Q: For what value of $k$ will the equations $3x - y - 5 = 0$ and $6x - 2y - k = 0$ have no solution?

For no solution: $\frac{a1}{a2} = \frac{b1}{b2} \neq \frac{c1}{c2}$.\nHere $a1=3, b1=-1, c1=-5$ and $a2=6, b2=-2, c2=-k$. $\frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{-k} \implies \frac{1}{2} \neq \frac{5}{k} \implies k \neq 10$.\nWhen $k \neq 10$, the equations have no solution.

Question 1

If a pair of linear equations $a1x + b1y + c1 = 0$ and $a2x + b2y + c2 = 0$ has a unique solution, then which of the following conditions is correct?

Accepted Answer

For a pair of linear equations to have a unique solution, the ratio of the coefficients of $x$ and $y$ must not be equal, i.e., $\frac{a1}{a2} 
eq \frac{b1}{b2}$.

Question 2

If the pair of linear equations $a1x + b1y + c1 = 0$ and $a2x + b2y + c2 = 0$ represents coincident lines, then:

Accepted Answer

Coincident lines have infinitely many solutions, which occurs when all three ratios are equal: $\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$.

Question 3

The pair of equations $x + y = 14$ and $x - y = 4$ has the solution:

Accepted Answer

Adding both equations: $(x+y) + (x-y) = 14 + 4 \implies 2x = 18 \implies x = 9$.
Substituting $x = 9$ into $x + y = 14 \implies 9 + y = 14 \implies y = 5$.
So, $x = 9, y = 5$.

Question 4

Graphical representation of the equations $x = a$ and $y = b$ represents two lines which are:

Accepted Answer

The line $x = a$ is vertical (parallel to the y-axis) and $y = b$ is horizontal (parallel to the x-axis). These two lines intersect each other at the point $(a, b)$.

Question 5

For what value of $k$ will the equations $3x - y - 5 = 0$ and $6x - 2y - k = 0$ have no solution?

Accepted Answer

For no solution: $\frac{a1}{a2} = \frac{b1}{b2} 
eq \frac{c1}{c2}$.
Here $a1=3, b1=-1, c1=-5$ and $a2=6, b2=-2, c2=-k$. $\frac{3}{6} = \frac{-1}{-2} 
eq \frac{-5}{-k} \implies \frac{1}{2} 
eq \frac{5}{k} \implies k 
eq 10$.
When $k 
eq 10$, the equations have no solution.

Pair of Linear Equations in Two Variables