You are given the system of linear equations
2x+ky=5, x+3y=7,
where k
k is a constant.
The system above has no solution when k=
2x+ky=5⇒2x+ky−5=0...(1)x+3y=7⇒x+3y−7=0...(2)Comparing (1) with a1x+b1y+c1=0 and (2) with a2x+b2y+c2=0a1=2,b1=k,c1=−5,a2=1,b2=3,c2=−72x+ky=5\Rightarrow 2x+ky-5=0 ...(1) \\ x+3y=7\Rightarrow x+3y-7=0 ...(2) \\Comparing\ (1)\ with \ a_1x+b_1y+c_1=0\ and\ (2)\ with \ a_2x+b_2y+c_2=0 \\a_1=2,b_1=k,c_1=-5,a_2=1,b_2=3,c_2=-72x+ky=5⇒2x+ky−5=0...(1)x+3y=7⇒x+3y−7=0...(2)Comparing (1) with a1x+b1y+c1=0 and (2) with a2x+b2y+c2=0a1=2,b1=k,c1=−5,a2=1,b2=3,c2=−7
For no solution,
a1a2=b1b2≠c1c221=k3≠−5−7⇒k=6,k≠157\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2} \\\frac{2}{1}=\frac{k}{3}\neq\frac{-5}{-7} \\\Rightarrow k=6, k\neq \frac {15}{7}a2a1=b2b1=c2c112=3k=−7−5⇒k=6,k=715
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