Question

For what value(s) of c, the pair of equations \(x - 2y = 8; 5x - 10y = c\) have infinite solution ?

• A. 16
• B. All real values
• C. 40
• D. All non-negative real values of c

Hint:

Notice that the second equation has its left-hand side coefficients (\(5x - 10y\)) exactly \(5\) times the first equation's coefficients (\(x - 2y\)).
For the equations to represent the same line (infinite solutions), the constant term must also be multiplied by the same factor of \(5\).
Thus, simply multiply \(8 \times 5 = 40\) to find the answer in seconds.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question is from the topic of Linear Equations in Two Variables.
We are given a system of two linear equations and we need to find the value of the constant \(c\) for which the system has infinitely many solutions.

Step 2: Key Formula or Approach:
For a pair of linear equations:
\[ a_1 x + b_1 y + c_1 = 0 \]
\[ a_2 x + b_2 y + c_2 = 0 \]
The condition for having infinitely many solutions (coincident lines) is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Step 3: Detailed Explanation:
First, write the given equations in standard form \(ax + by + c = 0\):
Equation 1: \(x - 2y - 8 = 0\)
Equation 2: \(5x - 10y - c = 0\)
Identify the coefficients from both equations:
\[ a_1 = 1, \quad b_1 = -2, \quad c_1 = -8 \]
\[ a_2 = 5, \quad b_2 = -10, \quad c_2 = -c \]
Now, apply the condition for infinitely many solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
Substitute the values:
\[ \frac{1}{5} = \frac{-2}{-10} = \frac{-8}{-c} \] Simplify the ratios:
\[ \frac{1}{5} = \frac{1}{5} = \frac{8}{c} \] To find \(c\), equate the ratios:
\[ \frac{1}{5} = \frac{8}{c} \] Cross-multiply to solve for \(c\):
\[ c = 5 \times 8 \]
\[ c = 40 \]

Step 4: Final Answer:
The value of \(c\) for which the pair of equations has infinite solutions is \(40\).