Question

What is the solution of the system of equations?

-3x-4y-3z=-7
2x-6y+2z=3
5x-2y+5z=9

Answer 1

If you have a TI-83 plus or higher, you could create a matrix and use rref to get the solution.

Let's begin with eliminating the y's in equations 1 and 3, then in equations 2 and 3:
$-3x-4y-3z=-7$ times -1⇒ $3x+4y+3z=7$ 
$5x-2y+5z=9$ times 2⇒ $10x-4y+10z=18$

ADD the two equations: 13x + 13z = 25

$2x-6y+2z=3$ times -1⇒$-2x+6y-2z=-3$
$5x-2y+5z=9$ times 3⇒ $15x-6y+15z=27$

ADD the two equations: 13x+13z=24

Subtract the two resulting equations and you get 0 = 1 which is false so there is no solution  for the system.

Answer 2

Answer:

Given equations have no solution

Step-by-step explanation:

Given -

Three set of equations -

$-3x-4y-3z=-7\\2x-6y+2z=3\\5x-2y+5z=9$

We will now equate "x" in the first equation in terms of "y" and "Z" and put it in second equation -

$-3x-4y-3z=-7\\x = \frac{1}{3} (-4y -3z+7)\\\frac{2}{3}(-4y -3z+7) -6y+2z=3\\-8.667 y = -1.667\\y = 0.192$

We will substitute the value of y in the under given equation-

$x = \frac{1}{3} (-4y -3z+7)\\x = \frac{1}{3} (-4*0.192 -3z+7)\\x= 2.077-z$

substituting the derived values in last equation, we get -

$5x-2y+5z=9\\5(2.077-z)-2(0.192)+5z=9\\10.385-5z-0.384+5z=9\\0=-1$

Hence, the  given equations have no solution