Question

algebra & e linear systems 25-26
question 4
4x - y = 10
2x + 3y = 12
given the system of equations, what method would you use to solve this system: substitution or elimination? explain why and what your first step in solving the system would be.

Explanation:

Step1: Analyze the system for methods

The system is \(

$$\begin{cases} 4x - y = 10 \\ 2x + 3y = 12 \end{cases}$$

\). For substitution, we can easily solve the first equation for \( y \) (since \( y \) has a coefficient of -1, making it simple to isolate). For elimination, we'd need to adjust coefficients, but substitution here is straightforward as one variable has a coefficient of \( \pm1 \).

Step2: Decide on method and first step

Substitution is good here because in \( 4x - y = 10 \), solving for \( y \) gives \( y = 4x - 10 \), which is simple. The first step would be to solve the first equation for \( y \): from \( 4x - y = 10 \), we get \( y = 4x - 10 \) by adding \( y \) to both sides and subtracting 10, or rearranging. Then substitute this \( y \) into the second equation \( 2x + 3(4x - 10) = 12 \) to solve for \( x \).

Alternatively, for elimination: If we multiply the first equation by 3, we get \( 12x - 3y = 30 \), then add to the second equation \( 2x + 3y = 12 \), the \( y \)-terms eliminate (\( (12x - 3y)+(2x + 3y)=30 + 12 \)), giving \( 14x = 42 \), so \( x = 3 \). Then substitute \( x = 3 \) into one of the original equations to find \( y \). Either method works, but substitution is straightforward due to the -1 coefficient on \( y \) in the first equation.

Answer:

Method: Substitution (or Elimination). Reason for Substitution: The first equation has \( y \) with coefficient -1, so solving for \( y \) is easy (\( y = 4x - 10 \)). First step: Solve \( 4x - y = 10 \) for \( y \) to get \( y = 4x - 10 \). (For Elimination: Multiply first equation by 3 to eliminate \( y \) when added to second equation, first step: \( 3\times(4x - y)=3\times10 \) i.e., \( 12x - 3y = 30 \))