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Tamiku [17]
3 years ago
11

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
Mathematics
1 answer:
Bond [772]3 years ago
8 0

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

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