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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