CBSE Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions Ex 3.2

For representing the situation graphically and algebraically, we need to form linear equations
Let number of girls \( =x \)
Let number of boys \( =y \)
According to the question, Total no of students is equal to 10 ,
\(x+y=10\)
\(\Rightarrow x=10-y \ldots(i)\)
Now we will find different points to plot the equation. We can take any value of \( y \) and put in eq (i) to obtain the value of \( x \) at that point
Putting \( \mathrm{y}=4,5 \) and 6. we get,
\(\text { at } x=4\)
\(X=10-4=6\)
\(\text { at } x=5\)
\(X=10-5=5\)
\(\text { at } x=6\)
\(X=10-6=4\)
\(\begin{array}{|l|l|l|l|}
\hline x & 4 & 5 & 6 \\
\hline y & 6 & 5 & 4 \\
\hline
\end{array}\)
Number of girls is 4 more than number of boys \(\ldots\) Given
So,
\(x=y+4\)
\(\Rightarrow y=x-4\ldots(ii)\)
Now for plotting the points on graph, take any values of \( x \) and put them in eq (ii) to obtain values of \( y \)
Putting \( x=3,5 \) and 7 we get
at \( x=3 y=3-4=-1 \) at \( x=4 y=5-4=1 \) at \( x=7 y=7-4=3 \)
\(\begin{array}{|l|l|l|l|}
\hline x & 3 & 5 & 7 \\
\hline Y & -1 & 1 & 3 \\
\hline
\end{array}\)
Graphical representation:
Plotting the points obtain on graph we get,

As, both lines intersect each other at \( (7,3) \)
Solution of this pair of equation is \( (7,3) \) i.e. No of girls, \( x=7 \) No of boys, \( y=3 \)

For representing the situation graphically and algebraically, we need to form linear equations
Let cost of one pencil \( = \) Rs. X
Let cost of one pen = Rs. Y
According to the question, 5 pencils and 7 pens together cost Rs 50
\(5 \mathrm{x}+7 \mathrm{y}=50\)
\(\Rightarrow 5 \mathrm{x}=50-7 \mathrm{y}\)
\(\Rightarrow x=10\frac{7}{5} y\)
Putting value of \( y=0,5,10 \) we get,
\(\mathrm{x}=10-0=10\)
\(x=\frac{50-35}{5}=\frac{15}{5}=3\)
\(x=\frac{50-70}{5}=\frac{-20}{5}=-4\)
\(\begin{array}{|l|l|l|l|}
\hline x & 10 & 3 & -4 \\
\hline Y & 0 & 5 & 10 \\
\hline
\end{array}\)
Now,
7 pencils and 5 pens together cost Rs. 46
\(7 x+5 y=46\)
\(\Rightarrow 5 y=46-7 x\)
\(y=\frac{46-7 x}{5}\)
Putting \( x=-2,3,8 \) we get
\(y=\frac{46-(7 \times-2)}{5}=\frac{46+14}{5}=\frac{60}{5}=12\)
\(y=\frac{46-7 \times 3}{5}=\frac{46-21}{5}=\frac{25}{5}=5\)
\(y=\frac{46-7 \times 8}{5}=\frac{46-56}{5}=\frac{-10}{5}=-2\)
\(\begin{array}{|l|l|l|l|}
\hline x & -2 & 3 & 8 \\
\hline y & 12 & 5 & -2 \\
\hline
\end{array}\)
Graphical Representation:
Plotting the points obtain on graph we get,

As, both lines intersect each other at \( (3,5) \)
Solution of this pair of equation is \( (3,5) \) i.e. cost of pencil, \( x=3 \operatorname{cost} \) of pen, \( y=5 \).

Comparing these equation with
\(a_{1} x+b_{1} y+c_{1}=0\)
\(a_{2} x+b_{2} y+c_{2}=0\)
We get
\(a_{1}=5, b_{1}=-4, c_{1}=8\)
\(a_{2}=7, b_{2}=6, c_{2}=-9\)
Hence,
\(=\frac{a_{1}}{a_{2}}=\frac{5}{7}, \frac{b_{1}}{b_{2}}=-\frac{4}{6} \text { and } \frac{c_{1}}{c_{2}}=\frac{8}{-9}\)
We find that \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)
Therefore, both lines intersect at one point.
Graph of the lines look like below

Comparing these equations with,
\(a_{1} x+b_{1} y+c_{1}=0\)
\(a_{2} x+b_{2} y+c_{2}=0\)
We get
\(=a_{1}=9, b_{1}=3, c_{1}=12\)
\(=a_{2}=18, b_{2}=6, c_{2}=24\)
Hence,
\(=\frac{a_{1}}{a_{2}}=\frac{9}{18}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2} \text { and } \frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}\)
We find that
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Therefore, both lines are coincident.

Comparing these equations with,
\(a_{1} x+b_{1} y+c_{1}=0\)
\(a_{2} x+b_{2} y+c_{2}=0\)
We get
\(=a_{1}=6, b_{1}=-3, c_{1}=10\)
\(=a_{2}=2, b_{2}=-1, c_{2}=9\)
Hence,
\(=\frac{a_{1}}{a_{2}}=\frac{6}{2}=3, \frac{b_{1}}{b_{2}}=-\frac{3}{-1}=3 \text { and } \frac{c_{1}}{c_{2}}=\frac{10}{9}
\)
We find that
\(=\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Therefore, both lines are parallel