Answer:
Option A - m=4
Step-by-step explanation:
Given : Equation [tex]y=4x+1,y=mx-5[/tex]
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 [tex]a_1x+b_1y+c_1=0, a_2x+b_2y+c_2=0[/tex] then the condition for no solutions is
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}[/tex]
Re-write equation as [tex]4x-y+1=0,mx-y-5=0[/tex]
Comparing with given equations,
[tex]a_1=4, b_1=-1,c_1=1\text{ and }a_2=m, b_2=-1,c_2=-5[/tex]
Substituting the values,
[tex]\frac{4}{m}=\frac{-1}{-1}\neq \frac{1}{-5}[/tex]
[tex]\frac{4}{m}=\frac{1}{1}\neq -\frac{1}{5}[/tex]
Taking first two equation,
[tex]\frac{4}{m}=\frac{1}{1}[/tex]
[tex]m=4[/tex]
Therefore, The value of m is 4.
So, Option A is correct.
