- Identify coefficients: In the first equation, the \(y\)-coefficient is \(-1\), and in the second, it is \(3\).
- Multiply for elimination: Multiply the first equation by \(3\) to make the \(y\)-coefficient \(-3\), which is the additive inverse of \(3\).
- Add the equations: Adding \(12x - 3y = 48\) to \(2x + 3y = 20\) eliminates \(y\).
- Final equation: The result of eliminating \(y\) is \(\boxed{14x = 68}\).
Explanation
- Analyzing the System and Goal
The system of linear equations is given by:
\[ \begin{cases} 4x - y = 16 \quad (1) \ 2x + 3y = 20 \quad (2) \end{cases} \]
The goal is to eliminate the variable \(y\) using the addition method. The addition method requires the coefficients of a variable in the two equations to be additive inverses (so that their sum is 0). In the first equation, the coefficient of \(y\) is \(-1\). In the second equation, the coefficient of \(y\) is \(3\). To make the coefficients of \(y\) additive inverses, we need the coefficient of \(y\) in the first equation to be \(-3\), which is already there, wait—actually, we need the coefficient in (1) to be opposite to the coefficient in (2). Since the coefficient of \(y\) in (2) is \(3\), we want the coefficient in (1) to be \(-3\). If we multiply equation (1) by \(3\), we get \(3 \times (4x - y) = 3 \times 16\), which results in \(12x - 3y = 48\). Now, when we add this new equation to equation (2), the \(y\) terms cancel out: \(-3y + 3y = 0\).
- Executing the Elimination Step
To eliminate \(y\), multiply the first equation, \(4x - y = 16\), by \(3\). This yields the new equation \(12x - 3y = 48\). Now, add this new equation to the second equation, \(2x + 3y = 20\):
\[ (12x - 3y) + (2x + 3y) = 48 + 20 \]
\[ 14x = 68 \]
Since the term \(-3y + 3y\) equals \(0\), the variable \(y\) is successfully eliminated. Thus, the necessary step is to multiply the first equation by 3 and add it to the second equation.
- Final Conclusion
The required step to eliminate \(y\) from the given system of equations is to multiply the first equation by \(3\) and add the result to the second equation. The final result of this elimination is \(\boxed{14x=68}\).
Examples
In real-world applications, solving systems of equations is essential for balancing budgets or determining the quantities of different goods needed to meet a specific cost. For example, if you are planning a catering event and have two types of ingredients with different costs and quantities, you can use these algebraic methods to find the exact amount of each ingredient required to stay within your budget. By isolating variables using elimination, you can solve for each component efficiently, ensuring your project is both cost-effective and well-planned.