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.