Answer:
Option A - m=4
Step-by-step explanation:
Given : Equation \(y=4x+1,y=mx-5\)
To find : For which of the following values of m would the system of equations have no solution ?
Solution :
When the system of equation is in form \(a_1x+b_1y+c_1=0, a_2x+b_2y+c_2=0\) then the condition for no solutions is
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq \frac{c_1}{c_2}\)
Re-write equation as \(4x-y+1=0,mx-y-5=0\)
Comparing with given equations,
\(a_1=4, b_1=-1,c_1=1\text{ and }a_2=m, b_2=-1,c_2=-5\)
Substituting the values,
\(\frac{4}{m}=\frac{-1}{-1}
eq \frac{1}{-5}\)
\(\frac{4}{m}=\frac{1}{1}
eq -\frac{1}{5}\)
Taking first two equation,
\(\frac{4}{m}=\frac{1}{1}\)
\(m=4\)
Therefore, The value of m is 4.
So, Option A is correct.
