- Substitute the expression for \(y\) from the first equation into the second: \(2(4x + 1) = -x - 1\).
- Apply the distributive property to simplify the left side: \(8x + 2 = -x - 1\).
- Rearrange terms to group \(x\) on one side and constants on the other: \(9x + 2 = -1\).
- Subtract 2 from both sides to obtain the final simplified equation: \(\boxed{9x = -3}\), which corresponds to option (c).
Explanation
- Problem Analysis and Objective
We are given a system of two linear equations:
1) \(y = 4x + 1\)
2) \(2y = -x - 1\)
Our objective is to use the substitution method, which involves replacing the variable \(y\) in the second equation with the expression given for \(y\) in the first equation. This will result in an equation that only contains the variable \(x\), allowing us to simplify the relationship between \(x\) and the constant terms.
- Performing the Substitution
We start with the second equation:
\[2y = -x - 1\]
Since the first equation defines \(y\) as \(4x + 1\), we substitute \((4x + 1)\) for every occurrence of \(y\) in the second equation:
\[2(4x + 1) = -x - 1\]
Next, we apply the distributive property to the left side of the equation:
\[2 \times 4x + 2 \times 1 = -x - 1\]
\[8x + 2 = -x - 1\]
- Simplifying the Equation and Selecting the Answer
Now that we have the equation \(8x + 2 = -x - 1\), we group the variable terms on one side and the constant terms on the other.
First, add \(x\) to both sides of the equation:
\[8x + x + 2 = -1\]
\[9x + 2 = -1\]
Next, subtract 2 from both sides of the equation:
\[9x = -1 - 2\]
\[9x = -3\]
Comparing this result to the given options:
a.) \(9x = 1\)
b.) \(7x = 1\)
c.) \(9x = -3\)
d.) \(7x = -3\)
The result matches option (c). Thus, the correct equation is \(9x = -3\).
Examples
The substitution method is highly useful in fields like engineering or economics when dealing with systems of interdependent variables. For instance, if you are modeling the cost of production (where total cost depends on variable costs and fixed costs) and the demand-supply equilibrium, you might have two equations representing different aspects of the system. By substituting one into the other, you can solve for a single variable—like the number of units to produce—to find the point where the system balances. This technique turns complex, multi-part problems into manageable, single-variable equations.