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. Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Find the value of k for which the following pair of linear equations has infinitely many solutions. We have, `2x + 3y = 7 ⇒ 2x + 3y - 7 = 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 Is there an error in this question or solution? |