Solve the system of equations

Question

x-3y+4z=13

2x+y+z=-2

-2x+3y-3z=-10

Answer 1

      x,  y,  z
R1: 1,-3, 4, 13
R2: 2, 1, 1, -2
R3: -2, 3,-3,-10

R1 > 2*R1
R2 > -R2

R1: 2,-6, 8, 26
R2: -2,-1,-1, 2
R3: -2, 3,-3,-10

R2 > R2 + R1
R3 > R3 + R1

R1: 2,-6, 8, 26
R2: 0,-7, 7, 28
R3: 0,-3, 5, 16

R1 > R1/2
R2 > R2*(-3/7)

R1: 1,-3, 4, 13
R2: 0, 3,-3,-12
R3: 0,-3, 5, 16

R1 > R1 + R2
R3 > R3 + R2

R1: 1, 0, 1, 1
R2: 0, 3,-3,-12
R3: 0, 0, 2, 4

R2 > R2/3
R3 > R3/2

R1: 1, 0, 1, 1
R2: 0, 1,-1, -4
R3: 0, 0, 1, 2

R1 > R1 - R3
R2 > R2 + R3

R1: 1, 0, 0, -1
R2: 0, 1, 0, -2
R3: 0, 0, 1, 2

Solution is:
(-1,-2,2)

check:

(-1) -3(-2) +4(2) =? 13
2(-1) + (-2) + (2) =? -2
-2(-1) +3(-2) -3(2) =? -10

-1 + 6 + 8 = 13 √
-2 - 2 + 2 = -2 √
2 - 6 - 6 = -10 √

Answer 2

Hi Paula;

FIRST...x-3y+4z=13
SECOND...2x+y+z=-2
THIRD...-2x+3y-3z=-10

Let's first note that in the FIRST and THIRD, the coefficient of y is -3 and 3.  Let's add these equations together and eliminate y...

FIRST.....x-3y+4z=13

THIRD...-2x+3y-3z=-10

..........-x+z=3

Let's isolate one variable.  I select z...

...........z=x+3

 

Let's second note the fact that in the SECOND equation, neither y nor z have a coefficient.  In the THIRD, the coefficients are 3 and -3.

SECOND...2x+y+z=-2

Multiply both sides by 3 and subtract THIRD from it to again eliminate y...

SECOND...6x+3y+3z=-6

THIRD.....-2x+3y-3z=-10

...............8x+6z=4

Divide both sides by 2...

...............4x+3z=2

Plug-in z=x+3...

.............4x+3(x+3)=2

............4x+3x+9=2

...............7x+9=2

..............7x=-7

.............x=-1

Let's plug this into z=x+3...

..........z=-1+3

..........z=2

 

FIRST...x-3y+4z=13

...........-1-3y+[(4)(2)]=13

..........-1-3y+8=13

............-3y+7=13

............-3y=6

.............y=-2

SECOND...2x+y+z=-2

.............[(2)(-1)]+(-2)+2=-2

.............-2-2+2=-2

.............-4+2=-2

.............-2=-2

THIRD...-2x+3y-3z=-10

............[(-2)(-1)]+[(3)(-2)]-[(3)(2)]=-10

............2-6-6=-10

............2-12=-10

............-10=-10