Use the method of substitution to solve the following system of equations. If the system is dependent, express the solution set in terms of one of the variables. Leave all fractional answers in fraction form.

{−3x−y=40−x=−12
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We are given the following system of equations:

1. \(-3x - y = 40\)
2. \(-x = -12\)

### Step 1: Solve one equation for a variable
The second equation, \(-x = -12\), is easy to solve for \(x\):

\[
-x = -12 \quad \Rightarrow \quad x = 12
\]

### Step 2: Substitute the value of \(x\) into the first equation
Now that we know \(x = 12\), substitute this into the first equation:

\[
-3(12) - y = 40
\]

Simplify the equation:

\[
-36 - y = 40
\]

### Step 3: Solve for \(y\)
Now, solve for \(y\):

\[
- y = 40 + 36
\]

\[
- y = 76 \quad \Rightarrow \quad y = -76
\]

### Final Answer:
The solution to the system is \(x = 12\) and \(y = -76\).

### Solution as an Ordered Pair:
The solution is \((12, -76)\).

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Would you like more details or further clarifications? Here are 5 related questions for further practice:

1. Solve the system \(2x + y = 10\) and \(x = 4\) using substitution.
2. If \(x = 5\), solve the equation \(3x + 4y = 17\).
3. Use substitution to solve \(4x + 2y = 14\) and \(x = 3\).
4. Solve the system of equations \(y = 2x - 1\) and \(x + y = 7\) by substitution.
5. Solve \(x - 3y = 9\) and \(x = 2y + 4\) using substitution.

**Tip**: Always simplify equations first if possible to make the substitution process easier.

Use the method of substitution to solve the following system of equations. If the system is dependent, express the solution set in terms of one of the variables. Leave all fractional answers in fraction form. {−3x−y=40, −x=−12}

Learn how to solve the system of linear equations -3x - y = 40 and -x = -12 using the substitution method. This step-by-step solution is perfect for students in Grades 8-10, covering key algebraic concepts and substitution techniques.

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