Question

solve the given system - or show that no solution exists:
\\(\

$$\begin{align*} x + 2y \\quad \\quad &= 1 \\\\ 3x + 2y + \\ 4z &= 7 \\\\ -2x + \\ \\ y - \\ 2z &= -1 \\end{align*}$$

\\)

Explanation:

Step1: Label equations

Let:

1. $x + 2y = 1$
2. $3x + 2y + 4z = 7$
3. $-2x + y - 2z = -1$

Step2: Eliminate $y$ from 1 and 2

Subtract equation1 from equation2:
$(3x + 2y + 4z) - (x + 2y) = 7 - 1$
$2x + 4z = 6$ → Simplify: $x + 2z = 3$ (Equation4)

Step3: Eliminate $y$ from 1 and 3

Multiply equation1 by 1: $x + 2y = 1$
Multiply equation3 by 2: $-4x + 2y - 4z = -2$ (Equation5)
Subtract equation5 from equation1:
$(x + 2y) - (-4x + 2y - 4z) = 1 - (-2)$
$5x + 4z = 3$ (Equation6)

Step4: Solve for $x,z$ using 4 and 6

From equation4: $x = 3 - 2z$
Substitute into equation6:
$5(3 - 2z) + 4z = 3$
$15 - 10z + 4z = 3$
$15 - 6z = 3$ → $-6z = -12$ → $z=2$? Wait correction:
Wait equation6 calculation: $(x+2y)-(-4x+2y-4z)=x+2y+4x-2y+4z=5x+4z=1+2=3$ correct.
Equation4: $x=3-2z$. Substitute into 5x+4z=3:
$5(3-2z)+4z=15-10z+4z=15-6z=3$ → $-6z=-12$ → $z=2$? Wait no: 15-3=12 → -6z=-12 → z=2. Then x=3-2*2=3-4=-1.

Wait correction step3: Multiply equation3 by 2: -4x+2y-4z=-2 (correct). Subtract equation5 from equation1: equation1 - equation5 = (x+2y) - (-4x+2y-4z) = x+2y+4x-2y+4z=5x+4z=1 - (-2)=3 (correct).

Step4: Equation4: x+2z=3; Equation6:5x+4z=3. Let's use elimination: multiply equation4 by 2:2x+4z=6 (Equation7). Subtract equation6 from equation7: (2x+4z)-(5x+4z)=6-3 → -3x=3 → x=-1. Then from equation4: -1+2z=3 → 2z=4 → z=2.

Step5: Find y from equation1: x+2y=1 → -1+2y=1 →2y=2→y=1? Wait check equation2:3(-1)+2(1)+4(2)= -3+2+8=7 (correct). Equation3: -2(-1)+1-2*(2)=2+1-4=-1 (correct).

Wait earlier mistake in step4 substitution: yes z=2, x=-1, y=1.

Final correction:
Step4: Multiply equation4 by 2: 2x+4z=6. Subtract equation6: (2x+4z)-(5x+4z)=6-3 → -3x=3 →x=-1. Then z=(3 -x)/2=(3 -(-1))/2=4/2=2. Step5: y=(1 -x)/2=(1 -(-1))/2=2/2=1.

So solution x=-1,y=1,z=2.

Wait verify all equations:

1. -1 +2*1=1 ✔️

2.3(-1)+21+4*2=-3+2+8=7 ✔️
3.-2(-1)+1-22=2+1-4=-1 ✔️

Yes correct.

Step1: Label equations

1. $x+2y=1$; 2. $3x+2y+4z=7$;3. $-2x+y-2z=-1$

Step2: Subtract 1 from 2

$2x+4z=6$ → $x+2z=3$ (Eq4)

Step3: Eliminate y (1-2*3)

$5x+4z=3$ (Eq6)

Step4: Solve Eq4 & Eq6

Multiply Eq4 by 2: $2x+4z=6$. Subtract Eq6: $-3x=3$ → $x=-1$

Step5: Find z

$z=(3 -x)/2=(3+1)/2=2$

Step6: Find y

$y=(1 -x)/2=(1+1)/2=1$

Final solution: $x=-1,y=1,z=2$

Answer:

$x=1$, $y=0$, $z=1$