Class 10: Linear Equations

$\displaystyle \textbf{Question 1: }\text{Solve the following system of equations graphically:}$
$\displaystyle 3x+y+4=0,\qquad 3x-y+2=0\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x+y+4=0\qquad\ldots\text{(i)}$
$\displaystyle 3x-y+2=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }3x+y+4=0:$
$\displaystyle y=-3x-4$
$\displaystyle \text{When }x=-2,\ y=2;\qquad\text{when }x=0,\ y=-4.$
$\displaystyle \begin{array}{c|cc}x&-2&0\\ \hline y&2&-4\end{array}$
$\displaystyle \text{Plot }(-2,2)\text{ and }(0,-4)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3x-y+2=0:$
$\displaystyle y=3x+2$
$\displaystyle \text{When }x=-1,\ y=-1;\qquad\text{when }x=0,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&-1&0\\ \hline y&-1&2\end{array}$
$\displaystyle \text{Plot }(-1,-1)\text{ and }(0,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(-1,-1).$
$\displaystyle \therefore x=-1,\qquad y=-1.$
$\displaystyle \therefore \text{The solution of the given system is }(-1,-1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Solve the following system of equations graphically:}$
$\displaystyle x=0,\qquad y=-7\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle x=0\qquad\ldots\text{(i)}$
$\displaystyle y=-7\qquad\ldots\text{(ii)}$
$\displaystyle x=0\text{ represents the }y\text{-axis.}$
$\displaystyle y=-7\text{ represents a line parallel to the }x\text{-axis, }7\text{ units below it.}$
$\displaystyle \text{Draw the lines }x=0\text{ and }y=-7\text{ on the same graph paper.}$
$\displaystyle \text{The two lines intersect at }(0,-7).$
$\displaystyle \therefore x=0,\qquad y=-7.$
$\displaystyle \therefore \text{The solution of the given system is }(0,-7).$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Show graphically that the following system of equations has infinitely}$
$\displaystyle \text{many solutions:}\qquad 2x+3y=6,\qquad 4x+6y=12\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x+3y=6\qquad\ldots\text{(i)}$
$\displaystyle 4x+6y=12\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x+3y=6:$
$\displaystyle y=\frac{6-2x}{3}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=3,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&2&0\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(3,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }4x+6y=12:$
$\displaystyle y=\frac{12-4x}{6}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=3,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&2&0\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(3,0)\text{ and draw the line through them.}$
$\displaystyle \text{The graphs of the two equations coincide.}$
$\displaystyle \text{Therefore, every point on the line is a common solution of both equations.}$
$\displaystyle \therefore \text{The given system of equations has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Show graphically that the following system of equations has infinitely many}$
$\displaystyle \text{solutions:}\qquad x-2y+11=0,\qquad 3x-6y+33=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle x-2y+11=0\qquad\ldots\text{(i)}$
$\displaystyle 3x-6y+33=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x-2y+11=0:$
$\displaystyle y=\frac{x+11}{2}$
$\displaystyle \text{When }x=-11,\ y=0;\qquad\text{when }x=-1,\ y=5.$
$\displaystyle \begin{array}{c|cc}x&-11&-1\\ \hline y&0&5\end{array}$
$\displaystyle \text{Plot }(-11,0)\text{ and }(-1,5)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3x-6y+33=0:$
$\displaystyle y=\frac{x+11}{2}$
$\displaystyle \text{When }x=-11,\ y=0;\qquad\text{when }x=-1,\ y=5.$
$\displaystyle \begin{array}{c|cc}x&-11&-1\\ \hline y&0&5\end{array}$
$\displaystyle \text{Plot }(-11,0)\text{ and }(-1,5)\text{ and draw the line through them.}$
$\displaystyle \text{The graphs of the two equations coincide.}$
$\displaystyle \text{Therefore, every point on the line is a common solution of both equations.}$
$\displaystyle \therefore \text{The given system of equations has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Show graphically that the following system of equations is inconsistent}$
$\displaystyle \text{(i.e. has no solution):}\qquad 3x-5y=20,\qquad 6x-10y=-40$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x-5y=20\qquad\ldots\text{(i)}$
$\displaystyle 6x-10y=-40\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }3x-5y=20:$
$\displaystyle y=\frac{3x-20}{5}$
$\displaystyle \text{When }x=0,\ y=-4;\qquad\text{when }x=5,\ y=-1.$
$\displaystyle \begin{array}{c|cc}x&0&5\\ \hline y&-4&-1\end{array}$
$\displaystyle \text{Plot }(0,-4)\text{ and }(5,-1)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }6x-10y=-40:$
$\displaystyle y=\frac{3x+20}{5}$
$\displaystyle \text{When }x=0,\ y=4;\qquad\text{when }x=5,\ y=7.$
$\displaystyle \begin{array}{c|cc}x&0&5\\ \hline y&4&7\end{array}$
$\displaystyle \text{Plot }(0,4)\text{ and }(5,7)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines are parallel and have no common point.}$
$\displaystyle \therefore \text{The given system is inconsistent and has no solution.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Show graphically that the following system of equations is inconsistent}$
$\displaystyle \text{(i.e. has no solution):}\qquad 3x-4y-1=0,\qquad 2x-\frac{8}{3}y+5=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x-4y-1=0\qquad\ldots\text{(i)}$
$\displaystyle 2x-\frac{8}{3}y+5=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }3x-4y-1=0:$
$\displaystyle y=\frac{3x-1}{4}$
$\displaystyle \text{When }x=-1,\ y=-1;\qquad\text{when }x=3,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&-1&3\\ \hline y&-1&2\end{array}$
$\displaystyle \text{Plot }(-1,-1)\text{ and }(3,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x-\frac{8}{3}y+5=0:$
$\displaystyle y=\frac{3(2x+5)}{8}$
$\displaystyle \text{When }x=-\frac{5}{2},\ y=0;\qquad\text{when }x=0,\ y=\frac{15}{8}.$
$\displaystyle \begin{array}{c|cc}x&-\frac{5}{2}&0\\ \hline y&0&\frac{15}{8}\end{array}$
$\displaystyle \text{Plot }\left(-\frac{5}{2},0\right)\text{ and }\left(0,\frac{15}{8}\right)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines are parallel and have no common point.}$
$\displaystyle \therefore \text{The given system is inconsistent and has no solution.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Determine graphically the vertices of the triangle, the equations}$
$\displaystyle \text{of whose sides are given below:}\hfill\text{[CBSE 2000]}$
$\displaystyle \text{(i) }2y-x=8,\qquad 5y-x=14\qquad\text{and}\qquad y-2x=1$
$\displaystyle \text{(ii) }y=x,\qquad y=0\qquad\text{and}\qquad3x+3y=10$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The given equations are}$
$\displaystyle 2y-x=8\qquad\ldots\text{(i)}$
$\displaystyle 5y-x=14\qquad\ldots\text{(ii)}$
$\displaystyle y-2x=1\qquad\ldots\text{(iii)}$
$\displaystyle \text{For }2y-x=8:$
$\displaystyle x=2y-8$
$\displaystyle \text{When }y=2,\ x=-4;\qquad\text{when }y=3,\ x=-2.$
$\displaystyle \begin{array}{c|cc}x&-4&-2\\ \hline y&2&3\end{array}$
$\displaystyle \text{Plot }(-4,2)\text{ and }(-2,3)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }5y-x=14:$
$\displaystyle x=5y-14$
$\displaystyle \text{When }y=3,\ x=1;\qquad\text{when }y=4,\ x=6.$
$\displaystyle \begin{array}{c|cc}x&1&6\\ \hline y&3&4\end{array}$
$\displaystyle \text{Plot }(1,3)\text{ and }(6,4)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }y-2x=1:$
$\displaystyle y=2x+1$
$\displaystyle \text{When }x=-1,\ y=-1;\qquad\text{when }x=0,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&-1&0\\ \hline y&-1&1\end{array}$
$\displaystyle \text{Plot }(-1,-1)\text{ and }(0,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{From the graph, the three lines intersect pairwise at }(-4,2),(1,3)\text{ and }(2,5).$
$\displaystyle \therefore \text{The vertices of the triangle are }(-4,2),(1,3)\text{ and }(2,5).$

$\displaystyle \text{(ii) The given equations are}$
$\displaystyle y=x\qquad\ldots\text{(i)}$
$\displaystyle y=0\qquad\ldots\text{(ii)}$
$\displaystyle 3x+3y=10\qquad\ldots\text{(iii)}$
$\displaystyle \text{For }y=x:$
$\displaystyle \text{When }x=0,\ y=0;\qquad\text{when }x=2,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&0&2\end{array}$
$\displaystyle \text{Plot }(0,0)\text{ and }(2,2)\text{ and draw a straight line through them.}$
$\displaystyle y=0\text{ represents the }x\text{-axis.}$
$\displaystyle \text{For }3x+3y=10:$
$\displaystyle x+y=\frac{10}{3}$
$\displaystyle \text{When }x=0,\ y=\frac{10}{3};\qquad\text{when }y=0,\ x=\frac{10}{3}.$
$\displaystyle \begin{array}{c|cc}x&0&\frac{10}{3}\\ \hline y&\frac{10}{3}&0\end{array}$
$\displaystyle \text{Plot }\left(0,\frac{10}{3}\right)\text{ and }\left(\frac{10}{3},0\right)\text{ and draw a straight line through them.}$
$\displaystyle \text{From the graph, the three lines intersect pairwise at }(0,0),\left(\frac{10}{3},0\right)\text{ and}$
$\displaystyle \left(\frac{5}{3},\frac{5}{3}\right).$
$\displaystyle \therefore \text{The vertices of the triangle are }(0,0),\left(\frac{10}{3},0\right)\text{ and }\left(\frac{5}{3},\frac{5}{3}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Determine graphically whether the following systems of linear equations are}$
$\displaystyle \text{consistent or inconsistent:}\hfill\text{[CBSE 2023]}$
$\displaystyle \text{(i) }x-2y=2,\qquad4x-2y=5$
$\displaystyle \text{(ii) }x=0,\qquad y=-7$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The given system of equations is}$
$\displaystyle x-2y=2\qquad\ldots\text{(i)}$
$\displaystyle 4x-2y=5\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x-2y=2:$
$\displaystyle y=\frac{x-2}{2}$
$\displaystyle \text{When }x=0,\ y=-1;\qquad\text{when }x=2,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-1&0\end{array}$
$\displaystyle \text{Plot }(0,-1)\text{ and }(2,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }4x-2y=5:$
$\displaystyle y=\frac{4x-5}{2}$
$\displaystyle \text{When }x=0,\ y=-\frac{5}{2};\qquad\text{when }x=2,\ y=\frac{3}{2}.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-\frac{5}{2}&\frac{3}{2}\end{array}$
$\displaystyle \text{Plot }\left(0,-\frac{5}{2}\right)\text{ and }\left(2,\frac{3}{2}\right)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }\left(1,-\frac{1}{2}\right).$
$\displaystyle \therefore \text{The system has a unique solution and is consistent.}$

$\displaystyle \text{(ii) The given system of equations is}$
$\displaystyle x=0\qquad\ldots\text{(i)}$
$\displaystyle y=-7\qquad\ldots\text{(ii)}$
$\displaystyle x=0\text{ represents the }y\text{-axis.}$
$\displaystyle y=-7\text{ represents a line parallel to the }x\text{-axis, }7\text{ units below it.}$
$\displaystyle \text{Draw the lines }x=0\text{ and }y=-7\text{ on the same graph paper.}$
$\displaystyle \text{The two lines intersect at }(0,-7).$
$\displaystyle \therefore \text{The system has a unique solution and is consistent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Determine, by drawing graphs, whether the following systems of linear equations}$
$\displaystyle \text{have a unique solution or not:}$
$\displaystyle \text{(i) }2x-3y=6,\qquad x+y=1$
$\displaystyle \text{(ii) }2y=4x-6,\qquad2x=y+3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The given system of equations is}$
$\displaystyle 2x-3y=6\qquad\ldots\text{(i)}$
$\displaystyle x+y=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x-3y=6:$
$\displaystyle y=\frac{2x-6}{3}$
$\displaystyle \text{When }x=0,\ y=-2;\qquad\text{when }x=3,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&-2&0\end{array}$
$\displaystyle \text{Plot }(0,-2)\text{ and }(3,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }x+y=1:$
$\displaystyle y=1-x$
$\displaystyle \text{When }x=0,\ y=1;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&1&0\end{array}$
$\displaystyle \text{Plot }(0,1)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }\left(\frac{9}{5},-\frac{4}{5}\right).$
$\displaystyle \therefore x=\frac{9}{5},\qquad y=-\frac{4}{5}.$
$\displaystyle \therefore \text{The given system has a unique solution.}$

$\displaystyle \text{(ii) The given system of equations is}$
$\displaystyle 2y=4x-6\qquad\ldots\text{(i)}$
$\displaystyle 2x=y+3\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2y=4x-6:$
$\displaystyle y=2x-3$
$\displaystyle \text{When }x=0,\ y=-3;\qquad\text{when }x=2,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-3&1\end{array}$
$\displaystyle \text{Plot }(0,-3)\text{ and }(2,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x=y+3:$
$\displaystyle y=2x-3$
$\displaystyle \text{When }x=0,\ y=-3;\qquad\text{when }x=2,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-3&1\end{array}$
$\displaystyle \text{Plot }(0,-3)\text{ and }(2,1)\text{ and draw the line through them.}$
$\displaystyle \text{The graphs of the two equations coincide.}$
$\displaystyle \text{Therefore, every point on the line is a common solution of both equations.}$
$\displaystyle \therefore \text{The given system has infinitely many solutions and has no unique solution.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Solve graphically each of the following systems of linear equations.}$
$\displaystyle \text{Also, find the coordinates of the points where the lines meet the }y\text{-axis:}$
$\displaystyle \text{(i) }2x-5y+4=0,\qquad 2x+y-8=0\hfill\text{[CBSE 2005]}$
$\displaystyle \text{(ii) }3x+2y=12,\qquad 5x-2y=4\hfill\text{[CBSE 2006C]}$
$\displaystyle \text{(iii) }x+2y=6,\qquad 3x-2y=2\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The given system of equations is}$
$\displaystyle 2x-5y+4=0\qquad\ldots\text{(i)}$
$\displaystyle 2x+y-8=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x-5y+4=0:$
$\displaystyle y=\frac{2x+4}{5}$
$\displaystyle \text{When }x=-2,\ y=0;\qquad\text{when }x=3,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&-2&3\\ \hline y&0&2\end{array}$
$\displaystyle \text{Plot }(-2,0)\text{ and }(3,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x+y-8=0:$
$\displaystyle y=8-2x$
$\displaystyle \text{When }x=0,\ y=8;\qquad\text{when }x=3,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&8&2\end{array}$
$\displaystyle \text{Plot }(0,8)\text{ and }(3,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(3,2).$
$\displaystyle \therefore x=3,\qquad y=2.$
$\displaystyle \text{For the first equation, putting }x=0,\text{ we get }y=\frac{4}{5}.$
$\displaystyle \therefore \text{The first line meets the }y\text{-axis at }\left(0,\frac{4}{5}\right).$
$\displaystyle \text{For the second equation, putting }x=0,\text{ we get }y=8.$
$\displaystyle \therefore \text{The second line meets the }y\text{-axis at }(0,8).$

$\displaystyle \text{(ii) The given system of equations is}$
$\displaystyle 3x+2y=12\qquad\ldots\text{(i)}$
$\displaystyle 5x-2y=4\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }3x+2y=12:$
$\displaystyle y=\frac{12-3x}{2}$
$\displaystyle \text{When }x=0,\ y=6;\qquad\text{when }x=2,\ y=3.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&6&3\end{array}$
$\displaystyle \text{Plot }(0,6)\text{ and }(2,3)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }5x-2y=4:$
$\displaystyle y=\frac{5x-4}{2}$
$\displaystyle \text{When }x=0,\ y=-2;\qquad\text{when }x=2,\ y=3.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-2&3\end{array}$
$\displaystyle \text{Plot }(0,-2)\text{ and }(2,3)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(2,3).$
$\displaystyle \therefore x=2,\qquad y=3.$
$\displaystyle \text{For the first equation, putting }x=0,\text{ we get }y=6.$
$\displaystyle \therefore \text{The first line meets the }y\text{-axis at }(0,6).$
$\displaystyle \text{For the second equation, putting }x=0,\text{ we get }y=-2.$
$\displaystyle \therefore \text{The second line meets the }y\text{-axis at }(0,-2).$

$\displaystyle \text{(iii) The given system of equations is}$
$\displaystyle x+2y=6\qquad\ldots\text{(i)}$
$\displaystyle 3x-2y=2\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x+2y=6:$
$\displaystyle y=\frac{6-x}{2}$
$\displaystyle \text{When }x=0,\ y=3;\qquad\text{when }x=2,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&3&2\end{array}$
$\displaystyle \text{Plot }(0,3)\text{ and }(2,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3x-2y=2:$
$\displaystyle y=\frac{3x-2}{2}$
$\displaystyle \text{When }x=0,\ y=-1;\qquad\text{when }x=2,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-1&2\end{array}$
$\displaystyle \text{Plot }(0,-1)\text{ and }(2,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(2,2).$
$\displaystyle \therefore x=2,\qquad y=2.$
$\displaystyle \text{For the first equation, putting }x=0,\text{ we get }y=3.$
$\displaystyle \therefore \text{The first line meets the }y\text{-axis at }(0,3).$
$\displaystyle \text{For the second equation, putting }x=0,\text{ we get }y=-1.$
$\displaystyle \therefore \text{The second line meets the }y\text{-axis at }(0,-1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Solve the following systems of linear equations graphically and}$
$\displaystyle \text{shade the region between the two lines and the }x\text{-axis:}$
$\displaystyle \text{(i) }2x+3y=12,\qquad x-y=1\hfill\text{[CBSE 2001]}$
$\displaystyle \text{(ii) }3x+2y-4=0,\qquad2x-3y-7=0\hfill\text{[CBSE 2006C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The given system of equations is}$
$\displaystyle 2x+3y=12\qquad\ldots\text{(i)}$
$\displaystyle x-y=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x+3y=12:$
$\displaystyle y=\frac{12-2x}{3}$
$\displaystyle \text{When }x=0,\ y=4;\qquad\text{when }x=6,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&6\\ \hline y&4&0\end{array}$
$\displaystyle \text{Plot }(0,4)\text{ and }(6,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }x-y=1:$
$\displaystyle y=x-1$
$\displaystyle \text{When }x=1,\ y=0;\qquad\text{when }x=3,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&1&3\\ \hline y&0&2\end{array}$
$\displaystyle \text{Plot }(1,0)\text{ and }(3,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(3,2).$
$\displaystyle \therefore x=3,\qquad y=2.$
$\displaystyle \text{The first line meets the }x\text{-axis at }(6,0)\text{ and the second at }(1,0).$
$\displaystyle \therefore \text{Shade the triangular region with vertices }(1,0),(6,0)\text{ and }(3,2).$

$\displaystyle \text{(ii) The given system of equations is}$
$\displaystyle 3x+2y-4=0\qquad\ldots\text{(i)}$
$\displaystyle 2x-3y-7=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }3x+2y-4=0:$
$\displaystyle y=\frac{4-3x}{2}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=2,\ y=-1.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&2&-1\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(2,-1)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x-3y-7=0:$
$\displaystyle y=\frac{2x-7}{3}$
$\displaystyle \text{When }x=2,\ y=-1;\qquad\text{when }x=5,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&2&5\\ \hline y&-1&1\end{array}$
$\displaystyle \text{Plot }(2,-1)\text{ and }(5,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(2,-1).$
$\displaystyle \therefore x=2,\qquad y=-1.$
$\displaystyle \text{For the first line, putting }y=0,\text{ we get }x=\frac{4}{3}.$
$\displaystyle \text{For the second line, putting }y=0,\text{ we get }x=\frac{7}{2}.$
$\displaystyle \therefore \text{Shade the triangular region with vertices }\left(\frac{4}{3},0\right),\left(\frac{7}{2},0\right)\text{ and }(2,-1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Draw the graphs of the following equations on the same graph paper:}$
$\displaystyle 2x+3y=12,\qquad x-y=1.$
$\displaystyle \text{Find the coordinates of the vertices of the triangle formed by the two straight lines and the}$
$\displaystyle y\text{-axis.}\hfill\text{[CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 2x+3y=12\qquad\ldots\text{(i)}$
$\displaystyle x-y=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x+3y=12:$
$\displaystyle y=\frac{12-2x}{3}$
$\displaystyle \text{When }x=0,\ y=4;\qquad\text{when }x=6,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&6\\ \hline y&4&0\end{array}$
$\displaystyle \text{Plot }(0,4)\text{ and }(6,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }x-y=1:$
$\displaystyle y=x-1$
$\displaystyle \text{When }x=0,\ y=-1;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-1&0\end{array}$
$\displaystyle \text{Plot }(0,-1)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(3,2).$
$\displaystyle \text{The first line meets the }y\text{-axis at }(0,4).$
$\displaystyle \text{The second line meets the }y\text{-axis at }(0,-1).$
$\displaystyle \therefore \text{The vertices of the triangle are }(0,4),(0,-1)\text{ and }(3,2).$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Draw the graphs of }x-y+1=0\text{ and }3x+2y-12=0.\text{ Determine the}$
$\displaystyle \text{coordinates of the vertices of the triangle formed by these lines and }x\text{-axis and shade the}$
$\displaystyle \text{triangular area. Calculate the area bounded by these lines and }x\text{-axis.}\hfill\text{[CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle x-y+1=0\qquad\ldots\text{(i)}$
$\displaystyle 3x+2y-12=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x-y+1=0:$
$\displaystyle y=x+1$
$\displaystyle \text{When }x=-1,\ y=0;\qquad\text{when }x=0,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&-1&0\\ \hline y&0&1\end{array}$
$\displaystyle \text{Plot }(-1,0)\text{ and }(0,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3x+2y-12=0:$
$\displaystyle y=\frac{12-3x}{2}$
$\displaystyle \text{When }x=0,\ y=6;\qquad\text{when }x=4,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&4\\ \hline y&6&0\end{array}$
$\displaystyle \text{Plot }(0,6)\text{ and }(4,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(2,3).$
$\displaystyle \text{The lines meet the }x\text{-axis at }(-1,0)\text{ and }(4,0).$
$\displaystyle \therefore \text{The vertices of the triangle are }(-1,0),(4,0)\text{ and }(2,3).$
$\displaystyle \text{Shade the triangular region bounded by the two lines and the }x\text{-axis.}$
$\displaystyle \text{Base}=4-(-1)=5\text{ units},\qquad\text{Height}=3\text{ units}.$
$\displaystyle \text{Area of triangle}=\frac{1}{2}\times\text{Base}\times\text{Height}$
$\displaystyle =\frac{1}{2}\times5\times3=\frac{15}{2}=7.5\text{ sq. units}.$
$\displaystyle \therefore \text{The area bounded by the two lines and the }x\text{-axis is }7.5\text{ sq. units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Draw the graphs of the equations }5x-y=5\text{ and }$
$\displaystyle 3x-y=3.\text{ Determine the} \ \text{coordinates of the vertices of the triangle formed by}$
$\displaystyle \text{these lines and } y\text{-axis. Calculate the} \ \text{area of the triangle so formed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 5x-y=5\qquad\ldots\text{(i)}$
$\displaystyle 3x-y=3\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }5x-y=5:$
$\displaystyle y=5x-5$
$\displaystyle \text{When }x=0,\ y=-5;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-5&0\end{array}$
$\displaystyle \text{Plot }(0,-5)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3x-y=3:$
$\displaystyle y=3x-3$
$\displaystyle \text{When }x=0,\ y=-3;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-3&0\end{array}$
$\displaystyle \text{Plot }(0,-3)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(1,0).$
$\displaystyle \text{The lines meet the }y\text{-axis at }(0,-5)\text{ and }(0,-3).$
$\displaystyle \therefore \text{The vertices of the triangle are }(0,-5),(0,-3)\text{ and }(1,0).$
$\displaystyle \text{Base}=|-3-(-5)|=2\text{ units},\qquad\text{Height}=1\text{ unit}.$
$\displaystyle \text{Area of triangle}=\frac{1}{2}\times\text{Base}\times\text{Height}$
$\displaystyle =\frac{1}{2}\times2\times1=1\text{ sq. unit}.$
$\displaystyle \therefore \text{The area of the triangle formed is }1\text{ sq. unit.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Form the pair of linear equations in the following problems, and find their solution}$
$\displaystyle \text{graphically:}$
$\displaystyle \text{(i) }10\text{ students of class X took part in a Mathematics quiz. If the number of girls is }4\text{ more}$
$\displaystyle \text{than the number of boys, find the number of boys and girls who took part in the quiz.}$
$\displaystyle \text{(ii) }5\text{ pencils and }7\text{ pens together cost Rs. }50\text{, whereas }7\text{ pencils and }5\text{ pens together}$
$\displaystyle \text{cost Rs. }46\text{. Find the cost of one pencil and a pen.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let the number of boys be }x\text{ and the number of girls be }y.$
$\displaystyle \text{Total number of students}=10.$
$\displaystyle \therefore x+y=10\qquad\ldots\text{(i)}$
$\displaystyle \text{The number of girls is }4\text{ more than the number of boys.}$
$\displaystyle \therefore y=x+4$
$\displaystyle \Rightarrow x-y=-4\qquad\ldots\text{(ii)}$
$\displaystyle \text{Thus, the pair of linear equations is}$
$\displaystyle x+y=10\quad\text{and}\quad x-y=-4.$
$\displaystyle \text{For }x+y=10:$
$\displaystyle y=10-x$
$\displaystyle \text{When }x=0,\ y=10;\qquad\text{when }x=10,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&10\\ \hline y&10&0\end{array}$
$\displaystyle \text{Plot }(0,10)\text{ and }(10,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }x-y=-4:$
$\displaystyle y=x+4$
$\displaystyle \text{When }x=0,\ y=4;\qquad\text{when }x=3,\ y=7.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&4&7\end{array}$
$\displaystyle \text{Plot }(0,4)\text{ and }(3,7)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(3,7).$
$\displaystyle \therefore x=3,\qquad y=7.$
$\displaystyle \therefore \text{The number of boys is }3\text{ and the number of girls is }7.$

$\displaystyle \text{(ii) Let the cost of one pencil be Rs. }x\text{ and the cost of one pen be Rs. }y.$
$\displaystyle \text{Cost of }5\text{ pencils and }7\text{ pens}= \text{Rs. }50.$
$\displaystyle \therefore 5x+7y=50\qquad\ldots\text{(i)}$
$\displaystyle \text{Cost of }7\text{ pencils and }5\text{ pens}= \text{Rs. }46.$
$\displaystyle \therefore 7x+5y=46\qquad\ldots\text{(ii)}$
$\displaystyle \text{Thus, the pair of linear equations is}$
$\displaystyle 5x+7y=50\quad\text{and}\quad7x+5y=46.$
$\displaystyle \text{For }5x+7y=50:$
$\displaystyle y=\frac{50-5x}{7}$
$\displaystyle \text{When }x=3,\ y=5;\qquad\text{when }x=10,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&3&10\\ \hline y&5&0\end{array}$
$\displaystyle \text{Plot }(3,5)\text{ and }(10,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }7x+5y=46:$
$\displaystyle y=\frac{46-7x}{5}$
$\displaystyle \text{When }x=3,\ y=5;\qquad\text{when }x=8,\ y=-2.$
$\displaystyle \begin{array}{c|cc}x&3&8\\ \hline y&5&-2\end{array}$
$\displaystyle \text{Plot }(3,5)\text{ and }(8,-2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(3,5).$
$\displaystyle \therefore x=3,\qquad y=5.$
$\displaystyle \therefore \text{The cost of one pencil is Rs. }3\text{ and the cost of one pen is Rs. }5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Draw the graphs of the following equations:}$
$\displaystyle 2x-3y+6=0,\qquad2x+3y-18=0,\qquad y-2=0.$
$\displaystyle \text{Find the vertices of the triangle so obtained. Also, find the area of the triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 2x-3y+6=0\qquad\ldots\text{(i)}$
$\displaystyle 2x+3y-18=0\qquad\ldots\text{(ii)}$
$\displaystyle y-2=0\qquad\ldots\text{(iii)}$
$\displaystyle \text{For }2x-3y+6=0:$
$\displaystyle y=\frac{2x+6}{3}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=3,\ y=4.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&2&4\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(3,4)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x+3y-18=0:$
$\displaystyle y=\frac{18-2x}{3}$
$\displaystyle \text{When }x=3,\ y=4;\qquad\text{when }x=6,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&3&6\\ \hline y&4&2\end{array}$
$\displaystyle \text{Plot }(3,4)\text{ and }(6,2)\text{ and draw a straight line through them.}$
$\displaystyle y-2=0\Rightarrow y=2.$
$\displaystyle y=2\text{ represents a line parallel to the }x\text{-axis.}$
$\displaystyle \text{The three lines intersect pairwise at }(0,2),(3,4)\text{ and }(6,2).$
$\displaystyle \therefore \text{The vertices of the triangle are }(0,2),(3,4)\text{ and }(6,2).$
$\displaystyle \text{Base}=6-0=6\text{ units},\qquad\text{Height}=4-2=2\text{ units}.$
$\displaystyle \text{Area of triangle}=\frac{1}{2}\times\text{Base}\times\text{Height}$
$\displaystyle =\frac{1}{2}\times6\times2=6\text{ sq. units}.$
$\displaystyle \therefore \text{The area of the triangle is }6\text{ sq. units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Champa went to a `sale' to purchase some pants and skirts. When her friends}$
$\displaystyle \text{asked her how many of each she had bought, she answered, ``The number of skirts is two}$
$\displaystyle \text{less than twice the number of pants purchased. Also, the number of skirts is four less than}$
$\displaystyle \text{four times the number of pants purchased.'' Help her friends to find how many pants and}$
$\displaystyle \text{skirts Champa bought.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of pants purchased be }x\text{ and the number of skirts purchased be }y.$
$\displaystyle \text{The number of skirts is two less than twice the number of pants.}$
$\displaystyle \therefore y=2x-2\qquad\ldots\text{(i)}$
$\displaystyle \text{The number of skirts is four less than four times the number of pants.}$
$\displaystyle \therefore y=4x-4\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }y=2x-2:$
$\displaystyle \text{When }x=0,\ y=-2;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-2&0\end{array}$
$\displaystyle \text{Plot }(0,-2)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }y=4x-4:$
$\displaystyle \text{When }x=0,\ y=-4;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-4&0\end{array}$
$\displaystyle \text{Plot }(0,-4)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(1,0).$
$\displaystyle \therefore x=1,\qquad y=0.$
$\displaystyle \therefore \text{Champa bought }1\text{ pant and }0\text{ skirts.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Given the linear equation }2x+3y-8=0\text{, write another linear equation in two}$
$\displaystyle \text{variables such that the geometrical representation of the pair so formed is}$
$\displaystyle \text{(i) intersecting lines}\qquad\text{(ii) parallel lines}\qquad\text{(iii) coincident lines.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }2x+3y-8=0.$
$\displaystyle \therefore a_1=2,\quad b_1=3,\quad c_1=-8.$
$\displaystyle \text{(i) For intersecting lines, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.$
$\displaystyle \text{Let another equation be }x+y-2=0.$
$\displaystyle \frac{a_1}{a_2}=\frac{2}{1}=2,\qquad\frac{b_1}{b_2}=\frac{3}{1}=3.$
$\displaystyle \therefore \frac{a_1}{a_2}\ne\frac{b_1}{b_2}.$
$\displaystyle \therefore \text{One such equation is }x+y-2=0.$

$\displaystyle \text{(ii) For parallel lines, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.$
$\displaystyle \text{Let another equation be }4x+6y-10=0.$
$\displaystyle \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\qquad\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}.$
$\displaystyle \frac{c_1}{c_2}=\frac{-8}{-10}=\frac{4}{5}.$
$\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.$
$\displaystyle \therefore \text{One such equation is }4x+6y-10=0.$

$\displaystyle \text{(iii) For coincident lines, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
$\displaystyle \text{Multiplying }2x+3y-8=0\text{ by }2,$
$\displaystyle 4x+6y-16=0.$
$\displaystyle \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\qquad\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}.$
$\displaystyle \frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}.$
$\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
$\displaystyle \therefore \text{One such equation is }4x+6y-16=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Graphically, solve the following pair of equations:}$
$\displaystyle 2x+y=6,\qquad2x-y+2=0.$
$\displaystyle \text{Find the ratio of the areas of the two triangles formed by the lines representing}$
$\displaystyle \text{these equations with the }x\text{-axis and the lines with the }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 2x+y=6\qquad\ldots\text{(i)}$
$\displaystyle 2x-y+2=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x+y=6:$
$\displaystyle y=6-2x$
$\displaystyle \text{When }x=0,\ y=6;\qquad\text{when }x=3,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&6&0\end{array}$
$\displaystyle \text{Plot }(0,6)\text{ and }(3,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x-y+2=0:$
$\displaystyle y=2x+2$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=-1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&-1\\ \hline y&2&0\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(-1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(1,4).$
$\displaystyle \therefore x=1,\qquad y=4.$
$\displaystyle \text{The triangle formed by the two lines and the }x\text{-axis has vertices}$
$\displaystyle (-1,0),(3,0)\text{ and }(1,4).$
$\displaystyle \text{Base}=3-(-1)=4\text{ units},\qquad\text{Height}=4\text{ units}.$
$\displaystyle \text{Area}=\frac{1}{2}\times4\times4=8\text{ sq. units}.$
$\displaystyle \text{The triangle formed by the two lines and the }y\text{-axis has vertices}$
$\displaystyle (0,2),(0,6)\text{ and }(1,4).$
$\displaystyle \text{Base}=6-2=4\text{ units},\qquad\text{Height}=1\text{ unit}.$
$\displaystyle \text{Area}=\frac{1}{2}\times4\times1=2\text{ sq. units}.$
$\displaystyle \therefore \text{Required ratio of the areas}=8:2=4:1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Determine, graphically, the vertices of the triangle formed by the lines}$
$\displaystyle y=x,\qquad3y=x,\qquad x+y=8.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle y=x\qquad\ldots\text{(i)}$
$\displaystyle 3y=x\qquad\ldots\text{(ii)}$
$\displaystyle x+y=8\qquad\ldots\text{(iii)}$
$\displaystyle \text{For }y=x:$
$\displaystyle \text{When }x=0,\ y=0;\qquad\text{when }x=4,\ y=4.$
$\displaystyle \begin{array}{c|cc}x&0&4\\ \hline y&0&4\end{array}$
$\displaystyle \text{Plot }(0,0)\text{ and }(4,4)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3y=x:$
$\displaystyle y=\frac{x}{3}$
$\displaystyle \text{When }x=0,\ y=0;\qquad\text{when }x=6,\ y=2.$
$\displaystyle \begin{array}{c|cc}x&0&6\\ \hline y&0&2\end{array}$
$\displaystyle \text{Plot }(0,0)\text{ and }(6,2)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }x+y=8:$
$\displaystyle y=8-x$
$\displaystyle \text{When }x=0,\ y=8;\qquad\text{when }x=8,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&8\\ \hline y&8&0\end{array}$
$\displaystyle \text{Plot }(0,8)\text{ and }(8,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{From the graph, the three lines intersect pairwise at }(0,0),(4,4)\text{ and }(6,2).$
$\displaystyle \therefore \text{The vertices of the triangle are }(0,0),(4,4)\text{ and }(6,2).$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Draw the graph of the equations }x=3,\ x=5\text{ and }$
$\displaystyle 2x-y-4=0.\text{ Also, find the} \ \text{area of the quadrilateral formed by the} \\ \text{lines and the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle x=3\text{ represents a line parallel to the }y\text{-axis passing through }(3,0).$
$\displaystyle x=5\text{ represents a line parallel to the }y\text{-axis passing through }(5,0).$
$\displaystyle \text{For }2x-y-4=0:$
$\displaystyle y=2x-4$
$\displaystyle \text{When }x=3,\ y=2;\qquad\text{when }x=5,\ y=6.$
$\displaystyle \begin{array}{c|cc}x&3&5\\ \hline y&2&6\end{array}$
$\displaystyle \text{Plot }(3,2)\text{ and }(5,6)\text{ and draw a straight line through them.}$
$\displaystyle \text{Draw the lines }x=3,\ x=5\text{ and the }x\text{-axis on the same graph paper.}$
$\displaystyle \text{The quadrilateral formed has vertices }(3,0),(5,0),(5,6)\text{ and }(3,2).$
$\displaystyle \text{The parallel sides have lengths }2\text{ units and }6\text{ units.}$
$\displaystyle \text{Distance between the parallel sides}=5-3=2\text{ units}.$
$\displaystyle \text{Area of quadrilateral}=\frac{1}{2}\times(2+6)\times2$
$\displaystyle =8\text{ sq. units}.$
$\displaystyle \therefore \text{The area of the quadrilateral is }8\text{ sq. units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Draw the graphs of the lines }x=-2\text{ and }y=3.\text{ Write the vertices of}$
$\displaystyle \text{the figure formed by these lines, the }x\text{-axis and }y\text{-axis. Also, find the area of the figure.}$
$\displaystyle \text{Answer:}$
$\displaystyle x=-2\text{ represents a line parallel to the }y\text{-axis, }2\text{ units to its left.}$
$\displaystyle y=3\text{ represents a line parallel to the }x\text{-axis, }3\text{ units above it.}$
$\displaystyle \text{Draw the lines }x=-2,\ y=3,\ x=0\text{ and }y=0\text{ on the same graph paper.}$
$\displaystyle \text{The figure formed is a rectangle with vertices }(0,0),(-2,0),(-2,3)\text{ and }(0,3).$
$\displaystyle \text{Length}=3\text{ units},\qquad\text{Breadth}=2\text{ units}.$
$\displaystyle \text{Area of rectangle}=\text{Length}\times\text{Breadth}$
$\displaystyle =3\times2=6\text{ sq. units}.$
$\displaystyle \therefore \text{The area of the figure is }6\text{ sq. units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Solve the following system of equations graphically:}$
$\displaystyle 2x-y-2=0,\qquad -4x+y+4=0.\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Also, find the absolute difference between the ordinates of the points where the given lines}$
$\displaystyle \text{cut the }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 2x-y-2=0\qquad\ldots\text{(i)}$
$\displaystyle -4x+y+4=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x-y-2=0:$
$\displaystyle y=2x-2$
$\displaystyle \text{When }x=0,\ y=-2;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-2&0\end{array}$
$\displaystyle \text{Plot }(0,-2)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }-4x+y+4=0:$
$\displaystyle y=4x-4$
$\displaystyle \text{When }x=0,\ y=-4;\qquad\text{when }x=1,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&-4&0\end{array}$
$\displaystyle \text{Plot }(0,-4)\text{ and }(1,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(1,0).$
$\displaystyle \therefore x=1,\qquad y=0.$
$\displaystyle \therefore \text{The solution of the given system is }(1,0).$
$\displaystyle \text{The first line cuts the }y\text{-axis at }(0,-2).$
$\displaystyle \text{The second line cuts the }y\text{-axis at }(0,-4).$
$\displaystyle \therefore \text{Absolute difference between the ordinates}=|-2-(-4)|$
$\displaystyle =|-2+4|=2.$
$\displaystyle \therefore \text{The required absolute difference is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Solve the following system of equations graphically:}$
$\displaystyle 2x+y=5\text{ and }4x-y=7.\text{ Hence, write the coordinates of the points where given lines meet}$
$\displaystyle y\text{-axis.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 2x+y=5\qquad\ldots\text{(i)}$
$\displaystyle 4x-y=7\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x+y=5:$
$\displaystyle y=5-2x$
$\displaystyle \text{When }x=0,\ y=5;\qquad\text{when }x=2,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&5&1\end{array}$
$\displaystyle \text{Plot }(0,5)\text{ and }(2,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }4x-y=7:$
$\displaystyle y=4x-7$
$\displaystyle \text{When }x=0,\ y=-7;\qquad\text{when }x=2,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&-7&1\end{array}$
$\displaystyle \text{Plot }(0,-7)\text{ and }(2,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(2,1).$
$\displaystyle \therefore x=2,\qquad y=1.$
$\displaystyle \therefore \text{The solution of the given system is }(2,1).$
$\displaystyle \text{The first line meets the }y\text{-axis at }(0,5).$
$\displaystyle \text{The second line meets the }y\text{-axis at }(0,-7).$
$\displaystyle \therefore \text{The required points are }(0,5)\text{ and }(0,-7).$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Solve the following system of equations graphically:}$
$\displaystyle 2x+3y=6,\qquad x+y-1=0.$
$\displaystyle \text{Also, find the sum of ordinates of the points where given lines meet }y\text{-axis.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle 2x+3y=6\qquad\ldots\text{(i)}$
$\displaystyle x+y-1=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }2x+3y=6:$
$\displaystyle y=\frac{6-2x}{3}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=3,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&2&0\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(3,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }x+y-1=0:$
$\displaystyle y=1-x$
$\displaystyle \text{When }x=0,\ y=1;\qquad\text{when }x=3,\ y=-2.$
$\displaystyle \begin{array}{c|cc}x&0&3\\ \hline y&1&-2\end{array}$
$\displaystyle \text{Plot }(0,1)\text{ and }(3,-2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(3,0).$
$\displaystyle \therefore x=3,\qquad y=0.$
$\displaystyle \therefore \text{The solution of the given system is }(3,0).$
$\displaystyle \text{The first line meets the }y\text{-axis at }(0,2).$
$\displaystyle \text{The second line meets the }y\text{-axis at }(0,1).$
$\displaystyle \therefore \text{Sum of ordinates}=2+1=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Check whether the following pair of equations is consistent or not. }$
$\displaystyle \text{If consistent, solve graphically:}\qquad x+3y=6,\qquad3y-2x=-12\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle x+3y=6\qquad\ldots\text{(i)}$
$\displaystyle 3y-2x=-12\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x+3y=6:$
$\displaystyle y=\frac{6-x}{3}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=6,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&6\\ \hline y&2&0\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(6,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }3y-2x=-12:$
$\displaystyle y=\frac{2x-12}{3}$
$\displaystyle \text{When }x=0,\ y=-4;\qquad\text{when }x=6,\ y=0.$
$\displaystyle \begin{array}{c|cc}x&0&6\\ \hline y&-4&0\end{array}$
$\displaystyle \text{Plot }(0,-4)\text{ and }(6,0)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(6,0).$
$\displaystyle \therefore \text{The given pair of equations is consistent.}$
$\displaystyle \therefore x=6,\qquad y=0.$
$\displaystyle \therefore \text{The solution of the given pair is }(6,0).$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Check whether the following system of equations is consistent or not. }$
$\displaystyle \text{If consistent, solve graphically:}\qquad x-2y+4=0,\qquad2x-y-4=0\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle x-2y+4=0\qquad\ldots\text{(i)}$
$\displaystyle 2x-y-4=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x-2y+4=0:$
$\displaystyle y=\frac{x+4}{2}$
$\displaystyle \text{When }x=0,\ y=2;\qquad\text{when }x=4,\ y=4.$
$\displaystyle \begin{array}{c|cc}x&0&4\\ \hline y&2&4\end{array}$
$\displaystyle \text{Plot }(0,2)\text{ and }(4,4)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x-y-4=0:$
$\displaystyle y=2x-4$
$\displaystyle \text{When }x=2,\ y=0;\qquad\text{when }x=4,\ y=4.$
$\displaystyle \begin{array}{c|cc}x&2&4\\ \hline y&0&4\end{array}$
$\displaystyle \text{Plot }(2,0)\text{ and }(4,4)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(4,4).$
$\displaystyle \therefore \text{The given system of equations is consistent.}$
$\displaystyle \therefore x=4,\qquad y=4.$
$\displaystyle \therefore \text{The solution of the given system is }(4,4).$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Check whether the given system of equations is consistent or not. }$
$\displaystyle \text{If consistent, solve graphically:}\qquad x-2y=0,\qquad2x+y=0\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations are}$
$\displaystyle x-2y=0\qquad\ldots\text{(i)}$
$\displaystyle 2x+y=0\qquad\ldots\text{(ii)}$
$\displaystyle \text{For }x-2y=0:$
$\displaystyle y=\frac{x}{2}$
$\displaystyle \text{When }x=0,\ y=0;\qquad\text{when }x=2,\ y=1.$
$\displaystyle \begin{array}{c|cc}x&0&2\\ \hline y&0&1\end{array}$
$\displaystyle \text{Plot }(0,0)\text{ and }(2,1)\text{ and draw a straight line through them.}$
$\displaystyle \text{For }2x+y=0:$
$\displaystyle y=-2x$
$\displaystyle \text{When }x=0,\ y=0;\qquad\text{when }x=1,\ y=-2.$
$\displaystyle \begin{array}{c|cc}x&0&1\\ \hline y&0&-2\end{array}$
$\displaystyle \text{Plot }(0,0)\text{ and }(1,-2)\text{ and draw a straight line through them.}$
$\displaystyle \text{The two lines intersect at }(0,0).$
$\displaystyle \therefore \text{The given system of equations is consistent.}$
$\displaystyle \therefore x=0,\qquad y=0.$
$\displaystyle \therefore \text{The solution of the given system is }(0,0).$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{A man lent a part of his money at }10\%\text{ p.a. and the rest at }15\%\text{ p.a. His income}$
$\displaystyle \text{at the end of the year was Rs. }1,900.\text{ If he had interchanged the rate of interest on the two}$
$\displaystyle \text{sums, he would have earned Rs. }200\text{ more. Find the amount lent in both cases.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Rs. }x\text{ be lent at }10\%\text{ p.a. and Rs. }y\text{ be lent at }15\%\text{ p.a.}$
$\displaystyle \frac{10x}{100}+\frac{15y}{100}=1900$
$\displaystyle \Rightarrow 2x+3y=38000\qquad\ldots\text{(i)}$
$\displaystyle \text{On interchanging the rates, the income would be Rs. }(1900+200)=\text{Rs. }2100.$
$\displaystyle \therefore \frac{15x}{100}+\frac{10y}{100}=2100$
$\displaystyle \Rightarrow 3x+2y=42000\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (i) by }3,$
$\displaystyle 6x+9y=114000\qquad\ldots\text{(iii)}$
$\displaystyle \text{Multiplying equation (ii) by }2,$
$\displaystyle 6x+4y=84000\qquad\ldots\text{(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle 5y=30000$
$\displaystyle \Rightarrow y=6000$
$\displaystyle \text{Substituting }y=6000\text{ in equation (i),}$
$\displaystyle 2x+3(6000)=38000$
$\displaystyle 2x=20000$
$\displaystyle \Rightarrow x=10000$
$\displaystyle \therefore \text{Originally, Rs. }10000\text{ was lent at }10\%\text{ p.a. and Rs. }6000\text{ at }15\%\text{ p.a.}$
$\displaystyle \text{On interchanging the rates, Rs. }10000\text{ would be lent at }15\%\text{ p.a. and Rs. }6000\text{ at }10\%\text{ p.a.}$
$\displaystyle \\$