In this problem, we have a public library with a reading area and a coffee area. The reading area is 5/8 of the coffee area. If the reading area increases by 6 square meters, it will be equal to the coffee area. We need to find the original areas of both sections. Let's start by setting variables: R for the reading area and C for the coffee area.
Now, let's set up the equations based on the problem conditions. First, we know that the reading area is 5/8 of the coffee area, so we can write: R equals 5/8 times C. Second, we know that if the reading area increases by 6 square meters, it will be equal to the coffee area. This gives us our second equation: R plus 6 equals C. With these two equations, we can solve for the original areas.
Now let's solve the system of equations. We have two equations: R equals 5/8 times C, and R plus 6 equals C. We can substitute the first equation into the second. This gives us: 5/8 times C plus 6 equals C. Rearranging the terms, we get: 6 equals C minus 5/8 times C, which simplifies to 6 equals 3/8 times C. Solving for C, we multiply both sides by 8/3: C equals 6 times 8/3, which equals 48/3, or 16. Now we can find R by substituting C equals 16 into our first equation: R equals 5/8 times 16, which equals 10. Therefore, the original area of the reading section was 10 square meters, and the coffee section was 16 square meters.
Let's verify our solution. First, we check if the reading area is 5/8 of the coffee area: 10 equals 5/8 times 16, which is indeed 10. Second, we verify that if the reading area increases by 6 square meters, it equals the coffee area: 10 plus 6 equals 16, which is correct. Therefore, our solution is verified. The original area of the reading section was 10 square meters, and the coffee section was 16 square meters. This problem demonstrates how to solve a system of linear equations with fractions, which is a common technique in algebra.
To summarize what we've learned: We solved a problem about finding the original areas of a library's reading and coffee sections. We knew that the reading area was 5/8 of the coffee area, and that if the reading area increased by 6 square meters, it would equal the coffee area. We approached this by setting up a system of two linear equations. Our solution process involved defining variables, writing equations based on the given conditions, substituting one equation into the other, and solving for the unknowns. We found that the original reading area was 10 square meters and the coffee area was 16 square meters. This problem demonstrates how algebraic methods can be applied to solve real-world scenarios involving proportions and areas.