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Find the value of k for which the following system of equations has a unique solution:$2x+3y−5=0$, $kx−6y−8=0$.
Given: Pair of equations: $2x+3y−5=0$,$kx−6y−8=0$.
To do: To find the value of $k$ for which the following system of equations has a unique solution.
Solution:
Given equations are:
$2x+3y−5=0$, $kx−6y−8=0$.
On comparing with General form of a pair of linear equations in two variables $x$ & $y$ is:
$a_1x + b_1y + c_1 = 0$
and $a_2x + b_2y + c_2= 0$
$a_1=2,\ b_1=3,\ c_1=-5$
$a_2= k,\ b_2=-6,\ c_2=-8$
For unique solution, we must have,
$\frac{a_1}{a_2} \
eq \frac{b_1}{b_2}$
eq \frac{b_1}{b_2}$
$\Rightarrow \frac{2}{k} \
eq\frac{3}{-6}$
eq\frac{3}{-6}$
$\Rightarrow 3k \
eq-12$
eq-12$
$\Rightarrow k \
eq-\frac{12}{3}$
eq-\frac{12}{3}$
$\Rightarrow k \
eq-4$
eq-4$
Thus, for $k\
eq-4$, the system of equations has a unique solution.
eq-4$, the system of equations has a unique solution.
Updated on: 10-Oct-2022
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