6. \( \begin{aligned} 4 x-3 y & =5 \\ 6 x+2 y & =14\end{aligned} \)
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To solve the system of equations, you can use either substitution or elimination. Here, I'll use the elimination method. First, we can multiply the first equation by 2 to align the \(y\) coefficients. This gives us: \[ \begin{aligned} 8x - 6y & = 10 \quad (1) \\ 6x + 2y & = 14 \quad (2) \end{aligned} \] Now, multiply (2) by 3 to match the coefficients of \(y\): \[ \begin{aligned} 6x + 2y & = 14 \quad \text{multiply by 3 gives} \\ 18x + 6y & = 42 \quad (3) \end{aligned} \] Next, now that we have the same coefficient of \(y\) in both equations: \[ \begin{aligned} 8x - 6y & = 10 \quad (1) \\ 18x + 6y & = 42 \quad (3) \end{aligned} \] Add (1) and (3) to eliminate \(y\): \[ 26x = 52 \] \[ x = 2 \] Substitute \(x = 2\) back into one of the original equations to find \(y\). Using the first equation: \[ 4(2) - 3y = 5 \] \[ 8 - 3y = 5 \] \[ -3y = -3 \implies y = 1 \] Thus, the solution to the system of equations is \( (2, 1) \).
