + 5z = 5
17y - 13z = 7

Note that at this point, we have a two equations, two unknowns problem. We can now solve this by elimination or substitution; I'll choose substitution. -7y + 5z = 5 implies 5z = 7y + 5, so
z = (1/5)(7y + 5)

Plugging this into the second equation, 17y - 13z = 7, we have

17y - 13[(1/5)(7y + 5)] = 7
17y - (13/5)(7y + 5) = 7

Multiplying both sides by 5 to get rid of the fraction,

85y - 13(7y + 5) = 7
85y - 91y - 65 = 7
-6y = 72
y = -12

Now that we have y = -12, we can get z.

z = (1/5)(7y + 5) = (1/5) (7(-12) + 5)
z = (1/5) (-84 + 5)
z = (1/5) (-79)
z = -79/5

Now that we have y and z, we can get x.

Answer 2

sorry i didn't know what grade, you are correct though the others are incorrect. i'll solve again differently:

multiply x + 4y - 3z = 0 by -2

-2(x + 4y - 3z = 0 ) --> -2x -8y +6z =0

add this equation to 2x + y -z = 5

2x + y -z = 5
-2x -8y +6z =0

result 1) -7y +5z = 5

now, this time multiply x + 4y - 3z = 0 by 5

5(x + 4y - 3z = 0) --> 5x +20y -15z =0

add this equation to -5x -3y + 2z = 7

5x +20y -15z =0
-5x -3y + 2z = 7

result 2) 17y -13z = 7

now use result 1 and 2 to find y and z:

-7y +5z = 5
17y -13z = 7

then go back and use any original equation to find x after you have y and z.

**********************************************
i'll solve using determinants (this method is known as Cramer's Rule):

Let D=
| 2 1 -1 |
|-5 -3 2 |
| 1 4 -3 |

Let Ny =
| 5 1 -1 |
| 7 -3 2 |
| 0 4 -3 |

Let Ny =
| 2 5 -1 |
| -5 7 2 |
| 1 0 -3 |

Let Nz =
| 2 1 5 |
|-5 -3 7|
| 1 4 0 |

the determinant of D = 6
the determinant of Nx = -2
the determinant of Ny = -100
the determinant of Nz = -134

so:

x = Nx/D = -1/3
y = Ny/D = -50/3
z = Nz/D = -67/3


the determinant of a 3X3 matrix is:

|a b c|
|d e f |
|g h i |

DET = a(e*i - h*f) -b(d*i - g*f) +c(d*h -g*e)

to solve systems of equations this way you substitute the solution column for each of the variable columns as i did in the example above, and then perform the final division.

Answer 3

don't get afraid of the question. Tackle it slowly. Now, let's remove Z from the 2nd and 3rd equations- By substituting.

Using the first equation: 2X + Y - Z = 5
2x + y - 5 = Z

So there we have it.

Now, look at the second equation.

-5X-3Y+2Z=7

take away the Z by putting whatever Z equals to, into the equation instead of z.

-5x-3y+ 2 ( 2x + y - 5)= 7
-5x- 3y + 4x +2y -10 = 7
-x - y = 17
-x - y = 17----Equation 4.

Now look at the third equation give to you in your question.

X + 4y- 3z = 0

Once again, replace the Z with what is equivalent to Z in terms of X and Y.

X + 4y- 3(2x + y - 5) = 0
X + 4y - 6x -3y +15 = 0
-4x + y = -15 ----equation 5.

Now look at equations 4 and 5.

-x - y = 17
-4x + y = -15

It's a normal simultaneous equation now!

Answer 4

We have
2x+y-z=5 --(i)
-5x-3y+2z=7 --(ii)
x+4y-3z=0 --(iii)

from (i), we have

2x+y=5+z
So, z=2x+y-5 --(iv)

Now, we put value of z from (ii)
(ii) becomes,
-5x-3y+2(2x+y-5)=7
So, -5x-3y+4x+2y-10=7
So, -x-y=7+10
So, y = -x-17 --(v)

Now, we put value of z in (iii)
(iii) becomes,
x+4y-3(2x+y-5)=0
So, x+4y-6x-3y+15=0
So, -5x+y=-15
So, y = 5x-15 --(vi)

From (v) & (vi), we have,
-x-17 = 5x-15
So, 6x = -2
So, x = -1/3

y = 1/3 - 17
y = -50/3

z = 2(-1/3)-50/3-5
z = -67/3

x=-1/3, y=-50/3, z=-67/3