Solve the pair \(\dfrac{x}{10} + \dfrac{y}{5} - 1 = 0\) and \(\dfrac{x}{8} + \dfrac{y}{6} = 15\). Hence, if \(y = \lambda x + 5\), find \(\lambda\).
Solution: \(x = 340\), \(y = -165\); hence \(\lambda = -\dfrac{1}{2}\).
Step 1: Write the equations clearly.
Equation (1): \(\dfrac{x}{10} + \dfrac{y}{5} - 1 = 0\)
Equation (2): \(\dfrac{x}{8} + \dfrac{y}{6} = 15\)
Step 2: Remove fractions in Equation (1).
Multiply the whole equation by 10:
\(x + 2y - 10 = 0\)
So, \(x = 10 - 2y\).
Step 3: Remove fractions in Equation (2).
The denominators are 8 and 6. The LCM is 24. Multiply the whole equation by 24:
\(3x + 4y = 360\)
Step 4: Substitute the value of \(x\) from Step 2.
Substitute \(x = 10 - 2y\) into \(3x + 4y = 360\):
\(3(10 - 2y) + 4y = 360\)
Step 5: Simplify.
\(30 - 6y + 4y = 360\)
\(30 - 2y = 360\)
Step 6: Solve for \(y\).
\(-2y = 360 - 30\)
\(-2y = 330\)
\(y = -165\)
Step 7: Solve for \(x\).
From Step 2, \(x = 10 - 2y\).
Put \(y = -165\):
\(x = 10 - 2(-165)\)
\(x = 10 + 330 = 340\)
Step 8: Use the given relation \(y = \lambda x + 5\).
Substitute \(x = 340\) and \(y = -165\):
\(-165 = 340\lambda + 5\)
Step 9: Solve for \(\lambda\).
\(340\lambda = -165 - 5\)
\(340\lambda = -170\)
\(\lambda = -\dfrac{170}{340}\)
\(\lambda = -\dfrac{1}{2}\)