Examples - Chapter 3 Pair Of Linear Equations In Two Variables class 10 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Examples (Revised) - Chapter 3 Pair Of Linear Equations In Two Variables class 10 ncert solutions Maths

Example 1 :

Check graphically whether the pair of equations
$
\begin{gathered}
x+3 y=6 \\
2 x-3 y=12
\end{gathered}
$
and
is consistent. If so, solve them graphically.
Solution :

Let us draw the graphs of the Equations (1) and (2). For this, we find two solutions of each of the equations, which are given in Table 3.2

Plot the points $\mathrm{A}(0,2), \mathrm{B}(6,0)$, $P(0,-4)$ and $Q(3,-2)$ on graph paper, and join the points to form the lines $A B$ and $P Q$ as shown in Fig. 3.1.

We observe that there is a point $B(6,0)$ common to both the lines $\mathrm{AB}$ and $\mathrm{PQ}$. So, the solution of the pair of linear equations is $x=6$ and $y=0$, i.e., the given pair of equations is consistent.

Example 2 :

Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
$
\begin{array}{r}
5 x-8 y+1=0 \\
3 x-\frac{24}{5} y+\frac{3}{5}=0
\end{array}
$

Solution :

Multiplying Equation (2) by $\frac{5}{3}$, we get
$
5 x-8 y+1=0
$

But, this is the same as Equation (1). Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.

Example 3 :

Champa went to a 'Sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased". Help her friends to find how many pants and skirts Champa bought.

Solution :

Let us denote the number of pants by $x$ and the number of skirts by $y$. Then the equations formed are :
$
\text { and } \quad \begin{aligned}
& y=2 x-2 \\
& y=4 x-4
\end{aligned}
$

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations. They are given in Table 3.3.

Plot the points and draw the lines passing through them to represent the equations, as shown in Fig. 3.2.

The two lines intersect at the point $(1,0)$. So, $x=1, y=0$ is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Verify the answer by checking whether it satisfies the conditions of the given problem.

Example 4 :

Solve the following pair of equations by substitution method:
$
\begin{array}{r}
7 x-15 y=2 \\
x+2 y=3
\end{array}
$

Solution :
Step 1 : We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) :
$
\begin{aligned}
x+2 y & =3 \\
x & =3-2 y
\end{aligned}
$
and write it as
Step 2 : Substitute the value of $x$ in Equation (1). We get
i.e.,
$
7(3-2 y)-15 y=2
$
i.e.,
$
21-14 y-15 y=2
$
$
-29 y=-19
$

Therefore,
$
y=\frac{19}{29}
$

Step 3 : Substituting this value of $y$ in Equation (3), we get
$
x=3-2\left(\frac{19}{29}\right)=\frac{49}{29}
$

Therefore, the solution is $x=\frac{49}{29}, y=\frac{19}{29}$.

Verification : Substituting $x=\frac{49}{29}$ and $y=\frac{19}{29}$, you can verify that both the Equations
(1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:

Step 1 : Find the value of one variable, say $y$ in terms of the other variable, i.e., $x$ from either equation, whichever is convenient.
Step 2 : Substitute this value of $y$ in the other equation, and reduce it to an equation in one variable, i.e., in terms of $x$, which can be solved. Sometimes, as in Examples 9 and 10 below, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.
Step 3 : Substitute the value of $x$ (or $y$ ) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Remark: We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.

Example 5 :

Solve the following question-Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isn't this interesting?) Represent this situation algebraically and graphically by the method of substitution.
Solution :

Let $s$ and $t$ be the ages (in years) of Aftab and his daughter, respectively. Then, the pair of linear equations that represent the situation is

$
\begin{aligned}
& s-7=7(t-7) \text {, i.e., } s-7 t+42=0 \\
& \text { and } \quad s+3=3(t+3) \text {, i.e., } s-3 t=6
\end{aligned}
$
Using Equation (2), we get $s=3 t+6$.
Putting this value of $s$ in Equation (1), we get
$
(3 t+6)-7 t+42=0,
$
i.e.,
$
4 t=48 \text {, which gives } t=12 \text {. }
$

Putting this value of $t$ in Equation (2), we get
$
s=3(12)+6=42
$

So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.

Example 6 :

In a shop the cost of 2 pencils and 3 erasers is ₹9 and the cost of 4 pencils and 6 erasers is $₹ 18$. Find the cost of each pencil and each eraser.
Solution :

The pair of linear equations formed were:
$
\begin{aligned}
& 2 x+3 y=9 \\
& 4 x+6 y=18
\end{aligned}
$

We first express the value of $x$ in terms of $y$ from the equation $2 x+3 y=9$, to get
$
x=\frac{9-3 y}{2}
$

Now we substitute this value of $x$ in Equation (2), to get
$
\frac{4(9-3 y)}{2}+6 y=18
$
i.e.,
$
\begin{aligned}
18-6 y+6 y & =18 \\
18 & =18
\end{aligned}
$

This statement is true for all values of $y$. However, we do not get a specific value of $y$ as a solution. Therefore, we cannot obtain a specific value of $x$. This situation has arisen because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.

Example 7 :

Two rails are represented by the equations $x+2 y-4=0$ and $2 x+4 y-12=0$. Will the rails cross each other?
Solution :

The pair of linear equations formed were:
$
\begin{aligned}
x+2 y-4 & =0 \\
2 x+4 y-12 & =0
\end{aligned}
$

We express $x$ in terms of $y$ from Equation (1) to get
$
x=4-2 y
$

Now, we substitute this value of $x$ in Equation (2) to get
$
2(4-2 y)+4 y-12=0
$

i.e.,
$
\begin{aligned}
8-12 & =0 \\
-4 & =0
\end{aligned}
$
which is a false statement.
Therefore, the equations do not have a common solution. So, the two rails will not cross each other.

Example 8: 3

The ratio of incomes of two persons is $9: 7$ and the ratio of their expenditures is $4: 3$. If each of them manages to save $Rs.2000$ per month, find their monthly incomes.
Solution :

Let us denote the incomes of the two person by Rs.$9 x$ and Rs. $7 x$ and their expenditures by Rs.$4 y$ and $Rs.3 y$ respectively. Then the equations formed in the situation is given by:
$
\begin{aligned}
& 9 x-4 y=2000 \\
& 7 x-3 y=2000
\end{aligned}
$
and
Step 1 : Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of $y$ equal. Then we get the equations:
$
\begin{aligned}
& 27 x-12 y=6000 \\
& 28 x-12 y=8000
\end{aligned}
$

Step 2 : Subtract Equation (3) from Equation (4) to eliminatey, because the coefficients of $y$ are the same. So, we get
$
\begin{aligned}
(28 x-27 x)-(12 y-12 y) & =8000-6000 \\
x & =2000
\end{aligned}
$
i.e.,

Step 3 : Substituting this value of $x$ in (1), we get
$
\begin{aligned}
9(2000)-4 y & =2000 \\
y & =4000
\end{aligned}
$
i.e.,

So, the solution of the equations is $x=2000, y=4000$. Therefore, the monthly incomes of the persons are $ Rs. 18,000$ and Rs.14,000 , respectively.

Verification : $18000: 14000=9: 7$. Also, the ratio of their expenditures $=$ $18000-2000: 14000-2000=16000: 12000=4: 3$

Remarks :
1. The method used in solving the example above is called the elimination method, because we eliminate one variable first, to get a linear equation in one variable.

In the example above, we eliminated $y$. We could also have eliminated $x$. Try doing it that way.
2. You could also have used the substitution, or graphical method, to solve this problem. Try doing so, and see which method is more convenient.
Let us now note down these steps in the elimination method:

Step 1 : First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either $x$ or $y$ ) numerically equal.
Step 2 : Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3 .

If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.

If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step 3 : Solve the equation in one variable ( $x$ or $y$ ) so obtained to get its value.
Step 4 : Substitute this value of $x$ (or $y$ ) in either of the original equations to get the value of the other variable.
Now to illustrate it, we shall solve few more examples.
Example 9 :

Use elimination method to find all possible solutions of the following pair of linear equations:
$
\begin{aligned}
& 2 x+3 y=8 \\
& 4 x+6 y=7
\end{aligned}
$

Solution :
Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of $x$ equal. Then we get the equations as:
$
\begin{aligned}
& 4 x+6 y=16 \\
& 4 x+6 y=7
\end{aligned}
$

Step 2 : Subtracting Equation (4) from Equation (3),
$
(4 x-4 x)+(6 y-6 y)=16-7
$
i.e.,
$0=9$, which is a false statement.

Therefore, the pair of equations has no solution.
Example 10 :

The sum of a two-digit number and the number obtained by reversing the digits is 66 . If the digits of the number differ by 2 , find the number. How many such numbers are there?

Solution : Let the ten's and the unit's digits in the first number be $x$ and $y$, respectively. So, the first number may be written as $10 x+y$ in the expanded form (for example, $56=10(5)+6)$.

When the digits are reversed, $x$ becomes the unit's digit and $y$ becomes the ten's digit. This number, in the expanded notation is $10 y+x$ (for example, when 56 is reversed, we get $65=10(6)+5$ ).
According to the given condition.
i.e.,
$
\begin{aligned}
(10 x+y)+(10 y+x) & =66 \\
11(x+y) & =66 \\
x+y & =6
\end{aligned}
$

We are also given that the digits differ by 2 , therefore,
either
$
\begin{aligned}
& x-y=2 \\
& y-x=2
\end{aligned}
$
or
If $x-y=2$, then solving (1) and (2) by elimination, we get $x=4$ and $y=2$.

In this case, we get the number 42 .
If $y-x=2$, then solving (1) and (3) by elimination, we get $x=2$ and $y=4$.

In this case, we get the number 24 .
Thus, there are two such numbers 42 and 24.
Verification : Here $42+24=66$ and $4-2=2$. Also $24+42=66$ and $4-2=2$.