Systems of linear equations for solving using algebraic methods.

Here are the step-by-step solutions for each system of equations.

a)

System:
1) $x + 2y = 5$
2) $y - 3z = 5$
3) $3x - z = 4$

Step 1: Solve equation (3) for $z$.
$z = 3x - 4$

Step 2: Substitute this expression for $z$ into equation (2).
$y - 3(3x - 4) = 5$
$y - 9x + 12 = 5$
$y = 9x - 7$

Step 3: Substitute this expression for $y$ into equation (1).
$x + 2(9x - 7) = 5$
$x + 18x - 14 = 5$
$19x = 19$
$x = 1$

Step 4: Find $y$ and $z$.
$y = 9(1) - 7 = 2$
$z = 3(1) - 4 = -1$

Check:
1) $1 + 2(2) = 5$ (Correct)
2) $2 - 3(-1) = 5$ (Correct)
3) $3(1) - (-1) = 4$ (Correct)

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b)

System:
1) $x + y = 13$
2) $y - z = 5$
3) $x - z = 2$

Step 1: From equation (3), express $x$ in terms of $z$: $x = z + 2$.
Step 2: From equation (2), express $y$ in terms of $z$: $y = z + 5$.
Step 3: Substitute both into equation (1).
$(z + 2) + (z + 5) = 13$
$2z + 7 = 13$
$2z = 6$
$z = 3$

Step 4: Find $x$ and $y$.
$x = 3 + 2 = 5$
$y = 3 + 5 = 8$

Check:
1) $5 + 8 = 13$ (Correct)
2) $8 - 3 = 5$ (Correct)
3) $5 - 3 = 2$ (Correct)

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c)

System:
1) $4x + 3y = 4$
2) $2x + 2y - 2z = 0 \Rightarrow x + y - z = 0 \Rightarrow z = x + y$
3) $5x + 3y + z = -2$

Step 1: Substitute $z = x + y$ into equation (3).
$5x + 3y + (x + y) = -2$
$6x + 4y = -2$
Divide by 2: $3x + 2y = -1$

Step 2: Now we have a system with equation (1):
1) $4x + 3y = 4$
New) $3x + 2y = -1$

Multiply the new equation by 3 and equation (1) by 2 to eliminate $y$:
$9x + 6y = -3$
$8x + 6y = 8$

Subtract the second from the first:
$(9x - 8x) + (6y - 6y) = -3 - 8$
$x = -11$

Step 3: Find $y$ using $3x + 2y = -1$.
$3(-11) + 2y = -1$
$-33 + 2y = -1$
$2y = 32$
$y = 16$

Step 4: Find $z$ using $z = x + y$.
$z = -11 + 16 = 5$

Check:
1) $4(-11) + 3(16) = -44 + 48 = 4$ (Correct)
2) $2(-11) + 2(16) - 2(5) = -22 + 32 - 10 = 0$ (Correct)
3) $5(-11) + 3(16) + 5 = -55 + 48 + 5 = -2$ (Correct)

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d)

System:
1) $x - 2y - 3z = 0$
2) $3x + 2y - z = 0$
3) $3y + z = 0 \Rightarrow z = -3y$

Step 1: Substitute $z = -3y$ into equations (1) and (2).
Into (1): $x - 2y - 3(-3y) = 0 \Rightarrow x - 2y + 9y = 0 \Rightarrow x + 7y = 0 \Rightarrow x = -7y$
Into (2): $3x + 2y - (-3y) = 0 \Rightarrow 3x + 5y = 0$

Step 2: Substitute $x = -7y$ into the modified equation (2).
$3(-7y) + 5y = 0$
$-21y + 5y = 0$
$-16y = 0$
$y = 0$

Step 3: Find $x$ and $z$.
$x = -7(0) = 0$
$z = -3(0) = 0$

Check: All equations equal 0 when variables are 0.

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e)

System:
1) $6x - 2z = 8 \Rightarrow 3x - z = 4 \Rightarrow z = 3x - 4$
2) $x + 2y = 5 \Rightarrow 2y = 5 - x \Rightarrow y = \frac{5-x}{2}$
3) $-y + 3z = -5$

Step 1: Substitute expressions for $y$ and $z$ into equation (3).
$-(\frac{5-x}{2}) + 3(3x - 4) = -5$
Multiply entire equation by 2 to clear fraction:
$-(5 - x) + 6(3x - 4) = -10$
$-5 + x + 18x - 24 = -10$
$19x - 29 = -10$
$19x = 19$
$x = 1$

Step 2: Find $y$ and $z$.
$z = 3(1) - 4 = -1$
$y = \frac{5-1}{2} = \frac{4}{2} = 2$

Check:
1) $6(1) - 2(-1) = 6+2=8$ (Correct)
2) $1 + 2(2) = 5$ (Correct)
3) $-2 + 3(-1) = -5$ (Correct)

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f)

System:
1) $2x - y + z = 0$
2) $x + 2y - 2z = 0$
3) $3x + y - z = 0$

Step 1: Add equation (1) and (3) to eliminate $y$ and $z$ partially? Let's just add (1) and (3).
$(2x - y + z) + (3x + y - z) = 0$
$5x = 0 \Rightarrow x = 0$

Step 2: Substitute $x=0$ into (1) and (2).
1) $-y + z = 0 \Rightarrow y = z$
2) $2y - 2z = 0 \Rightarrow 2y = 2z \Rightarrow y = z$

This implies any value where $y=z$ works? Let's check if there is a unique solution.
If $x=0, y=1, z=1$:
1) $0 - 1 + 1 = 0$
2) $0 + 2 - 2 = 0$
3) $0 + 1 - 1 = 0$
It seems there are infinite solutions along the line $x=0, y=z$. However, usually, these problems ask for the trivial solution or a specific one. Let's re-read carefully. Usually, homogeneous systems with determinant 0 have infinite solutions.
Let's check the determinant of coefficients:
$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -2 \\ 3 & 1 & -1 \end{vmatrix} = 2(-2+2) - (-1)(-1+6) + 1(1-6) = 0 - 5 - 5 = -10 \neq 0$.
Wait, my addition in Step 1 was:
Eq 1: $2x - y + z = 0$
Eq 3: $3x + y - z = 0$
Sum: $5x = 0 \rightarrow x=0$. This is correct.

If $x=0$, Eq 1 becomes $-y+z=0 \rightarrow y=z$.
Eq 2 becomes $2y-2z=0 \rightarrow y=z$.
Eq 3 becomes $y-z=0 \rightarrow y=z$.

So the solution is $x=0$, and $y$ can be any real number $t$, with $z=t$.
However, in many school contexts, if not specified, they might look for the integer solution $(0,0,0)$ or assume I made an arithmetic error. Let me re-check the determinant calculation.
Row 1: 2, -1, 1
Row 2: 1, 2, -2
Row 3: 3, 1, -1

Det = $2((2)(-1) - (-2)(1)) - (-1)((1)(-1) - (-2)(3)) + 1((1)(1) - (2)(3))$
$= 2(-2 + 2) + 1(-1 + 6) + 1(1 - 6)$
$= 2(0) + 1(5) + 1(-5)$
$= 0 + 5 - 5 = 0$.

The determinant IS zero. The system has infinitely many solutions.
Solution set: $x = 0, y = t, z = t$ for any real number $t$.
Commonly, the simplest answer is $x=0, y=0, z=0$. Given the other problems have unique integer solutions, let's look closer. Is it possible I copied a sign wrong?
Image:
f) $2x - y + z = 0$
$x + 2y - 2z = 0$
$3x + y - z = 0$
No, signs are correct.
Since it's a homework helper context, providing the general solution is most accurate, but often $(0,0,0)$ is the expected "trivial" answer if parameters aren't introduced. I will provide the parametric form as it is mathematically complete, but note $(0,0,0)$ is a valid solution.

Actually, looking at problem (d), it also resulted in $(0,0,0)$. Problem (f) is dependent.
Let's double check if I missed a constraint. No.
I will state the solution as $x=0, y=z$.

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g)

System:
1) $-x + 3y = -72 \Rightarrow x = 3y + 72$
2) $3x + 4y - 4z = -4$
3) $-20x - 12y + 5z = -50$

Step 1: Substitute $x = 3y + 72$ into (2) and (3).
Into (2):
$3(3y + 72) + 4y - 4z = -4$
$9y + 216 + 4y - 4z = -4$
$13y - 4z = -220$ (Eq A)

Into (3):
$-20(3y + 72) - 12y + 5z = -50$
$-60y - 1440 - 12y + 5z = -50$
$-72y + 5z = 1390$ (Eq B)

Step 2: Solve system of (A) and (B).
From (A): $4z = 13y + 220 \Rightarrow z = \frac{13y + 220}{4}$
Substitute into (B):
$-72y + 5(\frac{13y + 220}{4}) = 1390$
Multiply by 4:
$-288y + 5(13y + 220) = 5560$
$-288y + 65y + 1100 = 5560$
$-223y = 4460$
$y = -20$

Step 3: Find $x$ and $z$.
$x = 3(-20) + 72 = -60 + 72 = 12$
$z = \frac{13(-20) + 220}{4} = \frac{-260 + 220}{4} = \frac{-40}{4} = -10$

Check:
1) $-12 + 3(-20) = -12 - 60 = -72$ (Correct)
2) $3(12) + 4(-20) - 4(-10) = 36 - 80 + 40 = -4$ (Correct)
3) $-20(12) - 12(-20) + 5(-10) = -240 + 240 - 50 = -50$ (Correct)

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h)

System:
1) $16x + 16y + 17z = 10$
2) $-14x + 17y - 3z = 75$
3) $-5x - 11y - 18z = 43$

This looks messy. Let's try elimination.
Multiply (3) by something to match coefficients?
Let's eliminate $z$ first.
From (2), $3z = -14x + 17y - 75 \Rightarrow z = \frac{-14x + 17y - 75}{3}$

Substitute into (1):
$16x + 16y + 17(\frac{-14x + 17y - 75}{3}) = 10$
Multiply by 3:
$48x + 48y + 17(-14x + 17y - 75) = 30$
$48x + 48y - 238x + 289y - 1275 = 30$
$-190x + 337y = 1305$ (Eq A)

Substitute $z$ into (3):
$-5x - 11y - 18(\frac{-14x + 17y - 75}{3}) = 43$
$-5x - 11y - 6(-14x + 17y - 75) = 43$
$-5x - 11y + 84x - 102y + 450 = 43$
$79x - 113y = -407$ (Eq B)

Now solve:
A) $-190x + 337y = 1305$
B) $79x - 113y = -407$

From B: $113y = 79x + 407 \Rightarrow y = \frac{79x + 407}{113}$
This numbers are ugly. Let's guess integer solutions.
Look at Eq B: $79x - 113y = -407$.
If $x=2$, $158 - 113y = -407 \Rightarrow -113y = -565 \Rightarrow y=5$.
Let's test $x=2, y=5$ in Eq A.
$-190(2) + 337(5) = -380 + 1685 = 1305$. It matches!

So $x=2, y=5$.
Find $z$ using original eq (2):
$-14(2) + 17(5) - 3z = 75$
$-28 + 85 - 3z = 75$
$57 - 3z = 75$
$-3z = 18$
$z = -6$

Check:
1) $16(2) + 16(5) + 17(-6) = 32 + 80 - 102 = 112 - 102 = 10$ (Correct)
2) $-14(2) + 17(5) - 3(-6) = -28 + 85 + 18 = 75$ (Correct)
3) $-5(2) - 11(5) - 18(-6) = -10 - 55 + 108 = 43$ (Correct)

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i)

System:
1) $x + 2y - z = 1$
2) $2x + 3y + z = 2$
3) $x + 3y - 2z = 1$

Step 1: Add (1) and (2) to eliminate $z$.
$(x + 2y - z) + (2x + 3y + z) = 1 + 2$
$3x + 5y = 3$ (Eq A)

Step 2: Eliminate $z$ using (1) and (3).
Multiply (1) by 2: $2x + 4y - 2z = 2$
Subtract (3) from this:
$(2x + 4y - 2z) - (x + 3y - 2z) = 2 - 1$
$x + y = 1 \Rightarrow x = 1 - y$ (Eq B)

Step 3: Substitute (Eq B) into (Eq A).
$3(1 - y) + 5y = 3$
$3 - 3y + 5y = 3$
$2y = 0 \Rightarrow y = 0$

Step 4: Find $x$ and $z$.
$x = 1 - 0 = 1$
From (1): $1 + 2(0) - z = 1 \Rightarrow 1 - z = 1 \Rightarrow z = 0$

Check:
1) $1 + 0 - 0 = 1$ (Correct)
2) $2 + 0 + 0 = 2$ (Correct)
3) $1 + 0 - 0 = 1$ (Correct)

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j)

System:
1) $x + y + 2z = 4$
2) $x - 2y + z = 0$
3) $x - 5y = -4 \Rightarrow x = 5y - 4$

Step 1: Substitute $x = 5y - 4$ into (1) and (2).
Into (1):
$(5y - 4) + y + 2z = 4$
$6y + 2z = 8 \Rightarrow 3y + z = 4 \Rightarrow z = 4 - 3y$ (Eq A)

Into (2):
$(5y - 4) - 2y + z = 0$
$3y - 4 + z = 0 \Rightarrow z = 4 - 3y$ (Eq B)

Both give the same relationship. This means the system is dependent?
Let's check the determinant.
$\begin{vmatrix} 1 & 1 & 2 \\ 1 & -2 & 1 \\ 1 & -5 & 0 \end{vmatrix} = 1(0 - (-5)) - 1(0 - 1) + 2(-5 - (-2)) = 5 + 1 + 2(-3) = 6 - 6 = 0$.
Determinant is 0. Infinite solutions.
We have $x = 5y - 4$ and $z = 4 - 3y$.
Let $y = t$.
$x = 5t - 4$
$z = 4 - 3t$

Usually, if a single numeric answer is forced, there might be a typo in my reading or the problem expects a parameterized answer.
Let's re-read image j).
$x + y + 2z = 4$
$x - 2y + z = 0$
$x - 5y = -4$

Wait, look at equation 3 in the image: $x - 5y = -4$. There is no $z$.
My derivation holds.
Is there a specific integer solution intended?
If $y=1$, $x=1, z=1$.
Check:
1) $1+1+2=4$
2) $1-2+1=0$
3) $1-5=-4$
This works.
If $y=0$, $x=-4, z=4$.
1) $-4+0+8=4$
2) $-4-0+4=0$
3) $-4-0=-4$
This also works.

Since multiple answers exist, I will provide the parametric solution.
$x = 5y - 4, z = 4 - 3y$.

*Self-Correction*: In many online homework platforms, if the system is dependent, they might ask for the solution in terms of one variable, or there is a hidden constraint. Without further instruction, the general solution is the only correct mathematical answer. However, looking at the other problems, they all had unique integer solutions. Did I misread a sign?
j) $x+y+2z=4$
$x-2y+z=0$
$x-5y=-4$
No, signs look correct.
I will write the solution as a set of parametric equations.

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k)

System:
1) $x + 2y + 3z = 14$
2) $3x + 2y + z = 10$
3) $3x + y + 2z = 11$

Step 1: Subtract (2) from (3) to eliminate $x$.
$(3x + y + 2z) - (3x + 2y + z) = 11 - 10$
$-y + z = 1 \Rightarrow z = y + 1$

Step 2: Subtract (2) from 3*(1) to eliminate $x$? Or just substitute $z$.
Let's substitute $z = y + 1$ into (1) and (2).
Into (1):
$x + 2y + 3(y + 1) = 14$
$x + 5y + 3 = 14$
$x + 5y = 11 \Rightarrow x = 11 - 5y$ (Eq A)

Into (2):
$3x + 2y + (y + 1) = 10$
$3x + 3y = 9 \Rightarrow x + y = 3 \Rightarrow x = 3 - y$ (Eq B)

Step 3: Set Eq A and Eq B equal.
$11 - 5y = 3 - y$
$8 = 4y$
$y = 2$

Step 4: Find $x$ and $z$.
$x = 3 - 2 = 1$
$z = 2 + 1 = 3$

Check:
1) $1 + 2(2) + 3(3) = 1 + 4 + 9 = 14$ (Correct)
2) $3(1) + 2(2) + 3 = 3 + 4 + 3 = 10$ (Correct)
3) $3(1) + 2 + 2(3) = 3 + 2 + 6 = 11$ (Correct)

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l)

System:
1) $x + y + z = 4$
2) $x + 2y + 4z = 12$
3) $2x - 3y - z = 4$

Step 1: From (1), $z = 4 - x - y$. Substitute into (2) and (3).
Into (2):
$x + 2y + 4(4 - x - y) = 12$
$x + 2y + 16 - 4x - 4y = 12$
$-3x - 2y = -4 \Rightarrow 3x + 2y = 4$ (Eq A)

Into (3):
$2x - 3y - (4 - x - y) = 4$
$2x - 3y - 4 + x + y = 4$
$3x - 2y = 8$ (Eq B)

Step 2: Add (Eq A) and (Eq B).
$(3x + 2y) + (3x - 2y) = 4 + 8$
$6x = 12$
$x = 2$

Step 3: Find $y$ using (Eq A).
$3(2) + 2y = 4$
$6 + 2y = 4$
$2y = -2$
$y = -1$

Step 4: Find $z$ using (1).
$2 + (-1) + z = 4$
$1 + z = 4$
$z = 3$

Check:
1) $2 - 1 + 3 = 4$ (Correct)
2) $2 + 2(-1) + 4(3) = 2 - 2 + 12 = 12$ (Correct)
3) $2(2) - 3(-1) - 3 = 4 + 3 - 3 = 4$ (Correct)

Final Answer:
a) $x=1, y=2, z=-1$
b) $x=5, y=8, z=3$
c) $x=-11, y=16, z=5$
d) $x=0, y=0, z=0$
e) $x=1, y=2, z=-1$
f) $x=0, y=t, z=t$ (Infinitely many solutions, e.g., $0,0,0$)
g) $x=12, y=-20, z=-10$
h) $x=2, y=5, z=-6$
i) $x=1, y=0, z=0$
j) $x=5t-4, y=t, z=4-3t$ (Infinitely many solutions, e.g., $1,1,1$)
k) $x=1, y=2, z=3$
l) $x=2, y=-1, z=3$