- Simplify the given linear equations by clearing denominators to obtain a standard form: \(9x - 8y = 704\) and \(9x - 25y = 289\).
- Subtract the two equations to isolate \(y\), yielding \(y = \frac{415}{17}\).
- Substitute the value of \(y\) back into one of the original equations to solve for \(x\), resulting in \(x = \frac{5096}{51}\).
- The solution set is \(\boxed{\left( \frac{5096}{51}, \frac{1245}{51} \right)}\).
Explanation
- Simplifying the Equations
We are given the following system of linear equations:
\[\frac{x+8}{8} - \frac{y+16}{9} = 9\]
\[\frac{x+y}{17} = \frac{x-y}{8} - \frac{17}{8}\]
To find the solution, we first simplify each equation into the standard linear form \(Ax + By = C\).
For the first equation, multiply the entire equation by 72 (the least common multiple of 8 and 9) to eliminate the denominators:
\[72 \left( \frac{x+8}{8} - \frac{y+16}{9} \right) = 72 \times 9\]
\[9(x+8) - 8(y+16) = 648\]
\[9x + 72 - 8y - 128 = 648\]
\[9x - 8y - 56 = 648\]
\[9x - 8y = 704\]
For the second equation, multiply the entire equation by 136 (the least common multiple of 17 and 8) to eliminate the denominators:
\[136 \left( \frac{x+y}{17} = \frac{x-y}{8} - \frac{17}{8} \right)\]
\[8(x+y) = 17(x-y) - 17 \times 17\]
\[8x + 8y = 17x - 17y - 289\]
\[8y + 17y = 17x - 8x - 289\]
\[25y = 9x - 289\]
\[9x - 25y = 289\]
- Solving the System
Now that we have both equations in standard form, we can solve the system:
1) \(9x - 8y = 704\)
2) \(9x - 25y = 289\)
To solve this, we can subtract the second equation from the first to eliminate \(x\):
\[(9x - 8y) - (9x - 25y) = 704 - 289\]
\[9x - 8y - 9x + 25y = 415\]
\[17y = 415\]
\[y = \frac{415}{17}\]
Next, substitute the value of \(y\) back into the first equation to solve for \(x\):
\[9x - 8 \left( \frac{415}{17} \right) = 704\]
\[9x - \frac{3320}{17} = 704\]
\[9x = 704 + \frac{3320}{17}\]
\[9x = \frac{704 \times 17 + 3320}{17}\]
\[9x = \frac{11968 + 3320}{17}\]
\[9x = \frac{15288}{17}\]
\[x = \frac{15288}{17 \times 9} = \frac{1698.666...}{17} \text{ (or simply } \frac{1698.666...}{17} \text{ is } \frac{1698 \frac{2}{3}}{17} \text{ which is } \frac{5096}{51} \text{? Wait, let's re-calculate)}\]
Wait, \(15288 / 9 = 1698.666...\) is not quite right.
\(15288 / 9 = 1698.666...\) Let's divide 15288 by 9:
\(15288 \div 9 = 1698.666...\).
Wait, \(15288/153\): \(15288 \div 153 = 99.9215686...\)
Let's keep the fraction: \(x = \frac{15288}{153} = \frac{5096}{51}\).
So \(x = \frac{5096}{51}\) and \(y = \frac{415}{17} = \frac{1245}{51}\).
- Conclusion and Final Answer
The solution to the system of equations is the ordered pair \((x, y) = \left( \frac{5096}{51}, \frac{1245}{51} \right)\).
Therefore, the solution set is \(\left\{ \left( \frac{5096}{51}, \frac{1245}{51} \right) \right\}\).
This corresponds to Option A.
Final answer: \(\boxed{\left( \frac{5096}{51}, \frac{1245}{51} \right)}\)
Examples
This method of solving systems of linear equations is useful in various real-world applications, such as resource allocation or logistics, where multiple constraints need to be satisfied simultaneously. For example, if you are managing a warehouse with two types of products and need to balance their space requirements based on changing order volumes, setting up equations and finding the intersection point allows you to determine the exact quantities that optimize your storage capacity effectively.