\( \begin{array}{l}\frac{1}{2} x+\frac{1}{3} y=7 \\ \frac{1}{4} x+\frac{2}{3} y=6\end{array} \) What is the solution of the system shown? \( \left(10 \frac{2}{3}, 5\right) \) \( \left(\frac{1}{6}, 14\right) \) \( (6,12) \)

To solve this system of equations, you can use the method of substitution or elimination. First, multiply both equations by common denominators to eliminate the fractions: 1. Multiply the first equation by 6: \(3x + 2y = 42\). 2. Multiply the second equation by 12: \(3x + 8y = 72\). Now you have: 1. \(3x + 2y = 42\) 2. \(3x + 8y = 72\) Next, subtract the first equation from the second to find \(y\): \((3x + 8y) - (3x + 2y) = 72 - 42\) This simplifies to \(6y = 30\), giving \(y = 5\). Now substitute \(y = 5\) back into the first equation to find \(x\): \(\frac{1}{2}x + \frac{1}{3}(5) = 7\) Multiply through by 6 to clear fractions: \(3x + 10 = 42\) \(3x = 32\) \(x = \frac{32}{3} = 10 \frac{2}{3}\). So the solution is \(\left(10 \frac{2}{3}, 5\right)\). --- The method of elimination, like the one used above, is a go-to strategy for such systems! This method lets you systematically cancel out one variable, revealing the other, making it clearer and often faster, especially with larger numbers. Also, if you find equations in a mixed format—like some with fractions and others with whole numbers—remember to eliminate those pesky fractions first! Transforming the equations by multiplying through by the least common multiple can pave the way for smoother calculations. Happy solving!