Question

1. \

$$\begin{cases}x + 5y - z = 16 \\\\ 3x - 3y + 2z = 12 \\\\ 2x + 4y + z = 20\\end{cases}$$

1. \

$$\begin{cases}9x + 5y - z = -11 \\\\ 6x + 4y + 2z = 2 \\\\ 2x - 2y + 4z = 4\\end{cases}$$

1. \

$$\begin{cases}x + y + z = 4 \\\\ 5x + 5y + 5z = 12 \\\\ x - 4y + z = 9\\end{cases}$$

1. \

$$\begin{cases}x + y - 2z = 5 \\\\ x + 2y + z = 8 \\\\ 2x + 3y - z = 13\\end{cases}$$

Explanation:

For problem 3:

Step1: Eliminate z with Eq1+Eq3

$(x + 5y - z) + (2x + 4y + z) = 16 + 20$
$3x + 9y = 36$
Simplify: $x + 3y = 12$ (Eq4)

Step2: Eliminate z with 2*Eq1+Eq2

$2(x + 5y - z) + (3x - 3y + 2z) = 2*16 + 12$
$2x + 10y - 2z + 3x - 3y + 2z = 32 + 12$
$5x + 7y = 44$ (Eq5)

Step3: Solve x from Eq4

$x = 12 - 3y$

Step4: Substitute x into Eq5

$5(12 - 3y) + 7y = 44$
$60 - 15y + 7y = 44$
$-8y = -16$
$y = 2$

Step5: Find x with y=2

$x = 12 - 3*2 = 6$

Step6: Find z with x,y values

Substitute $x=6, y=2$ into Eq1: $6 + 5*2 - z = 16$
$6 + 10 - z = 16$
$z = 0$

Step1: Eliminate z with 2*Eq1+Eq2

$2(9x + 5y - z) + (6x + 4y + 2z) = 2*(-11) + 2$
$18x + 10y - 2z + 6x + 4y + 2z = -22 + 2$
$24x + 14y = -20$
Simplify: $12x + 7y = -10$ (Eq4)

Step2: Eliminate z with 4*Eq1+Eq3

$4(9x + 5y - z) + (2x - 2y + 4z) = 4*(-11) + 4$
$36x + 20y - 4z + 2x - 2y + 4z = -44 + 4$
$38x + 18y = -40$
Simplify: $19x + 9y = -20$ (Eq5)

Step3: Solve for x,y via elimination

Multiply Eq4 by 9: $108x + 63y = -90$
Multiply Eq5 by 7: $133x + 63y = -140$
Subtract: $(133x + 63y) - (108x + 63y) = -140 - (-90)$
$25x = -50$
$x = -2$

Step4: Find y with x=-2

Substitute into Eq4: $12*(-2) + 7y = -10$
$-24 + 7y = -10$
$7y = 14$
$y = 2$

Step5: Find z with x,y values

Substitute $x=-2, y=2$ into Eq1: $9*(-2) + 5*2 - z = -11$
$-18 + 10 - z = -11$
$-8 - z = -11$
$z = 3$

Step1: Analyze Eq1 and Eq2

Eq1: $x + y + z = 4$, multiply by 5: $5x + 5y + 5z = 20$
Eq2: $5x + 5y + 5z = 12$

Step2: Check consistency

$20
eq 12$, so no solution exists.

Answer:

$x=6$, $y=2$, $z=0$

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For problem 4: