Question

Solve the system using the elimination method.

$3x+2y-z=8$

$-3x+4y+5z=-14$

$x-3y+4z=-14$

The solution is $\left(x,y,z\right)=$ (
,
,
).
show work, make it simple and understandable

Answer 1

To solve the system of equations using the elimination method, let's start with the given equations:

1. \( 3x + 2y - z = 8 \) (Equation 1)
2. \( -3x + 4y + 5z = -14 \) (Equation 2)
3. \( x - 3y + 4z = -14 \) (Equation 3)

**Step 1: Eliminate \( z \) from equations**

Let's eliminate \( z \) from Equation 1 and Equation 3 first. We can express \( z \) in terms of \( x \) and \( y \) from Equation 1:

\[
z = 3x + 2y - 8 \tag{4}
\]

Now, substitute this expression for \( z \) in Equation 2 and Equation 3.

**Substituting into Equation 2:**

\[
-3x + 4y + 5(3x + 2y - 8) = -14
\]

Expanding this:

\[
-3x + 4y + 15x + 10y - 40 = -14
\]

Combining like terms:

\[
(15x - 3x) + (4y + 10y) - 40 = -14
\]
\[
12x + 14y - 40 = -14
\]

Adding 40 to both sides:

\[
12x + 14y = 26
\]

Dividing through by 2:

\[
6x + 7y = 13 \tag{5}
\]

**Substituting into Equation 3:**

\[
x - 3y + 4(3x + 2y - 8) = -14
\]

Expanding this:

\[
x - 3y + 12x + 8y - 32 = -14
\]

Combining like terms:

\[
(12x + x) + (-3y + 8y) - 32 = -14
\]
\[
13x + 5y - 32 = -14
\]

Adding 32 to both sides:

\[
13x + 5y = 18 \tag{6}
\]

Now we have a new system with two equations (5) and (6):

1. \( 6x + 7y = 13 \) (Equation 5)
2. \( 13x + 5y = 18 \) (Equation 6)

**Step 2: Eliminate \( y \)**

Let's eliminate \( y \) by multiplying Equation 5 by 5 and Equation 6 by 7:

**Multiply Equation 5 by 5:**
\[
30x + 35y = 65 \tag{7}
\]

**Multiply Equation 6 by 7:**
\[
91x + 35y = 126 \tag{8}
\]

**Step 3: Subtract Equation 7 from Equation 8:**
\[
(91x + 35y) - (30x + 35y) = 126 - 65
\]
\[
61x = 61
\]

Now divide both sides by 61:
\[
x = 1
\]

**Step 4: Substitute \( x = 1 \) back into Equation 5 to find \( y \)**:
\[
6(1) + 7y = 13
\]
\[
6 + 7y = 13
\]
Subtracting 6 from both sides, we get:
\[
7y = 7
\]
Dividing both sides by 7:
\[
y = 1
\]

**Step 5: Substitute \( x = 1 \) and \( y = 1 \) back into Equation (4) to find \( z \)**:
\[
z = 3(1) + 2(1) - 8
\]
\[
z = 3 + 2 - 8
\]
\[
z = -3
\]

Thus, the solution to the system is:
\[
(x, y, z) = (1, 1, -3)
\]

So the final answer is:
\[
\boxed{(1, 1, -3)}
\]