Question #9ab83

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Question #9ab83

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Answer 1 · 2017-03-23T03:26:51

$x = 4, y = 0, z = - 3$ $x=4,y=0,z=-3$

Explanation:

Given three equation and three unknown
$x + y - z = 7$ $x+y-z=7$ .......(1)
$- 2 x + 2 y - z = - 5$ $-2x+2y-z=-5$ .......(2)
$3 x - 3 y + 2 z = 6$ $3x-3y+2z=6$ ......(3)

Step 1. Eliminate one unknown from first two equations. Let us take $z$ $z$

Subtracting (2) from (1) we get
$(x + y - z) - (- 2 x + 2 y - z) = 7 - (- 5)$ $(x+y-z)-(-2x+2y-z)=7-(-5)$
$\Rightarrow x + y - z + 2 x - 2 y + z = 7 + 5$ $=>x+y-z+2x-2y+z=7+5$
$\Rightarrow 3 x - y = 12$ $=>3x-y=12$ ......(4)

Step 2. Eliminate same unknown from next two equations. (one can choose last two as well).

Multiplying (2) with $2$ $2$ and adding (3)
$2 \times (- 2 x + 2 y - z) + (3 x - 3 y + 2 z) = 2 \times (- 5) + 6$ $2xx(-2x+2y-z)+(3x-3y+2z)=2xx(-5)+6$
$\Rightarrow - 4 x + 4 y - 2 z + 3 x - 3 y + 2 z = - 10 + 6$ $=>-4x+4y-2z+3x-3y+2z=-10+6$
$\Rightarrow - x + y = - 4$ $=>-x+y=-4$ .....(5)

We now have two equations (4) and (5) with two unknowns.
Step 3. Eliminate one unknown from these two equations.

We see that if we add these two we eliminate $y$ $y$ . We get
$3 x - y + (- x + y) = 12 + (- 4)$ $3x-y+(-x+y)=12+(-4)$
$\Rightarrow 3 x - y - x + y = 12 - 4$ $=>3x-y-x+y=12-4$
$\Rightarrow 2 x = 8$ $=>2x=8$
$\Rightarrow x = \frac{8}{2}$ $=>x=8/2$
$\Rightarrow x = 4$ $=>x=4$

Step 4. Insert this value of $x$ $x$ in either (4) or (5) to obtain $y$ $y$ . Let us choose (5)

$- 4 + y = - 4$ $-4+y=-4$
$y = - 4 + 4$ $y=-4+4$
$y = 0$ $y=0$

Step 5. Insert these values of $x and y$ $x and y$ in any of (1), (2) or (3) to obtain $z$ $z$ . Let us choose (1)

$4 + 0 - z = 7$ $4+0-z=7$
$\Rightarrow z = 4 - 7$ $=>z=4-7$
$\Rightarrow z = - 3$ $=>z=-3$

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Question #9ab83

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