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: Isolate $y$ from first equation

$y = 4x + 1$

Step2: Substitute $y$ into second equation

$-16x + 5(4x + 1) = 8$

Step3: Simplify and solve for $x$

$-16x + 20x + 5 = 8$
$4x = 3$
$x = \frac{3}{4}$

Step4: Substitute $x$ back to find $y$

$y = 4\times\frac{3}{4} + 1 = 4$

Step5: Verify solution consistency

Substitute $x=\frac{3}{4}, y=4$ into both original equations:

1. $-4\times\frac{3}{4} + 4 = -3 + 4 = 1$ (matches first equation)
2. $-16\times\frac{3}{4} + 5\times4 = -12 + 20 = 8$ (matches second equation)

Answer:

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