Question

Solve the following system of equations for x, y, and z using the substitution method. What is the value of y?

4x+4y+4z=12
4
𝑥
+
4
𝑦
+
4
𝑧
=
12

2x+y+z=5
2
𝑥
+
𝑦
+
𝑧
=
5

3x+6y−3z=12
3
𝑥
+
6
𝑦
−
3
𝑧
=
12

(1 point)
Responses

1
1

3
3

2
2

0

Answer 1

To solve the system of equations using the substitution method, we'll start from the given equations:

1. \( 4x + 4y + 4z = 12 \)
2. \( 2x + y + z = 5 \)
3. \( 3x + 6y - 3z = 12 \)

Let's begin with the second equation and express \( z \) in terms of \( x \) and \( y \):

From equation 2:

\[
z = 5 - 2x - y
\]

Now we'll substitute this expression for \( z \) into equations 1 and 3.

**Substituting into equation 1:**

\[
4x + 4y + 4(5 - 2x - y) = 12
\]

Expanding this:

\[
4x + 4y + 20 - 8x - 4y = 12
\]

Combining like terms:

\[
-4x + 20 = 12
\]

Now, subtract 20 from both sides:

\[
-4x = -8
\]

Dividing both sides by -4 yields:

\[
x = 2
\]

**Substituting \( x = 2 \) back into the expression for \( z \):**

Using \( z = 5 - 2x - y \):

\[
z = 5 - 2(2) - y
\]
\[
z = 5 - 4 - y
\]
\[
z = 1 - y
\]

Next, substituting \( x = 2 \) and \( z = 1 - y \) into equation 3:

**Substituting into equation 3:**

\[
3(2) + 6y - 3(1 - y) = 12
\]

Now simplify:

\[
6 + 6y - 3 + 3y = 12
\]
\[
3 + 9y = 12
\]

Subtracting 3 from both sides:

\[
9y = 9
\]

Dividing by 9 gives:

\[
y = 1
\]

Now, we need to find \( z \):

Using \( z = 1 - y \):

\[
z = 1 - 1 = 0
\]

Finally, our solutions so far are:

\[
x = 2, \quad y = 1, \quad z = 0
\]

Thus, the value of \( y \) is:

\[
\boxed{1}
\]