- Simplify the first equation by multiplying by 72: \(9x - 8y = 704\).
- Simplify the second equation by multiplying by 136: \(-9x + 25y = -289\).
- Add the two resulting equations to eliminate \(x\): \(17y = 415\), which gives \(y = \frac{415}{17}\).
- Substitute \(y\) back into one of the simplified equations to find \(x = \frac{5096}{51}\).
- The solution set is \(\boxed{\left(\frac{5096}{51}, \frac{415}{17}\right)}\).
Explanation
- Problem Analysis
We are given a system of two linear equations in two variables, \(x\) and \(y\):
1) \(\frac{x+8}{8} - \frac{y+16}{9} = 9\)
2) \(\frac{x+y}{17} = \frac{x-y}{8} - \frac{17}{8}\)
Our goal is to simplify both equations to the standard linear form \(Ax + By = C\) and solve for the pair \((x, y)\).
- Simplifying the First Equation
To simplify the first equation \(\frac{x+8}{8} - \frac{y+16}{9} = 9\), we multiply both sides by the least common multiple of 8 and 9, which is 72:
\[72 \cdot \left(\frac{x+8}{8}\right) - 72 \cdot \left(\frac{y+16}{9}\right) = 72 \cdot 9\]
\[9(x+8) - 8(y+16) = 648\]
\[9x + 72 - 8y - 128 = 648\]
\[9x - 8y - 56 = 648\]
\[9x - 8y = 704\]
This is our first equation in standard form.
- Simplifying the Second Equation
To simplify the second equation \(\frac{x+y}{17} = \frac{x-y}{8} - \frac{17}{8}\), we multiply both sides by the least common multiple of 17 and 8, which is 136:
\[136 \cdot \left(\frac{x+y}{17}\right) = 136 \cdot \left(\frac{x-y}{8}\right) - 136 \cdot \left(\frac{17}{8}\right)\]
\[8(x+y) = 17(x-y) - 17 \cdot 17\]
\[8x + 8y = 17x - 17y - 289\]
Rearranging terms to put it into the form \(Ax + By = C\):
\[8x - 17x + 8y + 17y = -289\]
\[-9x + 25y = -289\]
This is our second equation in standard form.
- Solving the System
We now have the system:
1) \(9x - 8y = 704\)
2) \(-9x + 25y = -289\)
We can solve this system using the elimination method by adding the two equations:
\[(9x - 9x) + (-8y + 25y) = 704 - 289\]
\[17y = 415\]
\[y = \frac{415}{17}\]
Next, substitute \(y = \frac{415}{17}\) into the first equation:
\[9x - 8\left(\frac{415}{17}\right) = 704\]
\[9x - \frac{3320}{17} = 704\]
\[9x = 704 + \frac{3320}{17} = \frac{11968 + 3320}{17} = \frac{15288}{17}\]
\[x = \frac{15288}{17 \cdot 9} = \frac{1698.666...}{17} \text{ (wait, 15288 / 9 = 1698.666, no that is 1698.66...)} \]
Actually, \(15288 \div 9 = 1698.66\)... let's recheck \(15288/9\).
\(1+5+2+8+8 = 24\), so it's divisible by 3. \(15288 / 3 = 5096\).
\(5096 / 3\) is not an integer. Let's recheck the sum of the equations.
Eq1: \(9x - 8y = 704\)
Eq2: \(-9x + 25y = -289\)
\(17y = 415\), yes.
Maybe there was a typo in my simplification or the problem setup? Let's recheck.
\(\frac{x+8}{8} - \frac{y+16}{9} = 9 \implies 9x + 72 - 8y - 128 = 648 \implies 9x - 8y = 704\)
\(\frac{x+y}{17} = \frac{x-y}{8} - \frac{17}{8} \implies 8x+8y = 17x-17y - 289 \implies -9x+25y = -289\)
Everything seems correct.
Wait, let's re-calculate \(9x = 704 + \frac{3320}{17} = \frac{704 \cdot 17 + 3320}{17} = \frac{11968 + 3320}{17} = \frac{15288}{17}\).
\(15288 / 153 = 99.9215\).
Is there any chance the answer should be a simple fraction?
Wait, \(15288 / 9 = 1698.66\).
Maybe I made a mistake in the arithmetic \(9 \cdot 72\). Yes, \(9 \cdot 72 = 648\).
\(648 + 56 = 704\). Correct.
Everything seems fine. Let's express the answer as an ordered pair.
- Conclusion and Final Answer
The solution is the ordered pair \((x, y) = \left(\frac{1698.66...}{17}, \frac{415}{17}\right)\) which is \(\left(\frac{5096}{51}, \frac{415}{17}\right)\). Since the calculation led to this, we present the result clearly.
Final check:
\(9(5096/51) - 8(415/17) = 3(5096/17) - 3320/17 = (15288 - 3320)/17 = 11968/17 = 704\). Correct.
\(-9(5096/51) + 25(415/17) = -3(5096/17) + 10375/17 = (-15288 + 10375)/17 = -4913/17 = -289\). Correct.
The solution set is \(\boxed{\left(\frac{5096}{51}, \frac{415}{17}\right)}\).
Examples
Systems of linear equations are fundamental in fields like economics, engineering, and logistics to find optimal values for multiple interrelated variables. For example, if you are balancing the composition of two different chemical solutions to achieve a target concentration, you can represent the concentration requirements as a system of linear equations. Solving this system allows you to determine the precise amount of each ingredient needed to reach the desired outcome, ensuring precision in laboratory experiments or industrial production.