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

Accepted Solution

Answer:The system of linear equations has infinitely many solutionsStep-by-step explanation:Let's modified the equations and find the answer.Using the first equation:[tex]2x-y=5[/tex] we can multiply by 2 in both sides, obtaining:[tex]2*(2x-y)=2*5[/tex] which can by simplified as:[tex]4x-2y=10[/tex] which is equal to:[tex]4x=2y+10[/tex]Considering the second equation:[tex]=4x+ky=2[/tex]Taking into account that from the first equation we know that: [tex]4x=2y+10[/tex], we can express the second equation as:[tex]2y+10+ky=2[/tex], which can be simplified as: [tex](2+k)y=2-10[/tex][tex](2+k)y=-8[/tex][tex]y=-8/(2+k)[/tex]Because (-8) is being divided by (2+k), then (2+k) can't be equal to 0, so:[tex]2+k=0[/tex] if [tex]k=-2[/tex]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 [tex]y=-8/(2+0)[/tex] which give us y=-4, and, because:[tex]4x=2y+10[/tex] if y=-4 then [tex]x=(-8+10)/4=0.5[/tex]Now let's try with k=-1, then:[tex]y=-8/(2-1)[/tex] which give us y=-8, and, because:[tex]4x=2y+10[/tex] if y=-8 then [tex]x=(-16+10)/4=-1.5[/tex].Then, the system of linear equations has infinitely many solutions