| Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
Exercise 3.3
Question 1 A
Solve the following system of equations by elimination method:
3x – 5y – 4 = 09x = 2y + 7
Sol :
Given pair of linear equations is
3x – 5y – 4 = 0 …(i)
And 9x = 2y + 7 …(ii)
On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as
9x – 15y – 12 = 0 …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
9x – 15y – 12 – 9x = 0 – 2y – 7
⇒ – 15y + 2y = – 7 + 12
⇒ – 13y = 5
$\Rightarrow \mathrm{y}=-\frac{5}{13}$
On putting $y=-\frac{5}{13}$ in Eq. (ii), we get
$9 x=2\left(-\frac{5}{13}\right)+7$
$\Rightarrow 9 x=-\frac{10}{13}+7$
$\Rightarrow 9 x=\frac{-10+91}{13}$
$\Rightarrow 9 x=\frac{81}{13}$
$\Rightarrow x=\frac{81}{13 \times 9}$
$\Rightarrow x=\frac{9}{13}$
Hence, $x=\frac{9}{13}$ and $=-\frac{5}{13}$ , which is the required solution.
2x – 2y = 2
Sol :
Given pair of linear equations is
3x + 4y = 10 …(i)
And 2x – 2y = 2 …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as
6x + 8y = 20 …(iii)
6x – 6y = 6 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x – 6y – 6x – 8y = 6 – 20
⇒ – 14y = – 14
⇒ y = 1
On putting y = 1 in Eq. (ii), we get
2x – 2(1) = 2
⇒ 2x – 2 = 2
⇒ x = 2
Hence, x = 2 and y = 1 , which is the required solution.
2x – 3y = 4
Sol :
Given pair of linear equations is
x + y = 5 …(i)
And 2x – 3y = 4 …(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
2x + 2y = 10 …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
2x + 2y – 2x + 3y = 10 – 4
⇒ 5y = 6
$\Rightarrow \mathrm{y}=\frac{6}{5}$
$9 x=2\left(-\frac{5}{13}\right)+7$
$\Rightarrow 9 x=-\frac{10}{13}+7$
$\Rightarrow 9 x=\frac{-10+91}{13}$
$\Rightarrow 9 x=\frac{81}{13}$
$\Rightarrow x=\frac{81}{13 \times 9}$
$\Rightarrow x=\frac{9}{13}$
Hence, $x=\frac{9}{13}$ and $=-\frac{5}{13}$ , which is the required solution.
Question 1 B
Solve the following system of equations by elimination method:
3x + 4y = 102x – 2y = 2
Sol :
Given pair of linear equations is
3x + 4y = 10 …(i)
And 2x – 2y = 2 …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as
6x + 8y = 20 …(iii)
6x – 6y = 6 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x – 6y – 6x – 8y = 6 – 20
⇒ – 14y = – 14
⇒ y = 1
On putting y = 1 in Eq. (ii), we get
2x – 2(1) = 2
⇒ 2x – 2 = 2
⇒ x = 2
Hence, x = 2 and y = 1 , which is the required solution.
Question 1 C
Solve the following system of equations by elimination method:
x + y = 52x – 3y = 4
Sol :
Given pair of linear equations is
x + y = 5 …(i)
And 2x – 3y = 4 …(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
2x + 2y = 10 …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
2x + 2y – 2x + 3y = 10 – 4
⇒ 5y = 6
$\Rightarrow \mathrm{y}=\frac{6}{5}$
On putting $y=\frac{6}{5}$ in Eq. (i), we get
x + y = 5
$\Rightarrow x+\frac{6}{5}=5$
$\Rightarrow x=\frac{25-6}{5}$
$\Rightarrow x=\frac{19}{5}$
Hence, $x=\frac{19}{5}$ and $=\frac{6}{5}$ , which is the required solution.
4x + 6y = 7
Sol :
Given pair of linear equations is
2x + 3y = 8 …(i)
And 4x + 6y = 7 …(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
4x + 6y = 16 …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
4x + 6y – 4x – 6y = 16 – 7
⇒ 0 = 9
Which is a false equation involving no variable.
So, the given pair of linear equations has no solution i.e. this pair of linear equations is inconsistent.
3x + 2y = 4
Sol :
Given pair of linear equations is
8x + 5y = 9 …(i)
And 3x + 2y = 4 …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of y equal, we get the equation as
16x + 10y = 18 …(iii)
15x + 10y = 20 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
15x + 10y – 16x – 10y = 20 – 18
⇒ – x = 2
⇒ x = – 2
On putting x = – 2 in Eq. (ii), we get
3x + 2y = 4
⇒ 3( – 2) + 2y = 4
⇒ – 6 + 2y = 4
⇒ 2y = 4 + 6
⇒ 2y = 10
$\Rightarrow \mathrm{y}=\frac{10}{2}=5$
Hence, x=-2 and =5, which is the required solution.
3x + 5y = 74
Sol :
Given pair of linear equations is
2x + 3y = 46 …(i)
And 3x + 5y = 74 …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as
6x + 9y = 138 …(iii)
6x + 10y = 148 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x + 10y – 6x – 9y = 148 – 138
⇒ y = 10
On putting y = 10 in Eq. (ii), we get
3x + 5y = 74
⇒ 3x + 5(10) = 74
⇒ 3x + 50 = 74
⇒ 3x = 74 – 50
⇒ 3x = 24
⇒ x = 8
Hence, x = 8 and y = 10 , which is the required solution.
0.3x + 0.2y = 0.9
Sol :
Given pair of linear equations is
0.4x – 1.5y = 6.5 …(i)
And 0.3x + 0.2y = 0.9 …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as
1.2x – 4.5y = 19.5 …(iii)
1.2x + 0.8y = 3.6 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
1.2x + 0.8y – 1.2x + 4.5y = 3.6 – 19.5
⇒ 5.3y = – 15.9
$\Rightarrow \mathrm{y}=-\frac{15.9}{5.3}$
⇒ y = – 3
x + y = 5
$\Rightarrow x+\frac{6}{5}=5$
$\Rightarrow x=\frac{25-6}{5}$
$\Rightarrow x=\frac{19}{5}$
Hence, $x=\frac{19}{5}$ and $=\frac{6}{5}$ , which is the required solution.
Question 1 D
Solve the following system of equations by elimination method:
2x + 3y = 84x + 6y = 7
Sol :
Given pair of linear equations is
2x + 3y = 8 …(i)
And 4x + 6y = 7 …(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
4x + 6y = 16 …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
4x + 6y – 4x – 6y = 16 – 7
⇒ 0 = 9
Which is a false equation involving no variable.
So, the given pair of linear equations has no solution i.e. this pair of linear equations is inconsistent.
Question 1 E
Solve the following system of equations by elimination method:
8x + 5y = 93x + 2y = 4
Sol :
Given pair of linear equations is
8x + 5y = 9 …(i)
And 3x + 2y = 4 …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of y equal, we get the equation as
16x + 10y = 18 …(iii)
15x + 10y = 20 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
15x + 10y – 16x – 10y = 20 – 18
⇒ – x = 2
⇒ x = – 2
On putting x = – 2 in Eq. (ii), we get
3x + 2y = 4
⇒ 3( – 2) + 2y = 4
⇒ – 6 + 2y = 4
⇒ 2y = 4 + 6
⇒ 2y = 10
$\Rightarrow \mathrm{y}=\frac{10}{2}=5$
Hence, x=-2 and =5, which is the required solution.
Question 1 F
Solve the following system of equations by elimination method:
2x + 3y = 463x + 5y = 74
Sol :
Given pair of linear equations is
2x + 3y = 46 …(i)
And 3x + 5y = 74 …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as
6x + 9y = 138 …(iii)
6x + 10y = 148 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x + 10y – 6x – 9y = 148 – 138
⇒ y = 10
On putting y = 10 in Eq. (ii), we get
3x + 5y = 74
⇒ 3x + 5(10) = 74
⇒ 3x + 50 = 74
⇒ 3x = 74 – 50
⇒ 3x = 24
⇒ x = 8
Hence, x = 8 and y = 10 , which is the required solution.
Question 1 G
Solve the following system of equations by elimination method:
0.4x – 1.5y = 6.50.3x + 0.2y = 0.9
Sol :
Given pair of linear equations is
0.4x – 1.5y = 6.5 …(i)
And 0.3x + 0.2y = 0.9 …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as
1.2x – 4.5y = 19.5 …(iii)
1.2x + 0.8y = 3.6 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
1.2x + 0.8y – 1.2x + 4.5y = 3.6 – 19.5
⇒ 5.3y = – 15.9
$\Rightarrow \mathrm{y}=-\frac{15.9}{5.3}$
⇒ y = – 3
On putting y = – 3 in Eq. (ii), we get
0.3x + 0.2y = 0.9
⇒ 0.3x + 0.2( – 3) = 0.9
⇒ 0.3x – 0.6 = 0.9
⇒ 0.3x = 1.5
⇒ x = 1.5/0.3
⇒ x = 5
Hence, x = 5 and y = – 3 , which is the required solution.
Question 1 H
Solve the following system of equations by elimination method:
√2x – √3y = 0√5x + √2y = 0
Sol :
Given pair of linear equations is
√2 x – √3 y = 0 …(i)
And √5 x + √2 y = 0 …(ii)
On multiplying Eq. (i) by √2 and Eq. (ii) by √3 to make the coefficients of y equal, we get the equation as
2x – √6 y = 0 …(iii)
√15 x + √6 y = 0 …(iv)
On adding Eq. (iii) and (iv), we get
2x – √6 y + √15 x + √6 y = 0
⇒ 2x + √15 x = 0
⇒ x(2 + √15) = 0
⇒ x = 0
On putting x = 0 in Eq. (i), we get
√2 x – √3 y = 0
⇒ √2(0) – √3 y = 0
⇒ – √3 y = 0
⇒ y = 0
Hence, x = 0 and y = 0 , which is the required solution.
Question 1 I
Solve the following system of equations by elimination method:
2x + 5y = 12x + 3y = 3
Sol :
Given pair of linear equations is
2x + 5y = 1 …(i)
And 2x + 3y = 3 …(ii)
On subtracting Eq. (ii) from Eq. (i), we get
2x + 3y – 2x – 5y = 3 – 1
⇒ – 2y = 2
⇒ y = – 1
On putting y = – 1 in Eq. (ii), we get
2x + 3( – 1) = 3
⇒ 2x – 3 = 3
⇒ 2x = 6
⇒ x = 6/2
⇒ x = 3
Hence, x = 3 and y = – 1 , which is the required solution.
Question 2 A
Solve the following system of equations by elimination method:
$3 x-\frac{8}{y}=5$$x-\frac{y}{3}=3$
Sol :
Given pair of linear equations is
$\frac{x}{2}+\frac{2 y}{3}=-1$ …(i)
And $x-\frac{y}{3}=3$ …(ii)
On multiplying Eq. (ii) by 2 to make the coefficients of y equal, we get the equation as
$2 x-\frac{2 y}{3}=6$ …(iii)
On adding Eq. (i) and Eq. (ii), we get
$\frac{x}{2}+\frac{2 y}{3}+x-\frac{2 y}{3}=-1+6$
$\Rightarrow \frac{\mathrm{x}}{2}+2 \mathrm{x}=5$
$\Rightarrow \frac{x+4 x}{2}=5$
$\Rightarrow \frac{5 x}{2}=5$
⇒ x = 2
$\frac{x}{2}+\frac{2 y}{3}+x-\frac{2 y}{3}=-1+6$
$\Rightarrow \frac{\mathrm{x}}{2}+2 \mathrm{x}=5$
$\Rightarrow \frac{x+4 x}{2}=5$
$\Rightarrow \frac{5 x}{2}=5$
⇒ x = 2
On putting x=2 in Eq. (ii), we get
$x-\frac{y}{3}=3$
$\Rightarrow 2-\frac{y}{3}=3$
$\Rightarrow-\frac{y}{3}=3-2$
$\Rightarrow-\frac{y}{3}=1$
⇒ y = – 3
Hence, x = 2 and y = – 3 , which is the required solution.
$\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$
Sol :
Given pair of linear equations is
$\frac{x}{6}+\frac{y}{15}=4$ …(i)
And $\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$ …(ii)
On multiplying Eq. (ii) by $\frac{1}{2}$ to make the coefficients of x equal, we get the equation as
$\frac{x}{6}-\frac{y}{24}=\frac{19}{8}$ …(iii)
$x-\frac{y}{3}=3$
$\Rightarrow 2-\frac{y}{3}=3$
$\Rightarrow-\frac{y}{3}=3-2$
$\Rightarrow-\frac{y}{3}=1$
⇒ y = – 3
Hence, x = 2 and y = – 3 , which is the required solution.
Question 2 B
Solve the following system of equations by elimination method:
$\frac{x}{6}+\frac{y}{15}=4$$\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$
Sol :
Given pair of linear equations is
$\frac{x}{6}+\frac{y}{15}=4$ …(i)
And $\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$ …(ii)
On multiplying Eq. (ii) by $\frac{1}{2}$ to make the coefficients of x equal, we get the equation as
$\frac{x}{6}-\frac{y}{24}=\frac{19}{8}$ …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
$\frac{x}{6}-\frac{y}{24}-\frac{x}{6}-\frac{y}{15}=\frac{19}{8}-4$
$\Rightarrow-\frac{y}{24}-\frac{y}{15}=\frac{19-32}{8}$
$\Rightarrow \frac{-5 y-8 y}{120}=\frac{19-32}{8}$
$\Rightarrow \frac{-5 y-8 y}{120}=\frac{19-32}{8}$
$\Rightarrow \frac{-13 y}{120}=\frac{-13}{8}$
$\Rightarrow \mathrm{y}=\frac{13}{8} \times \frac{120}{13}$
⇒ y = 15
⇒ y = 15
On putting y = 15 in Eq. (ii), we get
$\frac{x}{6}+\frac{y}{15}=4$
$\Rightarrow \frac{x}{6}+\frac{15}{15}=4$
$\Rightarrow \frac{x}{6}=4-1$
$\Rightarrow \frac{x}{6}=3$
⇒ x = 18
Hence, x = 18 and y = 15 , which is the required solution.
$3 x-\frac{8}{y}=5$
Sol :
Given pair of linear equations is
$x+\frac{6}{y}=6$ …(i)
And $3 x-\frac{8}{y}=5$ …(ii)
On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as
$3 x+\frac{18}{y}=18$ …(iii)
$\frac{x}{6}+\frac{y}{15}=4$
$\Rightarrow \frac{x}{6}+\frac{15}{15}=4$
$\Rightarrow \frac{x}{6}=4-1$
$\Rightarrow \frac{x}{6}=3$
⇒ x = 18
Hence, x = 18 and y = 15 , which is the required solution.
Question 2 C
Solve the following system of equations by elimination method:
$x+\frac{6}{y}=6$$3 x-\frac{8}{y}=5$
Sol :
Given pair of linear equations is
$x+\frac{6}{y}=6$ …(i)
And $3 x-\frac{8}{y}=5$ …(ii)
On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as
$3 x+\frac{18}{y}=18$ …(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
$3 x+\frac{18}{y}-3 x+\frac{8}{y}=18-5$
$\Rightarrow \frac{26}{\mathrm{y}}=13$
⇒ y = 2
$3 x+\frac{18}{y}-3 x+\frac{8}{y}=18-5$
$\Rightarrow \frac{26}{\mathrm{y}}=13$
⇒ y = 2
On putting y = 2 in Eq. (i), we get
$x+\frac{6}{y}=6$
$\Rightarrow x+\frac{6}{2}=6$
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
43x + 37y = 117
Sol :
Given pair of linear equations is
37x + 43y = 123 …(i)
And 43x + 37y = 117 …(ii)
On multiplying Eq. (i) by 43 and Eq. (ii) by 37 to make the coefficients of x equal, we get the equation as
1591x + 1849y = 5289 …(iii)
1591x + 1369y = 4329 …(iv)
$x+\frac{6}{y}=6$
$\Rightarrow x+\frac{6}{2}=6$
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
Question 3 A
Solve the following equations by elimination method:
37x + 43y = 12343x + 37y = 117
Sol :
Given pair of linear equations is
37x + 43y = 123 …(i)
And 43x + 37y = 117 …(ii)
On multiplying Eq. (i) by 43 and Eq. (ii) by 37 to make the coefficients of x equal, we get the equation as
1591x + 1849y = 5289 …(iii)
1591x + 1369y = 4329 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 1591x + 1369y – 1591x – 1849y = 4329 – 5289
⇒ – 480y = – 960
$\Rightarrow \mathrm{y}=\frac{960}{480}$
⇒ y = 2
⇒ 1591x + 1369y – 1591x – 1849y = 4329 – 5289
⇒ – 480y = – 960
$\Rightarrow \mathrm{y}=\frac{960}{480}$
⇒ y = 2
On putting y = 2 in Eq. (ii), we get
⇒ 43x + 37(2) = 117 ⇒ 43x + 74 = 117
⇒ 43x = 117 – 74
⇒ 43x = 43
⇒ x = 1
Hence, x = 1 and y = 2 , which is the required solution.
131x + 217y = 827
Sol :
Given pair of linear equations is
217x + 131y = 913 …(i)
And 131x + 217y = 827 …(ii)
On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as
28427x + 17161y = 119603 …(iii)
28427x + 47089y = 179459 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 28427x + 47089y – 28427x – 17161y = 179459 – 119603
⇒ 47089y – 17161y = 179459 – 119603
⇒ 29928y = 59856
$\Rightarrow \mathrm{y}=\frac{59856}{29928}$
⇒ y = 2
On putting y = 2 in Eq. (ii), we get
⇒ 131x + 217(2) = 827 ⇒ 131x + 434 = 827
⇒ 131x = 393
$\Rightarrow x=\frac{393}{131}$
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
101x + 99y = 501
Sol :
Given pair of linear equations is
99x + 101y = 499 …(i)
And 101x + 99y = 501 …(ii)
On multiplying Eq. (i) by 101 and Eq. (ii) by 99 to make the coefficients of x equal, we get the equation as
9999x + 10201y = 50399 …(iii)
9999x + 9801y = 49599 …(iv)
⇒ 43x + 37(2) = 117 ⇒ 43x + 74 = 117
⇒ 43x = 117 – 74
⇒ 43x = 43
⇒ x = 1
Hence, x = 1 and y = 2 , which is the required solution.
Question 3 B
Solve the following equations by elimination method:
217x + 131y = 913131x + 217y = 827
Sol :
Given pair of linear equations is
217x + 131y = 913 …(i)
And 131x + 217y = 827 …(ii)
On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as
28427x + 17161y = 119603 …(iii)
28427x + 47089y = 179459 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 28427x + 47089y – 28427x – 17161y = 179459 – 119603
⇒ 47089y – 17161y = 179459 – 119603
⇒ 29928y = 59856
$\Rightarrow \mathrm{y}=\frac{59856}{29928}$
⇒ y = 2
On putting y = 2 in Eq. (ii), we get
⇒ 131x + 217(2) = 827 ⇒ 131x + 434 = 827
⇒ 131x = 393
$\Rightarrow x=\frac{393}{131}$
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
Question 3 C
Solve the following equations by elimination method:
99x + 101y = 499101x + 99y = 501
Sol :
Given pair of linear equations is
99x + 101y = 499 …(i)
And 101x + 99y = 501 …(ii)
On multiplying Eq. (i) by 101 and Eq. (ii) by 99 to make the coefficients of x equal, we get the equation as
9999x + 10201y = 50399 …(iii)
9999x + 9801y = 49599 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 9999x + 9801y – 9999x – 10201y = 49599 – 50399
⇒ 9801y – 10201y = 49599 – 50399
⇒ – 400y = – 800
$\Rightarrow \mathrm{y}=\frac{800}{400}$
⇒ y = 2
⇒ 9999x + 9801y – 9999x – 10201y = 49599 – 50399
⇒ 9801y – 10201y = 49599 – 50399
⇒ – 400y = – 800
$\Rightarrow \mathrm{y}=\frac{800}{400}$
⇒ y = 2
On putting y = 2 in Eq. (i), we get
⇒ 99x + 101(2) = 499 ⇒ 99x + 202 = 499
⇒ 99x = 297
⇒ x = 297/99
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
23x – 29y = 98
Sol :
Given pair of linear equations is
29x – 23y = 110 …(i)
And 23x – 29y = 98 …(ii)
On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as
667x – 529y = 2530 …(iii)
667x – 841y = 2842 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 667x – 841y – 667x + 529y = 2842 – 2530
⇒ – 312y = 312
⇒ y = – 1
On putting y = 2 in Eq. (ii), we get
⇒ 29x – 23( – 1) = 110 ⇒ 29x + 23 = 110
⇒ 29x = 110 – 23
⇒ 29x = 87
⇒ x = 3
Hence, x = 3 and y = – 1 , which is the required solution.
$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=7, \mathrm{x} \neq 0, \mathrm{y} \neq 0$
Sol :
Given pair of linear equations is
$\frac{1}{x}-\frac{1}{y}=1$ …(i)
And $\frac{1}{x}+\frac{1}{y}=7$ …(ii)
⇒ 99x + 101(2) = 499 ⇒ 99x + 202 = 499
⇒ 99x = 297
⇒ x = 297/99
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
Question 3 D
Solve the following equations by elimination method:
29x – 23y = 11023x – 29y = 98
Sol :
Given pair of linear equations is
29x – 23y = 110 …(i)
And 23x – 29y = 98 …(ii)
On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as
667x – 529y = 2530 …(iii)
667x – 841y = 2842 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
⇒ 667x – 841y – 667x + 529y = 2842 – 2530
⇒ – 312y = 312
⇒ y = – 1
On putting y = 2 in Eq. (ii), we get
⇒ 29x – 23( – 1) = 110 ⇒ 29x + 23 = 110
⇒ 29x = 110 – 23
⇒ 29x = 87
⇒ x = 3
Hence, x = 3 and y = – 1 , which is the required solution.
Question 4 A
Solve the following system of equations by elimination method:
$\frac{1}{x}-\frac{1}{y}=1$$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=7, \mathrm{x} \neq 0, \mathrm{y} \neq 0$
Sol :
Given pair of linear equations is
$\frac{1}{x}-\frac{1}{y}=1$ …(i)
And $\frac{1}{x}+\frac{1}{y}=7$ …(ii)
Adding Eq. (i) and Eq. (ii), we get
$\frac{1}{x}-\frac{1}{y}+\frac{1}{x}+\frac{1}{y}=1+7$
$\Rightarrow \frac{1}{x}+\frac{1}{x}=8$
$\Rightarrow \frac{2}{x}=8$
$\Rightarrow \frac{2}{8}=x$
$\Rightarrow x=\frac{1}{4}$
$\frac{1}{x}-\frac{1}{y}+\frac{1}{x}+\frac{1}{y}=1+7$
$\Rightarrow \frac{1}{x}+\frac{1}{x}=8$
$\Rightarrow \frac{2}{x}=8$
$\Rightarrow \frac{2}{8}=x$
$\Rightarrow x=\frac{1}{4}$
On putting $\mathrm{X}=\frac{1}{4}$ in Eq. (ii), we get
$\frac{1}{x}+\frac{1}{y}=7$
$\Rightarrow \frac{1}{\frac{1}{4}}+\frac{1}{y}=7$
$\frac{1}{x}+\frac{1}{y}=7$
$\Rightarrow \frac{1}{\frac{1}{4}}+\frac{1}{y}=7$
$\Rightarrow 4+\frac{1}{\mathrm{y}}=7$
$\Rightarrow \frac{1}{y}=7-4=3$
$\Rightarrow \mathrm{y}=\frac{1}{3}$
Hence, $x=\frac{1}{4}$ and $=\frac{1}{3}$ , which is the required solution.
$\frac{5}{x}-\frac{4}{y}=-2, x \neq 0, y \neq 0$
Sol :
Given pair of linear equations is
$\frac{2}{x}+\frac{3}{y}=13$ …(i)
And $\frac{5}{x}-\frac{4}{y}=-2$ …(ii)
$\Rightarrow \mathrm{y}=\frac{1}{3}$
Hence, $x=\frac{1}{4}$ and $=\frac{1}{3}$ , which is the required solution.
Question 4 B
Solve the following system of equations by elimination method:
$\frac{2}{x}+\frac{3}{y}=13$$\frac{5}{x}-\frac{4}{y}=-2, x \neq 0, y \neq 0$
Sol :
Given pair of linear equations is
$\frac{2}{x}+\frac{3}{y}=13$ …(i)
And $\frac{5}{x}-\frac{4}{y}=-2$ …(ii)
On multiplying Eq. (i) by 5 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as
$\frac{10}{x}+\frac{15}{y}=65$ …(iii)
$\frac{10}{x}-\frac{8}{y}=-4$ …(iv)
$\frac{10}{x}+\frac{15}{y}=65$ …(iii)
$\frac{10}{x}-\frac{8}{y}=-4$ …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
$\frac{10}{x}-\frac{8}{y}-\frac{10}{x}-\frac{15}{y}=-4-65$
$\Rightarrow-\frac{8}{y}-\frac{15}{y}=-69$
$\Rightarrow-\frac{23}{\mathrm{y}}=-69$
$\Rightarrow \frac{23}{69}=\mathrm{y}$
$\Rightarrow \mathrm{y}=\frac{1}{2}$
$\frac{10}{x}-\frac{8}{y}-\frac{10}{x}-\frac{15}{y}=-4-65$
$\Rightarrow-\frac{8}{y}-\frac{15}{y}=-69$
$\Rightarrow-\frac{23}{\mathrm{y}}=-69$
$\Rightarrow \frac{23}{69}=\mathrm{y}$
$\Rightarrow \mathrm{y}=\frac{1}{2}$
On putting $y=\frac{1}{2}$ in Eq. (ii), we get
$\frac{5}{x}-\frac{4}{y}=-2$
$\Rightarrow \frac{5}{x}-\frac{4}{\frac{1}{2}}=7$
$\frac{5}{x}-\frac{4}{y}=-2$
$\Rightarrow \frac{5}{x}-\frac{4}{\frac{1}{2}}=7$
$\Rightarrow \frac{5}{x}-8=7$
$\Rightarrow \frac{5}{x}=15$
$\Rightarrow x=\frac{1}{3}$
Hence, $x=\frac{1}{3}$ and $=\frac{1}{2}$ , which is the required solution.
Given pair of linear equations is
$\frac{4}{x}+\frac{7}{y}=11$ …(i)
And $\frac{3}{x}-\frac{5}{y}=-2$ …(ii)
Hence, $x=\frac{1}{3}$ and $=\frac{1}{2}$ , which is the required solution.
Question 4 C
Solve the following system of equations by elimination method:
$\frac{3 a}{x}-\frac{2 b}{y}+5=0$ , $\frac{a}{x}+\frac{3 b}{y}-2=0$, $(\mathrm{x} \neq 0, \mathrm{y} \neq 0)$
Sol :Given pair of linear equations is
$\frac{4}{x}+\frac{7}{y}=11$ …(i)
And $\frac{3}{x}-\frac{5}{y}=-2$ …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as
$\frac{12}{x}+\frac{21}{y}=33$ …(iii)
$\frac{12}{x}-\frac{20}{y}=-8$ …(iv)
$\frac{12}{x}+\frac{21}{y}=33$ …(iii)
$\frac{12}{x}-\frac{20}{y}=-8$ …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
$\frac{12}{x}-\frac{20}{y}-\frac{12}{x}-\frac{21}{y}=-8-33$
$\Rightarrow-\frac{20}{\mathrm{y}}-\frac{21}{\mathrm{y}}=-41$
$\Rightarrow-\frac{41}{\mathrm{y}}=-41$
$\Rightarrow \frac{41}{41}=\mathrm{y}$
$\Rightarrow-\frac{20}{\mathrm{y}}-\frac{21}{\mathrm{y}}=-41$
$\Rightarrow-\frac{41}{\mathrm{y}}=-41$
$\Rightarrow \frac{41}{41}=\mathrm{y}$
⇒y=1
On putting y=1 in Eq. (ii), we get
$\frac{3}{x}-\frac{5}{y}=-2$
$\Rightarrow \frac{3}{x}-5=-2$
$\Rightarrow \frac{3}{x}=-2+5$
$\Rightarrow \frac{3}{x}=3$
Hence, $x=\frac{1}{3}$ and $=\frac{1}{2}$ , which is the required solution.
Given pair of linear equations is
$\frac{3 a}{x}-\frac{2 b}{y}=-5$ …(i)
And $\frac{a}{x}+\frac{3 b}{y}=2$ …(ii)
⇒x=1
Question 4 D
Solve the following system of equations by elimination method:
$\frac{3 a}{x}-\frac{2 b}{y}+5=0$ , $\frac{a}{x}+\frac{3 b}{y}-2=0$, $(\mathrm{x} \neq 0, \mathrm{y} \neq 0)$
Sol :Given pair of linear equations is
$\frac{3 a}{x}-\frac{2 b}{y}=-5$ …(i)
And $\frac{a}{x}+\frac{3 b}{y}=2$ …(ii)
On multiplying Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as
$\frac{3 a}{x}+\frac{9 b}{y}=6$ …(iii)
$\frac{3 a}{x}+\frac{9 b}{y}=6$ …(iii)
On subtracting Eq. (i) from Eq. (iii), we get
$\frac{3 a}{x}+\frac{9 b}{y}-\frac{3 a}{x}+\frac{2 b}{y}=6-(-5)$
$\Rightarrow \frac{9 b}{y}+\frac{2 b}{y}=11$
$\Rightarrow \frac{11 \mathrm{b}}{\mathrm{y}}=11$
⇒ y = b
$\frac{3 a}{x}+\frac{9 b}{y}-\frac{3 a}{x}+\frac{2 b}{y}=6-(-5)$
$\Rightarrow \frac{9 b}{y}+\frac{2 b}{y}=11$
$\Rightarrow \frac{11 \mathrm{b}}{\mathrm{y}}=11$
⇒ y = b
On putting y = b in Eq. (ii), we get
$\frac{a}{x}+\frac{3 b}{y}=2$
$\Rightarrow \frac{a}{x}+\frac{3 b}{b}=2$
$\Rightarrow \frac{a}{x}=2-3$
$\Rightarrow \frac{\mathrm{a}}{\mathrm{x}}=-1$
⇒ x = – a
⇒ x = – a
Hence, x = – a and y = b , which is the required solution.
Question 5 A
Solve the following system of equations by elimination method:
$\frac{2 x+5 y}{x y}=6$, $\frac{4 x-5 y}{x y}=-3$, where x ≠ 0 and y ≠ 0
Sol :Given pair of linear equations is
$\frac{2 x+5 y}{x y}=6$
Or 2x + 5y = 6xy …(i)
And $\frac{4 x-5 y}{x y}=-3$
And $\frac{4 x-5 y}{x y}=-3$
Or 4x – 5y = – 3xy …(ii)
On adding Eq. (i) and Eq. (ii), we get
2x + 5y + 4x – 5y = 6xy – 3xy
⇒ 6x = 3xy
$\Rightarrow \frac{6 x}{3 x}=y$
⇒ y = 2 and x = 0
2x + 5y + 4x – 5y = 6xy – 3xy
⇒ 6x = 3xy
$\Rightarrow \frac{6 x}{3 x}=y$
⇒ y = 2 and x = 0
On putting y = 2 in Eq. (ii), we get
2x + 5(2) = 6xy
⇒ 2x + 10 = 6x(2)
⇒ 2x + 10 = 12x ⇒ 2x – 12x = – 10
⇒ – 10x = – 10
⇒ x = 1
2x + 5(2) = 6xy
⇒ 2x + 10 = 6x(2)
⇒ 2x + 10 = 12x ⇒ 2x – 12x = – 10
⇒ – 10x = – 10
⇒ x = 1
On putting x = 0 , we get y = 0
Hence, x = 0,1 and y = 0,2 , which is the required solution.
Question 5 B
Solve the following system of equations by elimination method:
x + y = 2xyx – y = 6xy
Sol :
Given pair of linear equations is
x + y = 2xy …(i)
And x – y = 6xy …(ii)
On adding Eq. (i) and Eq. (ii), we get
x + y + x – y = 2xy + 6xy
⇒ 2x = 8xy
$\Rightarrow \frac{2 \mathrm{x}}{8 \mathrm{x}}=\mathrm{y}$
$\Rightarrow \mathrm{y}=\frac{1}{4}$ and x=0
On putting $y=\frac{1}{4}$ in Eq. (ii), we get
$x-\frac{1}{4}=6 x y$
$\Rightarrow \frac{4 x-1}{4}=6 x\left(\frac{1}{4}\right)$
⇒ 4x – 1 = 6x ⇒ – 1 = 6x – 4x
⇒ – 1 = 2x
$\Rightarrow x=-\frac{1}{2}$
On putting $y=\frac{1}{4}$ in Eq. (ii), we get
$x-\frac{1}{4}=6 x y$
$\Rightarrow \frac{4 x-1}{4}=6 x\left(\frac{1}{4}\right)$
⇒ 4x – 1 = 6x ⇒ – 1 = 6x – 4x
⇒ – 1 = 2x
$\Rightarrow x=-\frac{1}{2}$
On putting x = 0 , we get y = 0
Hence, $x=-\frac{1}{2}, 0$ and $=\frac{1}{4}, 0$ , which is the required solution.
Question 5 C
Solve the following system of equations by elimination method:
5x + 3y = 19xy7x – 2y = 8xy
Sol :
Given pair of linear equations is
5x + 3y = 19xy …(i)
And 7x – 2y = 8xy …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of y equal, we get the equation as
10x + 6y = 38xy …(iii)
And 21x – 6y = 24xy …(iv)
On adding Eq. (i) and Eq. (ii), we get
10x + 6y + 21x – 6y = 38xy + 24xy
⇒ 31x = 62xy
$\Rightarrow \frac{31 \mathrm{x}}{62 \mathrm{x}}=\mathrm{y}$
$\Rightarrow \mathrm{y}=\frac{1}{2}$ and x=0
On putting $y=\frac{1}{2}$ in Eq. (ii), we get
7x – 2y = 8xy
$\Rightarrow 7 x-2\left(\frac{1}{2}\right)=8 x\left(\frac{1}{2}\right)$
⇒ 7x – 1 = 4x ⇒ – 1 = 4x – 7x
⇒ – 1 = – 3x
$\Rightarrow x=\frac{1}{3}$
7x – 2y = 8xy
$\Rightarrow 7 x-2\left(\frac{1}{2}\right)=8 x\left(\frac{1}{2}\right)$
⇒ 7x – 1 = 4x ⇒ – 1 = 4x – 7x
⇒ – 1 = – 3x
$\Rightarrow x=\frac{1}{3}$
On putting x = 0 , we get y = 0
Hence, $x=\frac{1}{3}, 0$ and $=\frac{1}{2}, 0$ , which is the required solution.
2x – 3y = – xy
Sol :
Given pair of linear equations is
x + y = 7xy …(i)
And 2x – 3y = – xy …(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
2x + 2y = 14xy …(iii)
On subtracting Eq. (ii) and Eq. (iii), we get
2x + 2y – 2x + 3y = 14xy + xy
⇒ 2y + 3y = 15xy
⇒ 5y = 15xy
$\Rightarrow \frac{5 y}{15 y}=x$
Question 5 D
Solve the following system of equations by elimination method:
x + y = 7xy2x – 3y = – xy
Sol :
Given pair of linear equations is
x + y = 7xy …(i)
And 2x – 3y = – xy …(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
2x + 2y = 14xy …(iii)
On subtracting Eq. (ii) and Eq. (iii), we get
2x + 2y – 2x + 3y = 14xy + xy
⇒ 2y + 3y = 15xy
⇒ 5y = 15xy
$\Rightarrow \frac{5 y}{15 y}=x$
$\Rightarrow x=\frac{1}{3}$ and y=0
On putting $x=\frac{1}{3}$ in Eq. (ii), we get
2x – 3y = – xy
$\Rightarrow 2\left(\frac{1}{3}\right)-3 y=6\left(\frac{1}{3}\right) y$
$\Rightarrow\left(\frac{2}{3}\right)-3 y=2 y \Rightarrow\left(\frac{2}{3}\right)=5 y$
$\Rightarrow \mathrm{y}=\frac{2}{15}$
2x – 3y = – xy
$\Rightarrow 2\left(\frac{1}{3}\right)-3 y=6\left(\frac{1}{3}\right) y$
$\Rightarrow\left(\frac{2}{3}\right)-3 y=2 y \Rightarrow\left(\frac{2}{3}\right)=5 y$
$\Rightarrow \mathrm{y}=\frac{2}{15}$
On putting x = 0 , we get x = 0
Hence, $x=\frac{1}{3}, 0$ and $=\frac{2}{15}, 0$ , which is the required solution.
Where (2x + 3y) ≠ 0 and (3x – 2y) ≠ 0
Sol :
Given pair of linear equations is
$\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2}$ …(i)
And $\frac{7}{(2 x+3 y)}+\frac{4}{(3 x-2 y)}=2$ …(ii)
Hence, $x=\frac{1}{3}, 0$ and $=\frac{2}{15}, 0$ , which is the required solution.
Question 6 A
Solve for x and y the following system of equations:
$\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2}$ , $\frac{7}{(2 x+3 y)}+\frac{4}{(3 x-2 y)}=2$
Where (2x + 3y) ≠ 0 and (3x – 2y) ≠ 0
Sol :
Given pair of linear equations is
$\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2}$ …(i)
And $\frac{7}{(2 x+3 y)}+\frac{4}{(3 x-2 y)}=2$ …(ii)
On multiplying Eq. (i) by 7 and Eq. (ii) by $\frac{1}{2}$ to make the coefficients equal of first term, we get the equation as
$\frac{7}{2(2 x+3 y)}+\frac{7 \times 12}{7(3 x-2 y)}=\frac{7}{2}$
$\Rightarrow \frac{7}{2(2 x+3 y)}+\frac{12}{(3 x-2 y)}=\frac{7}{2}$ …(iii)
$\frac{7}{2(2 x+3 y)}+\frac{4}{2(3 x-2 y)}=\frac{2}{2}$
$\frac{7}{2(2 x+3 y)}+\frac{7 \times 12}{7(3 x-2 y)}=\frac{7}{2}$
$\Rightarrow \frac{7}{2(2 x+3 y)}+\frac{12}{(3 x-2 y)}=\frac{7}{2}$ …(iii)
$\frac{7}{2(2 x+3 y)}+\frac{4}{2(3 x-2 y)}=\frac{2}{2}$
$\Rightarrow \frac{7}{2(2 x+3 y)}+\frac{2}{(3 x-2 y)}=1$ …(iv)
On substracting Eq. (iii) from Eq. (iv), we get
$\frac{7}{2(2 x+3 y)}+\frac{2}{(3 x-2 y)}-\frac{7}{2(2 x+3 y)}-\frac{12}{(3 x-2 y)}=1-\frac{7}{2}$
$\Rightarrow \frac{2}{(3 x-2 y)}-\frac{12}{(3 x-2 y)}=1-\frac{7}{2}$
$\Rightarrow \frac{-10}{(3 x-2 y)}=\frac{2-7}{2}$
$\Rightarrow \frac{-10}{(3 x-2 y)}=\frac{-5}{2}$
$\Rightarrow \frac{1}{(3 x-2 y)}=\frac{5}{2 \times 10}$
$\frac{7}{2(2 x+3 y)}+\frac{2}{(3 x-2 y)}-\frac{7}{2(2 x+3 y)}-\frac{12}{(3 x-2 y)}=1-\frac{7}{2}$
$\Rightarrow \frac{2}{(3 x-2 y)}-\frac{12}{(3 x-2 y)}=1-\frac{7}{2}$
$\Rightarrow \frac{-10}{(3 x-2 y)}=\frac{2-7}{2}$
$\Rightarrow \frac{-10}{(3 x-2 y)}=\frac{-5}{2}$
$\Rightarrow \frac{1}{(3 x-2 y)}=\frac{5}{2 \times 10}$
⇒3x-2y=4
⇒3x=4+2y
$\Rightarrow \mathrm{x}=\frac{4+2 \mathrm{y}}{3}$ …(a)
$\Rightarrow \mathrm{x}=\frac{4+2 \mathrm{y}}{3}$ …(a)
On multiplying Eq. (ii) by $\frac{3}{7}$ to make the coefficients equal of second term, we get the equation as
$\frac{7 \times 3}{7(2 x+3 y)}+\frac{4 \times 3}{7(3 x-2 y)}=\frac{2 \times 3}{7}$
$\Rightarrow \frac{3}{(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{6}{7}$ …(v)
$\frac{7 \times 3}{7(2 x+3 y)}+\frac{4 \times 3}{7(3 x-2 y)}=\frac{2 \times 3}{7}$
$\Rightarrow \frac{3}{(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{6}{7}$ …(v)
On substracting Eq. (i) from Eq. (iv), we get
$\frac{3}{(2 x+3 y)}+\frac{12}{7(3 x-2 y)}-\frac{1}{2(2 x+3 y)}-\frac{12}{7(3 x-2 y)}=\frac{6}{7}-\frac{1}{2}$
$\Rightarrow \frac{3}{(2 x+3 y)}-\frac{1}{2(2 x+3 y)}=\frac{6}{7}-\frac{1}{2}$
$\Rightarrow \frac{6-1}{2(2 x+3 y)}=\frac{12-7}{14}$
$\Rightarrow \frac{5}{2(2 x+3 y)}=\frac{5}{14}$
$\Rightarrow \frac{1}{(2 x+3 y)}=\frac{2}{14}$
$\frac{3}{(2 x+3 y)}+\frac{12}{7(3 x-2 y)}-\frac{1}{2(2 x+3 y)}-\frac{12}{7(3 x-2 y)}=\frac{6}{7}-\frac{1}{2}$
$\Rightarrow \frac{3}{(2 x+3 y)}-\frac{1}{2(2 x+3 y)}=\frac{6}{7}-\frac{1}{2}$
$\Rightarrow \frac{6-1}{2(2 x+3 y)}=\frac{12-7}{14}$
$\Rightarrow \frac{5}{2(2 x+3 y)}=\frac{5}{14}$
$\Rightarrow \frac{1}{(2 x+3 y)}=\frac{2}{14}$
⇒2x+3y=7
⇒$\mathrm{X}=\frac{7-3 \mathrm{y}}{2}$ …(b)
From Eq. (a) and (b), we get
$\frac{4+2 y}{3}=\frac{7-3 y}{2}$
⇒ 2(4 + 2y) = 3(7 – 3y)
⇒ 8 + 4y = 21 – 9y
⇒ 4y + 9y = 21 – 8
⇒ 13y = 13
⇒ y = 1
$\frac{4+2 y}{3}=\frac{7-3 y}{2}$
⇒ 2(4 + 2y) = 3(7 – 3y)
⇒ 8 + 4y = 21 – 9y
⇒ 4y + 9y = 21 – 8
⇒ 13y = 13
⇒ y = 1
On putting the value of y = 1 in Eq. (b), we get
$\Rightarrow x=\frac{7-3(1)}{2}$
$\Rightarrow x=\frac{4}{2}=2$
$\Rightarrow x=\frac{7-3(1)}{2}$
$\Rightarrow x=\frac{4}{2}=2$
Hence, x = 2 and y = 1 , which is the required solution.
$\frac{3}{x-1}+\frac{2}{y+1}=\frac{13}{6}, x \neq 1, y \neq 1$
Sol :
Given pair of linear equations is
$\frac{2}{x-1}+\frac{3}{y+1}=2$ …(i)
And $\frac{3}{x-1}+\frac{2}{y+1}=\frac{13}{6}$ …(ii)
Question 6 B
Solve for x and y the following system of equations:
$\frac{6}{x-1}-\frac{3}{y-2}=1, x \neq 1, y \neq 1$$\frac{3}{x-1}+\frac{2}{y+1}=\frac{13}{6}, x \neq 1, y \neq 1$
Sol :
Given pair of linear equations is
$\frac{2}{x-1}+\frac{3}{y+1}=2$ …(i)
And $\frac{3}{x-1}+\frac{2}{y+1}=\frac{13}{6}$ …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients equal of first term, we get the equation as
$\frac{6}{x-1}+\frac{9}{y+1}=6$ …(iii)
$\frac{6}{x-1}+\frac{9}{y+1}=6$ …(iii)
$\frac{6}{x-1}+\frac{4}{y+1}=\frac{13}{3}$ …(iv)
On substracting Eq. (iii) from Eq. (iv), we get
$\frac{6}{x-1}+\frac{4}{y+1}-\frac{6}{x-1}-\frac{9}{y+1}=\frac{13}{3}-6$
$\Rightarrow \frac{4}{y+1}-\frac{9}{y+1}=\frac{13}{3}-6$
$\Rightarrow \frac{-5}{y+1}=\frac{13-18}{3}$
$\Rightarrow \frac{-5}{y+1}=\frac{-5}{3}$
$\Rightarrow \frac{1}{y+1}=\frac{1}{3}$
⇒ y + 1 = 3
⇒ y = 3 – 1
⇒ y = 2
$\frac{6}{x-1}+\frac{4}{y+1}-\frac{6}{x-1}-\frac{9}{y+1}=\frac{13}{3}-6$
$\Rightarrow \frac{4}{y+1}-\frac{9}{y+1}=\frac{13}{3}-6$
$\Rightarrow \frac{-5}{y+1}=\frac{13-18}{3}$
$\Rightarrow \frac{-5}{y+1}=\frac{-5}{3}$
$\Rightarrow \frac{1}{y+1}=\frac{1}{3}$
⇒ y + 1 = 3
⇒ y = 3 – 1
⇒ y = 2
On putting the value of y = 2 in Eq. (ii), we get
$\frac{6}{x-1}+\frac{4}{2+1}=\frac{13}{3}$
$\Rightarrow \frac{6}{x-1}+\frac{4}{3}=\frac{13}{3}$
$\Rightarrow \frac{6}{x-1}=\frac{13}{3}-\frac{4}{3}$
$\Rightarrow \frac{6}{x-1}=\frac{9}{3}$
$\Rightarrow \frac{1}{x-1}=\frac{9}{3 \times 6}$
⇒ x – 1 = 2
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
$\frac{55}{x+y}+\frac{40}{x-y}=13, x+y \neq 0, x-y \neq 0$
Sol :
Given pair of linear equations is
$\frac{44}{x+y}+\frac{30}{x-y}=10$ …(i)
And $\frac{55}{x+y}+\frac{40}{x-y}=13$ …(ii)
$\frac{6}{x-1}+\frac{4}{2+1}=\frac{13}{3}$
$\Rightarrow \frac{6}{x-1}+\frac{4}{3}=\frac{13}{3}$
$\Rightarrow \frac{6}{x-1}=\frac{13}{3}-\frac{4}{3}$
$\Rightarrow \frac{6}{x-1}=\frac{9}{3}$
$\Rightarrow \frac{1}{x-1}=\frac{9}{3 \times 6}$
⇒ x – 1 = 2
⇒ x = 3
Hence, x = 3 and y = 2 , which is the required solution.
Question 6 C
Solve for x and y the following system of equations:
$\frac{44}{x+y}+\frac{30}{x-y}=10$$\frac{55}{x+y}+\frac{40}{x-y}=13, x+y \neq 0, x-y \neq 0$
Sol :
Given pair of linear equations is
$\frac{44}{x+y}+\frac{30}{x-y}=10$ …(i)
And $\frac{55}{x+y}+\frac{40}{x-y}=13$ …(ii)
On multiplying Eq. (i) by 4 and Eq. (ii) by 3 to make the coefficients equal of second term, we get the equation as
$\frac{176}{x+y}+\frac{120}{x-y}=40$ …(iii)
$\frac{176}{x+y}+\frac{120}{x-y}=40$ …(iii)
$\frac{165}{x+y}+\frac{120}{x-y}=39$ …(iv)
On substracting Eq. (iii) from Eq. (iv), we get
$\frac{176}{x+y}+\frac{120}{x-y}-\frac{165}{x+y}-\frac{120}{x-y}=40-39$
$\Rightarrow \frac{176-165}{x+y}=1$
$\Rightarrow \frac{11}{x+y}=1$
⇒ x + y = 11 …(a)
$\frac{176}{x+y}+\frac{120}{x-y}-\frac{165}{x+y}-\frac{120}{x-y}=40-39$
$\Rightarrow \frac{176-165}{x+y}=1$
$\Rightarrow \frac{11}{x+y}=1$
⇒ x + y = 11 …(a)
On putting the value of x + y = 11 in Eq. (1), we get
$\Rightarrow \frac{44}{11}+\frac{30}{x-y}=10$
$\Rightarrow \frac{30}{x-y}=10-4$
⇒ 6(x – y) = 30
⇒ x – y = 5 …(b)
$\Rightarrow \frac{44}{11}+\frac{30}{x-y}=10$
$\Rightarrow \frac{30}{x-y}=10-4$
⇒ 6(x – y) = 30
⇒ x – y = 5 …(b)
Adding Eq. (a) and (b), we get
⇒ 2x = 16
⇒ x = 8
⇒ 2x = 16
⇒ x = 8
On putting value of x = 8 in eq. (a), we get
8 + y = 11
⇒ y = 3
Hence, x = 8 and y = 3 , which is the required solution.
Sol :
Given pair of linear equations is
$\frac{5}{x-1}+\frac{1}{y-2}=2$ …(i)
And $\frac{6}{x-1}-\frac{3}{y-2}=1$ …(ii)
8 + y = 11
⇒ y = 3
Hence, x = 8 and y = 3 , which is the required solution.
Question 6 D
Solve for x and y the following system of equations:
$\frac{5}{x-1}+\frac{1}{y-2}=2$ ,$\frac{6}{x-1}-\frac{3}{y-2}=1$Sol :
Given pair of linear equations is
$\frac{5}{x-1}+\frac{1}{y-2}=2$ …(i)
And $\frac{6}{x-1}-\frac{3}{y-2}=1$ …(ii)
On multiplying Eq. (i) by 3 to make the coefficients equal of second term, we get the equation as
$\frac{15}{x-1}+\frac{3}{y-2}=6$ …(iii)
$\frac{15}{x-1}+\frac{3}{y-2}=6$ …(iii)
On adding Eq. (ii) and Eq. (iii), we get
$\frac{6}{x-1}-\frac{3}{y-2}+\frac{15}{x-1}+\frac{3}{y-2}=6+1$
$\Rightarrow \frac{6}{x-1}-\frac{15}{x-1}=7$
$\Rightarrow \frac{21}{x-1}=7$
⇒ x – 1 = 3
⇒ x = 3 + 1
⇒ x = 4
$\frac{6}{x-1}-\frac{3}{y-2}+\frac{15}{x-1}+\frac{3}{y-2}=6+1$
$\Rightarrow \frac{6}{x-1}-\frac{15}{x-1}=7$
$\Rightarrow \frac{21}{x-1}=7$
⇒ x – 1 = 3
⇒ x = 3 + 1
⇒ x = 4
On putting the value of x = 4 in Eq. (ii), we get
$\frac{6}{4-1}-\frac{3}{y-2}=1$
$\Rightarrow 2-\frac{3}{y-2}=1$
$\Rightarrow-\frac{3}{y-2}=1-2$
$\Rightarrow-\frac{3}{y-2}=-1$
⇒ (y – 2) = 3
⇒ y = 3 + 2
⇒ y = 5
Hence, x = 4 and y = 5 , which is the required solution.
Sol :
Let the present age of father i.e. Aftab = x yr
And the present age of his daughter = y yr
Seven years ago,
Aftab’s age = (x – 7)yr
Daughter’s age = (y – 7)yr
According to the question,
(x – 7) = 7(y – 7)
⇒ x – 7 = 7y – 49
⇒ x – 7y = – 42 …(i)
After three years,
Aftab’s age = (x + 3)yr
Daughter’s age = (y + 3)yr
According to the question,
(x + 3) = 3(y + 3)
⇒ x + 3 = 3y + 9
⇒ x – 3y = 6 …(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x – 3y – x + 7y = 6 – ( – 42)
⇒ – 3y + 7y = 6 + 42
⇒ 4y = 48
⇒ y = 12
On putting y = 12 in Eq. (ii) we get
x – 3(12) = 6
⇒ x – 36 = 6
⇒ x = 42
Hence, the age of Aftab is 42years and age of his daughter is 12years.
Sol :
Let the present age of Nuri = x yr
And the present age of Sonu = y yr
Five years ago,
Nuri’s age = (x – 5)yr
Sonu’s age = (y – 5)yr
According to the question,
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = – 10 …(i)
After ten years,
Aftab’s age = (x + 10)yr
Daughter’s age = (y + 10)yr
$\frac{6}{4-1}-\frac{3}{y-2}=1$
$\Rightarrow 2-\frac{3}{y-2}=1$
$\Rightarrow-\frac{3}{y-2}=1-2$
$\Rightarrow-\frac{3}{y-2}=-1$
⇒ (y – 2) = 3
⇒ y = 3 + 2
⇒ y = 5
Hence, x = 4 and y = 5 , which is the required solution.
Question 7 A
Form the pair of linear equations for the following problems and find their solution by elimination method:
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." Find their present ages.Sol :
Let the present age of father i.e. Aftab = x yr
And the present age of his daughter = y yr
Seven years ago,
Aftab’s age = (x – 7)yr
Daughter’s age = (y – 7)yr
According to the question,
(x – 7) = 7(y – 7)
⇒ x – 7 = 7y – 49
⇒ x – 7y = – 42 …(i)
After three years,
Aftab’s age = (x + 3)yr
Daughter’s age = (y + 3)yr
According to the question,
(x + 3) = 3(y + 3)
⇒ x + 3 = 3y + 9
⇒ x – 3y = 6 …(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x – 3y – x + 7y = 6 – ( – 42)
⇒ – 3y + 7y = 6 + 42
⇒ 4y = 48
⇒ y = 12
On putting y = 12 in Eq. (ii) we get
x – 3(12) = 6
⇒ x – 36 = 6
⇒ x = 42
Hence, the age of Aftab is 42years and age of his daughter is 12years.
Question 7 B
Form the pair of linear equations for the following problems and find their solution by elimination method:
Five years 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?Sol :
Let the present age of Nuri = x yr
And the present age of Sonu = y yr
Five years ago,
Nuri’s age = (x – 5)yr
Sonu’s age = (y – 5)yr
According to the question,
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = – 10 …(i)
After ten years,
Aftab’s age = (x + 10)yr
Daughter’s age = (y + 10)yr
According to the question,
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 10 …(ii)
Now, we can solve this by an elimination method
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 10 …(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x – 2y – x + 3y = 10 – ( – 10)
⇒ – 2y + 3y = 10 + 10
⇒ y = 20
x – 2y – x + 3y = 10 – ( – 10)
⇒ – 2y + 3y = 10 + 10
⇒ y = 20
On putting y = 20 in Eq. (i) we get
x – 3(20) = – 10
⇒ x – 60 = – 10
⇒ x = 50
Hence, the age of Nuri is 50 years and age of Sonu is 20 years.
Sol :
Let the one number = x
And the other number = y
According to the question,
x – y = 26 …(i)
and x = 3y
or x – 3y = 0 …(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x – 3y – x + y = 0 – 26
⇒ – 3y + y = – 26
⇒ – 2y = – 26
⇒ y = 13
On putting y = 13 in Eq. (ii) we get
x – 3(13) = 0
⇒ x – 39 = 0
⇒ x = 39
Hence, the two numbers are 39 and 13.
x – 3(20) = – 10
⇒ x – 60 = – 10
⇒ x = 50
Hence, the age of Nuri is 50 years and age of Sonu is 20 years.
Question 7 C
Form the pair of linear equations for the following problems and find their solution by elimination method:
The difference between two numbers is 26 and one number is three times the other. Find them.Sol :
Let the one number = x
And the other number = y
According to the question,
x – y = 26 …(i)
and x = 3y
or x – 3y = 0 …(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x – 3y – x + y = 0 – 26
⇒ – 3y + y = – 26
⇒ – 2y = – 26
⇒ y = 13
On putting y = 13 in Eq. (ii) we get
x – 3(13) = 0
⇒ x – 39 = 0
⇒ x = 39
Hence, the two numbers are 39 and 13.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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