Q.1 Form the pair of linear equations in the following problems, and find their solutions
graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4
more than the number of boys, find the number of boys and girls who took part in
the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together
cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution (i)
Let number of Girls =\(x\)
No of boys = \(y\)
Substituting value of a from equation-(1) to equation-(2)
$$\begin{aligned} y+4+y&=10\\ 2y+4&=10\\ 2y&=10-4\\ &=6\\ 2y&=6\\ y&=\dfrac{6}{2}\\ &=3\end{aligned}$$Putting value of \(y\) in equation-(2)
$$\begin{aligned}x+y&=10\\ x+3&=10\\ x&=10-3\\ &=7\end{aligned}$$
So,
No. of Girls = 7 No of Boys =3
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Solution (ii)
Let cost of pencil = \(x\)
and cost of pen = \(y\)
Equating Eqn. (1) and eqn. (2)
$$\begin{aligned}\dfrac{1}{7}\left( 50-5x\right) &=\dfrac{1}{5}\left( 46-7x\right) \\ \Rightarrow 5\left( 30-5x\right) &=7\left( 46-7x\right) \\ \Rightarrow 250-25x&=322-49x\\ \Rightarrow 49x-25x&=322-250\\ \Rightarrow 24x&=72\\ \Rightarrow x&=\dfrac{72}{24}\\ x&=3\end{aligned}$$Putting value of a equation-(2),
$$\begin{aligned}y&=\dfrac{1}{7}\left( 50-5x\right) \\ &=\dfrac{1}{7}\left( 50-5\times 3\right) \\ &=\dfrac{1}{7}\left( 50-15\right) \\ &=\dfrac{1}{7}\times 35\\ y&=5\end{aligned}$$Hence Cost of one pencil = ₹3 and cost of one pen = ₹5
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Q.2 On comparing the ratios \(\frac{a_1}{a_2},\ \frac{b_1}{b_2} \text{ and }\frac{c_1}{c_2}\) ,
find out whether the lines representing the
following pairs of linear equations intersect at a point, are parallel or coincident:
-
5x – 4y + 8 = 0
7x + 6y – 9 = 0 -
9x + 3y + 12 = 0
18x + 6y + 24 = 0 -
6x – 3y + 10 = 0
2x – y + 9 = 0
Solution-(i)
$$\begin{aligned} 5x-4y+8&=0\\ 7x+6y-9&=0\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{5}{7}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{-4}{6}=\dfrac{-2}{3}\\\\ \dfrac{a_{1}}{a_{2}}&\neq \dfrac{b_{1}}{b_{2}}\end{aligned}$$Line is an intersecting line with unique Solution
Solution-(ii)
$$\begin{aligned} 9x+3y+12&=0\\ 18x+6y+24&=0\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{9}{18}=\dfrac{1}{2}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{3}{6}=\dfrac{1}{2}\\ \dfrac{c_{1}}{c_{2}}&=\dfrac{12}{24}=\dfrac{1}{2}\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}=\frac{1}{2}\end{aligned}$$Coincident line, infinite Solutions
Solution-(iii)
$$\begin{aligned} 6x-3y+10&=0\\ 2x-y+9&=0\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{6}{2}=3\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{-3}{-1}=3\\ \dfrac{c_{1}}{c_{2}}&=\dfrac{10}{9}\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}\end{aligned}$$Hence Parallel line, no solution
Q.3 On comparing the ratios \(\frac{a_1}{a_2},\ \frac{b_1}{b_2} \text{ and }\frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent, or inconsistent.
- 3x + 2y = 5 ; 2x – 3y = 7
- 2x – 3y = 8 ; 4x – 6y = 9
- \(\frac{3}{2}x + \frac{5}{3}y =7\); 9x – 10y = 14
- 5x – 3y = 11 ;– 10x + 6y = –22
- \(\frac{4}{3}x + 2y=8\);2x + 3y = 12
Solution-(i)
$$\begin{aligned} 3x+2y&=5\\ 2x-3y&=7\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{3}{2}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{2}{-3}=-\dfrac{2}{3}\\\\ \dfrac{a_{1}}{a_{2}}&\neq \dfrac{b_{1}}{b_{2}}\end{aligned}$$ Consistent line (unique solution)Solution-(ii)
$$\begin{aligned}2x-3y=8\\ 4x-6y=9\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{2}{4}=\dfrac{1}{2}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{-3}{-6}=\dfrac{1}{2}\\ \dfrac{C_{1}}{c_{2}}&=\dfrac{8}{9}\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}\end{aligned}$$ Parallel line, no solution, Hence InconsistentSolution-(iii)
$$\begin{aligned} \dfrac{3}{2}x+\dfrac{5}{3}y&=7\\ 9x-10y&=14\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{3}{2\times 9}=\dfrac{1}{6}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{5}{-3}\times 10=-\dfrac{1}{6}\\\\ \dfrac{a_{1}}{a_{2}}&\neq \dfrac{b_{1}}{b_{2}}\end{aligned}$$ Consistent line with unique solutionSolution-(iv)
$$\begin{aligned}5x-3y&=11\\ -10x+6y&=-22\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{-5}{10}=-\dfrac{1}{2}\\ \dfrac{b_{1}}{b_{2}}&=-\dfrac{3}{6}=-\dfrac{1}{2}\\ \dfrac{c_{1}}{c_{2}}&=-\dfrac{11}{22}=-\dfrac{1}{2}\\\\ \dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}&=\dfrac{c_{1}}{c_{2}}=-\dfrac{1}{2}\end{aligned}$$ Consistent line with infinite solutionSolution-(v)
$$\begin{aligned} \dfrac{4}{3}x+2y&=8\\ 2x+3y&=12\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{4}{3\times 2}=\dfrac{2}{3}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{2}{3}\\ \dfrac{c_{1}}{c_{2}}&=\dfrac{8}{12}=\dfrac{2}{3}\\\\ \dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}&=\dfrac{c_{1}}{c_{2}}=\dfrac{2}{3}\end{aligned}$$ consistent line with infinte solutionQ.4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
- x + y = 5, \(\quad\)2x + 2y = 10
- x – y = 8, \(\quad\)3x – 3y = 16
- 2x + y – 6 = 0, \(\quad\)4x – 2y – 4 = 0
- 2x – 2y – 2 = 0, \(\quad\)4x – 4y – 5 = 0
Solution- (i)
$$\begin{aligned} x+y&=5\\ 2x+2y&=10\\\\ \dfrac{a_{1}}{92}&=\dfrac{1}{2}\\ \dfrac{b1}{b_{2}}&=\dfrac{1}{2}\\ \dfrac{c_{1}}{c_{2}}&=\dfrac{5}{10}=\dfrac{1}{2}\\\\ \dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}&=\dfrac{c_{1}}{c_{2}}=\dfrac{1}{2}\end{aligned}$$ Consistent, same line, infinite solutionSolution- (ii)
$$\begin{aligned} x-y&=8\\ 3x-3y&=16\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{1}{3}\\ \dfrac{b_{1}}{b_{2}}&=\dfrac{-1}{-3}=\dfrac{1}{3}\\ \dfrac{c_{1}}{c_{2}}&=\dfrac{8}{16}=\dfrac{1}{2}\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}\end{aligned}$$ Hence, Parallel line (Inconsistent)Solution- (iii)
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Solution- (iv)
$$\begin{aligned} 2x-2y-2&=0\\ 4x-4y-5&=0\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{2}{4}=\dfrac{1}{2}\\ \dfrac{b,}{b_{2}}&=\dfrac{-2}{-4}=\dfrac{1}{2}\\ \dfrac{c_{1}}{c_{2}}&=\dfrac{-2}{-5}=\dfrac{2}{5}\\\\ \dfrac{a_{1}}{a_{2}}&=\dfrac{b,}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}\end{aligned}$$ parallel lines, inconsistent, no solutionQ.5 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution
Let width of the rectangular garden = \(x\)
Length which 4 m more of width = \(y\)
Half perimeter of rectangular garden = 36 m
$$\implies x+y=36\tag{2}$$Substituting Value of y from equation-(1), to equation-(2)
$$\begin{aligned} x+x+4&=36\\ 2x+4&=36\\ 2x&=36-4\\ 2x&=32\\ x&=\dfrac{32}{2}\\ &=16\\\\ y&=x+4\\ &=16+4\\ &=20\end{aligned}$$Length of rectangular Garden = 20m and width = 16m
Q.6 Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables
such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution
Given Equation$$2x+3y-8=0$$
For intersecting lines
$$\dfrac{a_{1}}{a_{2}}\neq \dfrac{b_{1}}{b_{2}}$$Hence, Equation of intersenting line
$$5x+6y-7=0$$For Parallel lines
$$\dfrac{a_{1}}{92}=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$$Hence, Equation of parallel line
$$2x+3y+5=0$$For Coincident line
$$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$$Hence, Equation of coincedent line
$$4x+6y-16=0$$Q.7 Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution
Coordinates of equation \(x – y + 1 = 0\)
\[ \begin{array}{|c|c|c|c|} \hline x&-1&0&1\\\hline y&0&1&2\\\hline \end{array} \]Coordinates of equation \(3x + 2y – 12 = 0\)
\[ \begin{array}{|c|c|c|c|} \hline x&0&2&4\\\hline y&6&3&0\\\hline \end{array} \] Coordinates of the vertices of the triangle formed by lines and the x-axis are (-1,0), (2,3), (4,0)