For what value of $k$ will the following pair of linear equations have infinitely many solutions? $$2x + 3y = 7$$ $$(k -

Question

For what value of $k$ will the following pair of linear equations have infinitely many solutions? $$2x + 3y = 7$$ $$(k - 1)x + (k + 2)y = 3k$$

Accepted Answer

Given equations are: $2x + 3y - 7 = 0$ $(k - 1)x + (k + 2)y - 3k = 0$ 
Here, $a1 = 2$, $b1 = 3$, $c1 = -7$ $a2 = k - 1$, $b2 = k + 2$, $c2 = -3k$ 
For infinitely many solutions, the required condition is: $$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$$ 
Substituting the values: $$\frac{2}{k - 1} = \frac{3}{k + 2} = \frac{-7}{-3k}$$ 
Taking the first two ratios: $$\frac{2}{k - 1} = \frac{3}{k + 2}$$ $$2(k + 2) = 3(k - 1)$$ $$2k + 4 = 3k - 3$$ $$3k - 2k = 4 + 3$$ $$k = 7$$ 
Now, verifying with the second and third ratios for $k = 7$: $$\frac{3}{7 + 2} = \frac{3}{9} = \frac{1}{3}$$ $$\frac{-7}{-3(7)} = \frac{7}{21} = \frac{1}{3}$$
Since both ratios are equal to $\frac{1}{3}$, $k = 7$ is correct. Final Answer: $k = 7$.

Pair of Linear Equations in Two Variables

❓ Sibling Questions in this Chapter