Find the value of k for which the system of equations x+2y=3 and 5x+ky+7=0 has no solution.
Answers
The value of k is k=10k=10 and k\neq-\frac{14}{3}k≠−
3
14
Step-by-step explanation:
Given : Equation x+2y-3=0,\ 5x+ky+7=0x+2y−3=0, 5x+ky+7=0
To find : The value of k for which the system of equation has no solution. ?
Solution :
When the system of equation is in form ax+by+c=0, dx+ey+f=0ax+by+c=0,dx+ey+f=0 then the condition for no solutions is
\frac{a}{d}=\frac{b}{e}\neq \frac{c}{f}
d
a
=
e
b
≠
f
c
Comparing with equations,
a=1,b=2,c=-3,d=5,e=k,f=7
Substituting the values,
\frac{1}{5}=\frac{2}{k}\neq \frac{-3}{7}
5
1
=
k
2
≠
7
−3
Taking 1, \frac{1}{5}=\frac{2}{k}
5
1
=
k
2
k=2\times 5k=2×5
k=10k=10
Taking 2, \frac{2}{k}\neq \frac{-3}{7}
k
2
≠
7
−3
2\times 7\neq-3\times k2×7≠−3×k
14\neq-3k14≠−3k
k\neq-\frac{14}{3}k≠−
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14
Therefore, The value of k is k=10k=10 and k\neq-\frac{14}{3}k≠−
3
14