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Find the value of $k$ for which the system of equations $kx+2y=5$, $3x+y=1$ has no solution.
Given: Equations: $kx+2y=5$, $3x+y=1$
To do: To find the value of $k$.
Solution:
If a pair of linear equations is given by
$a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$
When a system of linear equations will represent two Parallel Lines there will be no point of intersection and consequently, there will be no pair of values of $x$ and $y$ which satisfy both equations. Thus, the system has no solution and such pair of linear equation are known as inconsistent pair of linear equations
For parallel lines(inconsistent)
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\
eq\frac{c_1}{c_2}$
eq\frac{c_1}{c_2}$
The given pair of linear equations are
$5x+ky+7=0 ........( 1)$
$x+2y−3=0 .........( 2)$
On comparing with General form of a pair of linear equations in two variables x & y is:
we get
$a_1=5,\ b_1=k,\ c_1=7,\ a_2=1,\ b_2=2,\ c_2=−3$
$\Rightarrow \frac{5}{1}=\frac{k}{2}\
eq\frac{7}{-3}$
eq\frac{7}{-3}$
$\Rightarrow k=5\times2$
$\Rightarrow k=10$
Thus for $k=10$ the given system of equations have no solutions.
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