Question

1. determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.

$-4x + y = 1$
$-16x + 5y = 8$

Explanation:

Step1: Eliminate x variable

Multiply first equation by 4:
$$4*(4x + y) = 4*1 \implies 16x + 4y = 4$$

Step2: Add to second equation

Add to $-16x + 5y = 8$:
$$(16x + 4y) + (-16x + 5y) = 4 + 8$$
$$9y = 12$$

Step3: Solve for y

Isolate y:
$$y = \frac{12}{9} = \frac{4}{3}$$

Step4: Substitute y to find x

Plug $y=\frac{4}{3}$ into $4x + y = 1$:
$$4x + \frac{4}{3} = 1$$
$$4x = 1 - \frac{4}{3} = -\frac{1}{3}$$
$$x = -\frac{1}{12}$$

Answer:

The system has exactly one solution: $x=-\frac{1}{12}$, $y=\frac{4}{3}$