For what value of k, the following pair of linear equation has infinitely many solutions?
Question:

For what value of k, the following pair of linear equation has infinitely many solutions?

$10 x+5 y-(k-5)=0$

$20 x+10 y-k=0$

Solution:

The given equations are

$10 x+5 y-(k-5)=0$

$20 x+10 y-k=0$

$\frac{a_{1}}{a_{2}}=\frac{10}{20}, \frac{b_{1}}{b_{2}}=\frac{5}{10}, \frac{c_{1}}{c_{2}}=\frac{k-5}{k}$

For the equations to have infinite number of solutions

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Let us take

$\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\frac{5}{10}=\frac{k-5}{k}$

$5 \times k=10 \times(k-5)$

$50=10 k-5 k$

$50=5 k$

$\frac{50}{5}=k$

$10=k$

Hence, the value of $k=10$ when the pair of linear equations has infinitely many solutions.

 

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