NCERT · Class 10 · Mathematics · Pair of Linear Equations in Two VariablesFor what value of $k$ will the equations $2x + 3y = 7$ and $(k-1)x + (k+2)y = 3k$ have infinitely many solutions?

The given linear equations are $2x + 3y - 7 = 0$ and $(k-1)x + (k+2)y - 3k = 0$.\nHere, $a_1 = 2, b_1 = 3, c_1 = -7$ and $a_2 = k-1, b_2 = k+2, c_2 = -3k$. \nFor infinitely many solutions, the ratio condition is: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{2}{k-1} = \frac{3}{k+2} = \frac{-7}{-3k}$$ \nTaking the first two fractions: $$\frac{2}{k-1} = \frac{3}{k+2}$$ $$2(k+2) = 3(k-1) \implies 2k + 4 = 3k - 3$$ $$3k - 2k = 4 + 3 \implies k = 7$$ \nVerifying with the third fraction: $\frac{3}{7+2} = \frac{3}{9} = \frac{1}{3}$ and $\frac{7}{3(7)} = \frac{7}{21} = \frac{1}{3}$.\nHence, the required value of $k$ is $7$.