Question

Solve the following system of equations for all three variables.
x, plus, 5, y, minus, 3, z, equals, minus, 5
x+5y−3z=
−5
minus, x, plus, y, plus, 4, z, equals, 10
−x+y+4z=
10
minus, x, minus, 4, y, plus, 2, z, equals, 0
−x−4y+2z=
0
Answer
Attempt 1 out of 2

x, equalsx=
y, equalsy=
z, equalsz=

Answer 1

To solve the system of equations, we have:

1. \( x + 5y - 3z = -5 \) (Equation 1)
2. \( -x + y + 4z = 10 \) (Equation 2)
3. \( -x - 4y + 2z = 0 \) (Equation 3)

Let's solve this step by step.

### Step 1: Isolate x in Equation 2
From Equation 2, we can express \( x \):
\[
-x + y + 4z = 10 \implies x = y + 4z - 10 \tag{Equation 4}
\]

### Step 2: Substitute Equation 4 into Equations 1 and 3
Substituting \( x \) from Equation 4 into Equations 1 and 3.

**For Equation 1:**
\[
(y + 4z - 10) + 5y - 3z = -5
\]
\[
y + 4z - 10 + 5y - 3z = -5
\]
\[
6y + z - 10 = -5
\]
\[
6y + z = 5 \tag{Equation 5}
\]

**For Equation 3:**
\[
-(y + 4z - 10) - 4y + 2z = 0
\]
\[
-y - 4z + 10 - 4y + 2z = 0
\]
\[
-5y - 2z + 10 = 0
\]
\[
5y + 2z = 10 \tag{Equation 6}
\]

### Step 3: Solve the new system of equations (Equations 5 and 6)
Now we have:
1. \( 6y + z = 5 \) (Equation 5)
2. \( 5y + 2z = 10 \) (Equation 6)

We can express \( z \) in terms of \( y \) using Equation 5:
\[
z = 5 - 6y \tag{Equation 7}
\]

Now substitute Equation 7 into Equation 6:
\[
5y + 2(5 - 6y) = 10
\]
\[
5y + 10 - 12y = 10
\]
\[
-7y + 10 = 10
\]
\[
-7y = 0 \implies y = 0
\]

### Step 4: Find z using y = 0
Substituting \( y = 0 \) back into Equation 7:
\[
z = 5 - 6(0) = 5
\]

### Step 5: Find x using y = 0 and z = 5
Substituting \( y = 0 \) and \( z = 5 \) into Equation 4:
\[
x = 0 + 4(5) - 10 = 20 - 10 = 10
\]

### Solution
Thus, the solution to the system of equations is:
\[
x = 10, \quad y = 0, \quad z = 5
\]

So, the final answer is:
\[
x = 10, \quad y = 0, \quad z = 5
\]