6) \( 3 x+5 y=20 \) \( 7 x+4 y=39 \)

To solve this system of equations, let's first use substitution or elimination. By multiplying the first equation by 4 and the second one by 5, we can align the coefficients of \(y\) for easy elimination: - \(4(3x + 5y) = 4(20) \Rightarrow 12x + 20y = 80\) - \(5(7x + 4y) = 5(39) \Rightarrow 35x + 20y = 195\) Now, subtract the first modified equation from the second: \( (35x + 20y) - (12x + 20y) = 195 - 80 \) leads to \(23x = 115\), thus \(x = 5\). Plugging \(x\) back into the first equation gives \(3(5) + 5y = 20 \Rightarrow 15 + 5y = 20\), so \(5y = 5\) or \(y = 1\). The solution to the system is \( (x, y) = (5, 1) \). If your brain is buzzing after solving these equations, fear not! It’s common to lose track of negatives or misalign coefficients; always double-check your calculations to avoid little slips that can lead to big mistakes. A great way to double-check your solution is by plugging the values of \(x\) and \(y\) back into the original equations to see if both hold true! For further exploration of systems of equations, check out resources like Khan Academy or Purplemath. They provide awesome step-by-step guides, practice problems, and interactive tools to sharpen your skills! Plus, these sites often make math feel a lot more engaging. Happy learning!