Question

7x - 56y = 14
one of the two equations in a system of linear equations is given. the system has no solution. which equation could be the second equation in this system?
a (\frac{1}{2}x - 4y = 1)
b (x - 8y = 2)
c (\frac{1}{2}x - 4y = 0)
d (x - 14y = 0)

Explanation:

Step1: Simplify given equation

Divide $7x-56y=14$ by 7:
$\frac{7x}{7}-\frac{56y}{7}=\frac{14}{7}$
$x-8y=2$

Step2: Identify no-solution condition

A system $a_1x+b_1y=c_1$, $a_2x+b_2y=c_2$ has no solution if $\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}$.

Step3: Test each option

Option A: $\frac{1}{2}x-4y=1$

Multiply by 2: $x-8y=2$. Here $\frac{1}{1}=\frac{-8}{-8}=\frac{2}{2}$, so infinite solutions.

Option B: $x-8y=2$

Matches simplified given equation, so infinite solutions.

Option C: $\frac{1}{2}x-4y=0$

Multiply by 2: $x-8y=0$. Here $\frac{1}{1}=\frac{-8}{-8}
eq\frac{2}{0}$ (since $\frac{2}{0}$ is undefined, $\frac{2}{0}
eq1$), so no solution.

Option D: $x-14y=0$

$\frac{1}{1}
eq\frac{-8}{-14}$, so system has one solution.

Answer:

C. $\frac{1}{2}x - 4y = 0$