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

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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Find the value of k for which the following pair of linear equations has infinitely many solutions. 
2x + 3y = 7, (k +1) x+ (2k -1) y = 4k + 1

We have,

`2x + 3y = 7 ⇒ 2x + 3y - 7 = 0`
`(k + 1) x + (2k - 1)y = 4k + 1 ⇒ (k + 1)x (2k -1)y - (4k + 1) = 0`

For infinitely many solutions

`a_1/a_2 = b_1/b_2 = c_1/c_2`

⇒ `(2)/(k+1) = (3)/((2k -1)) = (-7)/-(4k +1)`

⇒ `(2)/(k+1) = (3)/(2k -1)`

⇒ `2(2k + 1) = 3 (k+1)`

⇒`4k - 2 = 3k + 3`

⇒`4k - 3k = 3 +2`

`k = 5`

or

⇒ `(2)/(k+1) = (-7)/-(4k +1)`

⇒ `2(4k + 1) = 7 (k+1)`

⇒ `8k + 2 = 7k + 2`

⇒`8k - 7k = 7- 2`

`k = 5`

Hence, the value of k is 5 for which given equations have infinitely many solutions.

Concept: Pair of Linear Equations in Two Variables

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