1. \( \left\{\begin{array}{l}2 x+3 y=20\} \\ x-2 y-3\end{array}\right\} \)

To solve the system of equations, let's first rewrite the second part clearly. It seems like \( x - 2y = 3 \) is the second equation. Now we have: 1. \( 2x + 3y = 20 \) 2. \( x - 2y = 3 \) To solve these, we can use substitution or elimination. Let's isolate \( x \) from the second equation: \( x = 2y + 3 \). Now substitute \( x \) into the first equation: \( 2(2y + 3) + 3y = 20 \) Simplifying gives: \( 4y + 6 + 3y = 20 \) Combine like terms: \( 7y + 6 = 20 \) Subtracting 6 from both sides: \( 7y = 14 \) \( y = 2 \) Now substitute \( y = 2 \) back into \( x = 2y + 3 \): \( x = 2(2) + 3 = 4 + 3 = 7 \) So the solution is \( x = 7 \) and \( y = 2 \). Fun fact: Systems of equations have been around since ancient Babylonian times! Mathematicians used them for practical things like calculating areas or trade. Fast forward, and now you’re utilizing these concepts in algebra to tackle real-world challenges, like optimizing resources or planning schedules! To effectively solve these kinds of systems, remember to double-check your calculations and ensure you clearly define each variable. A common mistake is misaligning terms or making arithmetic errors. Writing each step down systematically can help you catch potential errors before they become a problem.