Mahabir Singh Solutions for Chapter: Linear Equations in Two Variables, Exercise 3: ACHIEVERS SECTION (HOTS)

A linear equation in two variables has $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{ }\mathbf{ }}$solutions(s).

The graph of $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{Q}\mathbf{ }\mathbf{ }\mathbf{ }}$line has an equation of the form $x=k$.

A line parallel to the $x-$axis cuts the $y-$axis at $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{R}\mathbf{ }\mathbf{ }\mathbf{ }}$ point(s).

Distance between the graph of equation $y=2$ and $y=-4$ is $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{S}\mathbf{ }\mathbf{ }\mathbf{ }}$units.

Then $\mathbf{P}\mathbf{,}\mathbf{ }\mathbf{Q}\mathbf{,}\mathbf{ }\mathbf{R}$ and $\mathbf{S}$ are

The equation representing the given graph is ________

Match the linear equations in Column-I with their solutions in Column-II
         Column-I            Column-II
($\mathrm{P}$) $4x+3y=24$ $\left(i\right)$ $\left(2, -3\right)$
($\mathrm{Q}$) $\frac{x}{2}-\frac{y}{3}=2$     ($ii$) $(\mathrm{2,} 3)$
($\mathrm{R}$) $3x+5y=15$ ($iii$) $\left(\mathrm{3,} 4\right)$
($\mathrm{S}$) $\frac{x-2}{3}=y-3$    ($iv$) $\frac{9}{3}, \frac{6}{5}$