Verified Answer :

  • Identify the system of linear equations: \(3x + 4y = -3\) and \(x + 2y = -1\).
  • Use the elimination method: multiply the second equation by 2 to get \(2x + 4y = -2\).
  • Subtract the new equation from the first to isolate \(x\): \(x = -1\).
  • Back-substitute \(x = -1\) into the second equation to find \(y\): \(y = 0\).
  • Conclude that the solution is \(\boxed{{(x, y) = (-1, 0)}}\).

Explanation

  1. Problem Analysis and Strategy

We are given a system of two linear equations in two variables, \(x\) and \(y\):
\[ \begin{cases} 3x + 4y = -3 & \text{(1)} \ x + 2y = -1 & \text{(2)} \end{cases} \]
Our goal is to find the unique pair \((x, y)\) that satisfies both equations simultaneously. Since both are linear equations, we can solve this system using either the substitution method or the elimination method. We will proceed with the elimination method for its efficiency in this case.

  1. Elimination of the Variable y

To eliminate one of the variables, we look at the coefficients of \(x\) and \(y\). Equation (2) has \(x\) with a coefficient of 1, and equation (1) has \(x\) with a coefficient of 3. Alternatively, equation (2) has \(y\) with a coefficient of 2, and equation (1) has \(y\) with a coefficient of 4.

Multiplying equation (2) by 2, we get:
\[2(x + 2y) = 2(-1)\]
\[2x + 4y = -2 \quad \text{(3)}\]

Now, we subtract equation (3) from equation (1) to eliminate \(y\):
\[(3x + 4y) - (2x + 4y) = -3 - (-2)\]
\[3x - 2x + 4y - 4y = -3 + 2\]
\[x = -1\]

  1. Solving for y and Verification

Now that we have the value of \(x = -1\), we substitute this back into equation (2) to solve for \(y\):
\[(-1) + 2y = -1\]
Adding 1 to both sides:
\[2y = -1 + 1\]
\[2y = 0\]
Dividing by 2:
\[y = 0\]

Thus, the solution to the system is \(x = -1\) and \(y = 0\). We verify this by substituting the values into equation (1):
\[3(-1) + 4(0) = -3 + 0 = -3\]
The solution holds true for both equations. The pair \((-1, 0)\) is the unique solution to the system.

Examples

Systems of equations are used in many real-life situations, such as budgeting. Suppose you are buying two types of office supplies: notebooks and pens. If 3 notebooks and 4 pens cost a total of \(\$3\), and 1 notebook and 2 pens cost a total of \(\$1\), you can represent this as a system of equations. Solving the system tells you exactly how much each item costs, allowing you to manage your finances effectively. The algebraic approach of substitution or elimination is a foundational skill for solving these types of problems.