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Pair of Linear Equations in Two Variables : Exercise - 3.4 (Mathematics NCERT Class 10th)

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Q.1      Solve the following pair of linear equation by the elimination method and the substituti method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) ${x \over 2} + {{2y} \over 3} = - 1$ and $x - {y \over 3} = 3$
Sol.       (i) By elimination method :
The given system of equation is
x + y = 5
and, 2x – 3y = 4
Multiplying (1) by 3, we get
3x + 3y = 15
Adding (2) and (3), we get
5x = 19
$\Rightarrow$ $x = {{19} \over 5}$
Putting $x = {{19} \over 5}$ in (1), we get
${{19} \over 5}$ + y = 5
$\Rightarrow$ y = 5 – ${{19} \over 5} = {{25 - 19} \over 5}$
Hence, $x = {{19} \over 5}$, $y = {6 \over 5}$
By substitution method :
The given system of equations is
x + y = 5 ...(1)
and, 2x – 3y = 4 ...(2)
From (1), y = 5 – x
Substituting y = 5 – x in (2), we get
2x – 3 (5 – x) = 4
$\Rightarrow$ 2x – 15 + 3x = 4
$\Rightarrow$ 5x = 4 + 15
$\Rightarrow$ 5x = 19
$\Rightarrow$ $x = {{19} \over 5}$
Putting $x = {{19} \over 5}$ in (1), we get
${{19} \over 5} + y = 5$
$\Rightarrow$ y = 5 – ${{19} \over 5} = {{25 - 19} \over 5} = {6 \over 5}$
Hence, $x = {{19} \over 5},y = {6 \over 5}$
By geometrical method :
Graph of x + y = 5 :
We have, x + y = 5
$\Rightarrow$ y = 5 – x
When x = 0 , y = 5; When x = 5, y = 0
Thus, we have the following table : Plotting the points A(0, 5) and B(5, 0) and drawing a line joining them, we get the graph of the equation x + y = 5 as shown.
Graph of 2x – 3y = 4
We have, 2x – 3y = 4
$\Rightarrow$ 3x = 2x – 4
$\Rightarrow$ $y = {{2x - 4} \over 3}$
When x = 2, $y = {{4 - 4} \over 3} = 0$; when x = – 1, $y = {{ - 2 - 4} \over 3}$$= {{ - 6} \over 2} = - 2$
Thus, we have the following table : Plotting the points C(2, 0) and D(–1,–2) on the same graph paper and drawing a line joining them, we obtain the graph of the equation. Clearly, two lines intersect at the point $P\left( {{{19} \over 5},{6 \over 5}} \right)$.
Hence, $x = {{19} \over 5},y = {6 \over 5}$ is the solution of the given system.
The method of elimination is the most efficient in this case.

(ii) By elimination method :
The given system of equation is
3x + 4y = 10 ...(1)
and, 2x – 2y = 2 ...(2)
Multiplying (2) by 2 and adding to (1), we get
3x + 4y + 4x – 4y = 10 +4
$\Rightarrow$ 7x = 14
$\Rightarrow$ x = 2
Putting x = 2 in (1), we get
3(2) + 4y = 10
$\Rightarrow$ 4y = 10 – 6 = 4
$\Rightarrow$ y = 1
Hence, x = 2 y = 1
By substitution method :
The given system of equaitons is
3x + 4y = 10 ...(1)
and, 2x – 2y = 2
$\Rightarrow$ x – y = 1 ...(2)
From (2), y = x – 1
Substituting y = x – 1 in (1), we get
3x + 4(x – 1) = 10
$\Rightarrow$ 3x + 4x – 4 = 10
$\Rightarrow$ 7x = 14
$\Rightarrow$x = 2
Putting x = 2 in (1), we get
3(2) + 4y = 10
$\Rightarrow$ 4y = 10 – 6 = 4
$\Rightarrow$ y = 1
Hence, x = 2, y = 1
By geometrical method :
The given system of equation is
3x + 4y = 10 ...(1)
and, 2x – 2y = 2
$\Rightarrow$ x – y = 1 ...(2)
For the graph of 3x + 4y = 10
We have, 3x + 4y = 10
$\Rightarrow$ 4y = 10 – 3x
$\Rightarrow$ $y = {{10 - 3x} \over 4}$
When x = 2, $y = {{10 - 6} \over 4} = {4 \over 4} = 1$; when x = – 2, $y = {{10 + 6} \over 4} = {{16} \over 4} = 4$
Thus, we have the following table : Plotting the points A(2, 1) and B(–2, 4) and drawing a line joining them, we get the graph of the equation 3x + 4y = 10.
For the graph of x – y = 1 :
We have, x – y = 1
$\Rightarrow$ y = x – 1
When x = 3, y = 2; When x = 0, y = – 1
Thus, we have the following table : Plotting the points C(3, 2) and D(0, – 1) on the same graph paper and drawing a line joining them, we obtain the graph of the equation. Clearly, two lines intersect at the point A(2, 1).
Hence, x = 2, y = 1
In this case, almost all the methods are equally efficient, however, geometrical method takes more time.

(iii) By elimination method :
The given system of equations is
3x – 5y – 4 = 0
$\Rightarrow$ 3x – 5y = 4 ....(1)
and 9x = 2y + 7
$\Rightarrow$ 9x – 2y = 7 ...(2)
Multiplying (1) by 3, we get
9x – 15y = 12 ...(3)
Sibtracting (2) from (3), we get
– 13y = 5
$\Rightarrow$ $y = - {5 \over {13}}$
Putting $y = - {5 \over {13}}$ in (1), we get
3x –5$\left( {{{ - 5} \over {13}}} \right)$ = 4
$\Rightarrow$ ${{3x + 25} \over {13}} = 4$
3x = 4 – ${{25} \over {13}}$
$\Rightarrow$ $3x = {{52 - 25} \over {13}}$
$\Rightarrow 3x = {{27} \over {13}}$
$\Rightarrow$ $x = {9 \over {13}}$
Hence, $x = {9 \over {13}}$, $y = - {5 \over {13}}$
By, substitution method :
The given system of equations is
3x – 5y – 4 = 0
$\Rightarrow$ 3x – 5y = 4 ...(1)

and, 9x = 2y + 7
$\Rightarrow$ 9x – 2y = 7....(2)

From (2), 2y = 9x – 7
$\Rightarrow$ y = ${{9x - 7} \over 2}$

Substituting y = ${{9x - 7} \over 2}$ in (1), we get
$3x - 5\left( {{{9x - 7} \over 2}} \right) = 4$
$\Rightarrow$ 6x – 45x + 35 = 8

$\Rightarrow$ – 39x = 8 – 35
$\Rightarrow$ – 39x = – 27

$\Rightarrow$ x = ${{ - 27} \over { - 39}} = {9 \over {13}}$
Putting x = ${9 \over {13}}$ in (2), we get
$3 \times {9 \over {13}}-5y = 4$
$\Rightarrow 5y = {{27} \over {13}}-4$

$\Rightarrow 5y = {{27 - 52} \over {13}} = {{ - 25} \over {13}}$
$\Rightarrow y = {{ - 5} \over {13}}$
Hence, $x = {9 \over {13}}$ $y = - {5 \over {13}}$
By geometrical method :
The given system of equations is
3x – 5y – 4 = 0
$\Rightarrow$ 3x – 5y = 4 ...(1)

and, 9x = 2y + 7
$\Rightarrow$ 9x – 2y = 7...(2)

For the graph of 3x – 5y = 4 :
We have, 3x – 5y = 4
$\Rightarrow$ 5y = 3x – 4
$\Rightarrow$ $y = {{3x - 4} \over 5}$]

When x = 3 y = ${{9 - 4} \over 5} = {5 \over 5}$ = 1;
When x = – 2, y = ${{ - 6 - 4} \over 5} = {{ - 10} \over 5} = - 2$
Thus, we have the following table : Plotting the points A(3, 1) and B(–2,–2) and drawing a line joining them, we get the graph of the equation 3x – 5y = 4.
For the graph of 9x – 2y = 7 :
We have, 9x – 2y = 7
$\Rightarrow$ 2y = 9x – 7

$\Rightarrow$ $y = {{9x - 7} \over 2}$
When x = 1, $y = {{9 - 7} \over 2} = {2 \over 2} = 1$ ;
When x = 3, $y = {{27 - 7} \over 2} = {20 \over 2} = 10$ ;

Thus, we have the following table : Plotting the points C(1, 1) and D(3, 10) on the same graph paper and drawing a line joining them, we obtain the graph of equation. Clearly, two lines intersect at the point $P\left( {{9 \over {13}},{{ - 5} \over {13}}} \right)$
Hence, $x = {9 \over {13}},y = {{ - 5} \over {13}}$
In this case, the method of elimination is the most efficient.

(iv) By elimination method ;
The given system of equations is
${x \over 2} + {{2y} \over 3} = - 1$
$\Rightarrow$ 3x + 4y = – 6 ...(1)

and, $x - {y \over 3} = 3$
$\Rightarrow$ 3x – y = 9...(2)

Multiplying (2) by 4 and adding to (1), we get
15x = 30
$\Rightarrow$ x = 2

Putting x = 2 in (2), we get
3(2) – y = 9
$\Rightarrow$ –y = 9 – 6 = 3

$\Rightarrow$ y = – 3
Hence, x = 2, y = – 3
By substitution method :
${x \over 2} + {{2y} \over 3} = -1$
$\Rightarrow$ 3x + 4y = – 6 ...(1)

and, $x - {y \over 3} = 3$
$\Rightarrow$ 3x – y = 9 ...(2)

From (2), y = 3x – 9
Putting y = 3x – 9 in (1), we get
3x + 4(3x – 9) = – 6
$\Rightarrow$ 3x + 12x – 36 = – 6

$\Rightarrow$ 15x = 30
$\Rightarrow$ x = 2

Putting x = 2 in (2), we get
3(2) – y = 9
$\Rightarrow$ – y = 9 – 6 = 3

$\Rightarrow$ y = – 3
Hence, x = 2, y = – 3
By geometrical method :
The given system of equations is
${x \over 2} + {{2y} \over 3} = - 1$
$\Rightarrow$ 3x + 4y = – 6 ...(1)

and, $x - {y \over 3} = 3$
$\Rightarrow$ 3x – y = 9  ...(2)
For the graph of 3x + 4y = – 6 :
We have,
3x + 4y = – 6
$\Rightarrow$ 4y = – 6 – 3x

$\Rightarrow$ y = ${{ - 6 - 3x} \over 4}$
When x = 2, y = ${{ - 6 - 6} \over 4} = {{ - 12} \over 4} = - 3;$
when x = – 2, y = ${{ - 6 + 6} \over 4} = {0 \over 4} = 0$
Thus, we have the following table : Plotting the points A(2,–3) and B(–2, 0) and drawing a line joining them, we get the graph of the equation 3x +4y = – 6.
For the graph of 3x – y = 9 :
We have, 3x – y = 9
$\Rightarrow$ y = 3x – 9

When x = 3, y = 6
$\Rightarrow$ y = 3x – 9

When x = 3, y = 9 – 9 = 0;
When x = 4, y = 12 – 9 = 3
Thus, we have the following table : Plotting the points C(3, 0) and D(4, 3) on the same graph paper and drawing a line joining them, we obtain the graph of the equation. Clearly, the two lines intersect at A(2, –3).
Hence, x = 2, y = – 3
In this case, almost all the methods are equally efficient, however, geometrical method takes more time.

Q.2      Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ${1 \over 2}$ if we only add 1 to the denominator. What is the fraction ?
(ii) Five year ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?

(iii) The sum of the digits of a two - digit number 9. Also, nine times this number is twice the number obtained by reversingthe order of the digits. Find the number.

(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs.100 notes only. Meena got 25 notes in all. Find how man notes of Rs. 50 and RS 100 she received.

(v) A lending library has a fixed charge for first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book she kept for seven days, while Susy paid Rs. 21 for book she kept for five days. Find the fixed charge and the charge for each extra day.
Sol.       (i) Let x be the numerator and y be the denominator of the fraction. So, the fraction is ${x \over y}$.
By given conditions : ${{x + 1} \over {y - 1}} = 1$
$\Rightarrow x + 1 = y - 1$

$\Rightarrow$ x – y = – 2...(1)
and, ${x \over {y + 1}} = {1 \over 2}$
$\Rightarrow$ 2x = y + 1 ...(2)

$\Rightarrow$ 2x – y = 1
Subtracting (2) from (1), we get
(x – y) – (2x – y) = – 2 – 1
$\Rightarrow$ x – y – 2x + y = – 3
$\Rightarrow$ – x = – 3
$\Rightarrow$ x = 3

Substituting x = 3 in (1), we get
3 – y = –2
$\Rightarrow$ y = 5

Hence, the required fraction is ${3 \over 5}$.

(ii) Let the present age of Nuri = x years
and, the present age of Sonu = y years
Five years ago, Nuri's age = (x – 5) years
Sonu's age = (y – 5) years
As per conditions, x – 5 = 3(y – 5)
$\Rightarrow$ x – 5 = 3y – 15
$\Rightarrow$ x – 3y = – 15 + 5
$\Rightarrow$ x – 3y = – 10 ...(1)
Ten years later, Nuri's age = (x + 10) years
Sonu's age = (y + 10)
As per condition, x + 10 = 2(y + 10)
$\Rightarrow$ x + 10 = 2y + 20
$\Rightarrow$ x – 2y = 20 – 10
$\Rightarrow$ x – 2y = 10 ...(2)
Subtracting (2) from (1), we get
– y = – 20
$\Rightarrow$ y = 20
Putting y = 20 in (2), we get
x – 2(20) = 10
$\Rightarrow$ x = 10 + 40 = 50
Therefore, Nuri's present age = 50 years
and Sonu's present age = 20 years

(iii) Let the digits in the units's place and ten's place be x and y respectively.
Therefore, Number = 10y + x
If the digits age reversed, the new number = 10x + y
As per conditions : x + y = 9 ....(1)
and, 9(10y + x) = 2(10x + y)
$\Rightarrow$ 90y + 9x = 20x +2y
$\Rightarrow$ 20x – 9x + 2y – 90y = 0
$\Rightarrow$ 11x – 88 y = 0 ...(2)
From (1), y = 9 – x
Putting y = 9 – x in (2), we get
11x –88(9 – x) = 0
$\Rightarrow$ 11x – 88 × 9 + 88x = 0
$\Rightarrow$ 99x = 88 × 9
$\Rightarrow$ x = ${{88 \times 9} \over {99}} = 8$
Putting x = 8 in (1), we get
8 + y = 9
$\Rightarrow$ y = 1

Hence, the number = 10y + x = 10 × 1 + 8 = 18

(iv) Let the number of Rs. 50 notes be x and number of Rs. 100 notes be y. Then,
x + y = 25
and, 50x + 100y = 2000
$\Rightarrow$ x + 2y = 40

On subtracting (1), from (2), we get
y = 15
Putting y = 15 in (1), we get
x + 15 = 25
$\Rightarrow$ x = 10
Hence, number of Rs. 50 notes = 10
and, number of Rs. 100 notes = 15

(v) Let the fixed charge for 3 days be Rs. x and charge per gday be Rs. y.
Therefore, By the given conditions,
x + 4y = 27
and, x + 2y = 21
Subtracting (2) form (1), we get
2y = 6
$\Rightarrow$ y = 3

Put y = 3 in (1), we get
x + 4 × 3 = 27
$\Rightarrow$ x = 27 – 12 = 15
Fixed charges = Rs. 15
and, charges per day = Rs. 3