Question

Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. 2x-y=5 and 4x+ky=2

Answer 1

Answer:

The system of linear equations has infinitely many solutions

Step-by-step explanation:

Let's modified the equations and find the answer.

Using the first equation:

$2x-y=5$ we can multiply by 2 in both sides, obtaining:

$2*(2x-y)=2*5$ which can by simplified as:

$4x-2y=10$ which is equal to:

$4x=2y+10$

Considering the second equation:

$=4x+ky=2$

Taking into account that from the first equation we know that: $4x=2y+10$, we can express the second equation as:

$2y+10+ky=2$, which can be simplified as:

$(2+k)y=2-10$

$(2+k)y=-8$

$y=-8/(2+k)$

Because (-8) is being divided by (2+k), then (2+k) can't be equal to 0, so:

$2+k=0$ if $k=-2$

This means that k can be any number different than -2, and for each of these solutions, there is a different solution for y, allowing also, different solutions for x.

For example, if k=0 then

$y=-8/(2+0)$ which give us y=-4, and, because:

$4x=2y+10$ if y=-4 then $x=(-8+10)/4=0.5$

Now let's try with k=-1, then:

$y=-8/(2-1)$ which give us y=-8, and, because:

$4x=2y+10$ if y=-8 then $x=(-16+10)/4=-1.5$.

Then, the system of linear equations has infinitely many solutions