Question

for what value of k, the pair of equation 2kx-5y=9, 3x+7y=11 has unique solutions​

Answer 1

GiveN :-

• Two linear equations are given to us .
• 2kx - 5y = 9.
• 3x + 7y = 11 .

To FinD :-

• The value of k for which the system of equations have unique solution .

SolutioN :-

Two linear equations are given to us and we need to find the value of k for which the system of equations have unique solution . The standard form of system of Linear Equations in two variables is ,

$\red{\frak{Standard\ Form}}\begin{cases} \sf a_1x+b_1y+c_1=0 \\ \sf a_2x + b_2y + c_2 = 0 \end{cases}$

With respect to Standard Form we see that ,

$\boxed{\begin{array}{c|c|c} \sf \odot\:\: a_1 = 2k &\sf \:\: \odot b_1=5 &\sf \odot \:\: c_1= -9 \\\sf \odot\:\: a_2= 3 &\sf \:\: \odot b_2=7 &\sf \odot \:\: c_2=-11 \end{array}}$

Now we know that the condition for unique solution is ,

$\sf:\implies\pink{\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2} }\\\\\sf:\implies \dfrac{2k}{3} \neq \dfrac{-5}{7}\\\\\sf:\implies k \neq \dfrac{-5\times 3}{7\times 2 } \\\\\sf:\implies k \neq \dfrac{-15}{14} \\\\\sf:\implies \underset{\blue{\sf Required\ Answer}}{\underbrace{\boxed{\pink{\frak{k \in \mathbb{R}-\bigg\{\dfrac{-15}{14}\bigg\}}}}}}$

In simple words k can have any other real Value b than -15/14 .