Polynomial word problems worksheet for algebra practice.

Problem Analysis:

The problem involves solving a system of linear equations using two methods: substitution and elimination. Let's break it down step by step.

#### Given Equations:
1. \( 3x - 2y = 6 \)
2. \( 6x + y = 27 \)

#### Task:
1. Solve the system of equations using substitution.
2. Solve the system of equations using elimination.
3. Verify the solution.

---

Solution:

#### Step 1: Solve using Substitution

##### Step 1.1: Express one variable in terms of the other
From the second equation \( 6x + y = 27 \), solve for \( y \):
\[
y = 27 - 6x
\]

##### Step 1.2: Substitute into the first equation
Substitute \( y = 27 - 6x \) into the first equation \( 3x - 2y = 6 \):
\[
3x - 2(27 - 6x) = 6
\]

##### Step 1.3: Simplify and solve for \( x \)
Distribute the \(-2\):
\[
3x - 54 + 12x = 6
\]
Combine like terms:
\[
15x - 54 = 6
\]
Add 54 to both sides:
\[
15x = 60
\]
Divide by 15:
\[
x = 4
\]

##### Step 1.4: Solve for \( y \)
Substitute \( x = 4 \) back into \( y = 27 - 6x \):
\[
y = 27 - 6(4)
\]
\[
y = 27 - 24
\]
\[
y = 3
\]

##### Solution from Substitution:
\[
(x, y) = (4, 3)
\]

---

#### Step 2: Solve using Elimination

##### Step 2.1: Align the equations
We have:
\[
3x - 2y = 6 \quad \text{(Equation 1)}
\]
\[
6x + y = 27 \quad \text{(Equation 2)}
\]

##### Step 2.2: Make coefficients of \( y \) opposites
To eliminate \( y \), we can multiply Equation 2 by 2:
\[
2(6x + y) = 2(27)
\]
\[
12x + 2y = 54 \quad \text{(New Equation 2)}
\]

Now we have:
\[
3x - 2y = 6 \quad \text{(Equation 1)}
\]
\[
12x + 2y = 54 \quad \text{(New Equation 2)}
\]

##### Step 2.3: Add the equations
Add Equation 1 and New Equation 2:
\[
(3x - 2y) + (12x + 2y) = 6 + 54
\]
\[
3x + 12x - 2y + 2y = 60
\]
\[
15x = 60
\]
Solve for \( x \):
\[
x = 4
\]

##### Step 2.4: Solve for \( y \)
Substitute \( x = 4 \) into Equation 2 (\( 6x + y = 27 \)):
\[
6(4) + y = 27
\]
\[
24 + y = 27
\]
\[
y = 3
\]

##### Solution from Elimination:
\[
(x, y) = (4, 3)
\]

---

#### Step 3: Verify the Solution

Substitute \( x = 4 \) and \( y = 3 \) into both original equations to verify:

1. Check \( 3x - 2y = 6 \):
\[
3(4) - 2(3) = 12 - 6 = 6 \quad \text{(True)}
\]

2. Check \( 6x + y = 27 \):
\[
6(4) + 3 = 24 + 3 = 27 \quad \text{(True)}
\]

The solution satisfies both equations.

---

Final Answer:

\[
\boxed{(4, 3)}
\]