Question

Consider the system of linear equations given below.
\[ ax + y = b \] \[ 16x + ay = 24 \] Suppose the values of \(a\) and \(b\) are chosen such that the system of linear equations produce multiple solutions. Then the product of \(a\) and \(b\) is _____. (answer in integer)

Hint:

Two linear equations have infinitely many solutions when they are scalar multiples of each other; equate the ratios of x-coefficients, y-coefficients, and constants.

Answer 1

Correct Answer: 24

Step 1: Recall the condition for infinitely many solutions.
For a system of two linear equations in two variables,
\[ a_1x + b_1y = c_1 \]
\[ a_2x + b_2y = c_2 \]
the system has infinitely many solutions, that is "multiple solutions", exactly when the two equations represent the same line, which happens when all three ratios of corresponding coefficients are equal:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Step 2: Write down the coefficients from the given system.
The equations are
\[ ax + y = b \]
\[ 16x + ay = 24 \]
Matching terms, \(a_1 = a\), \(b_1 = 1\), \(c_1 = b\), and \(a_2 = 16\), \(b_2 = a\), \(c_2 = 24\).

Step 3: Set up the ratio condition.
\[ \frac{a}{16} = \frac{1}{a} = \frac{b}{24} \]

Step 4: Solve the first equality for \(a\).
\[ \frac{a}{16} = \frac{1}{a} \]
Cross multiplying,
\[ a^2 = 16 \]
\[ a = 4 \quad \text{or} \quad a = -4 \]

Step 5: Solve for \(b\) in the case \(a = 4\).
Using \(\dfrac{1}{a} = \dfrac{b}{24}\),
\[ \frac{1}{4} = \frac{b}{24} \]
\[ b = \frac{24}{4} = 6 \]
So in this case \(a = 4\), \(b = 6\), and the product is
\[ ab = 4 \times 6 = 24 \]

Step 6: Solve for \(b\) in the case \(a = -4\).
Using \(\dfrac{1}{a} = \dfrac{b}{24}\),
\[ \frac{1}{-4} = \frac{b}{24} \]
\[ b = \frac{24}{-4} = -6 \]
So in this case \(a = -4\), \(b = -6\), and the product is
\[ ab = (-4) \times (-6) = 24 \]

Step 7: Final conclusion.
Both valid cases give the same product.
\[ \boxed{ab = 24} \]