How do you determine whether a linear system has one solution, many solutions, or no solution when given y=-x+6 and y=3/4x-1?

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Answer 1 · 2018-03-10T06:03:33

Given, #y=-x+6#

or, #x+y=6.....................1#

and, #y=3/4 x-1#

or, #3x-4y=4....................................2#

Now,we know for two equations , #ax +by=c# and #px +qy=k#

if, #a/p ne b/q# then they will have unique solution.

if, #a/p=b/q ne c/k# then they will have no solution.

if, #a/p=b/q=c/k# then they will have multiple solutions.

here in this case we get, #1/3 ne -(1/4)# i.e #a/p ne b/q#,so it will have one solution only.