Question

(04.02 mc)
solve the following system of equations:
$x - 2y = 6$
$2x - 4y = 10$

$\bigcirc$ infinitely many solutions
$\bigcirc$ no solutions
$\bigcirc$ $(0, 0)$
$\bigcirc$ $(6, 10)$

Explanation:

Step1: Simplify the second equation

Divide the second equation \(2x - 4y = 10\) by 2, we get \(x - 2y = 5\).

Step2: Compare with the first equation

The first equation is \(x - 2y = 6\). Now we have two equations: \(x - 2y = 6\) and \(x - 2y = 5\). But a number (\(x - 2y\)) can't be both 6 and 5 at the same time. So, there are no solutions for this system of equations.

Answer:

B. No solutions