10类NCERT解决方案-第3章两个变量的线性方程对–练习3.6

问题1.通过将以下方程对简化为一对线性方程来求解它们：

(一世) $\frac{1}{2x} + \frac{1}{3y} = 2$

$\frac{1}{3x} + \frac{1}{2y}= \frac{13}{6}$

解决方案：

Lets, take 1/x = a and 1/y = b

Here, the two given equation will be as follows:

$\frac{a}{2}$ + $\frac{b}{3}$ = 2 

Multiply it by 6, we get

3a + 2b = 12            -(1)

and,

$\frac{a}{3}$ + $\frac{b}{2}$ = $\frac{13}{6}$           

Multiply it by 6, we get

2a + 3b = 13            -(2)

Now, by using Elimination method,

Multiply eq(1) by 2 and multiply eq(2) by 3, and then subtract them

5b = 15

b = 3

Now putting b = 3 in eq(1), we get

3a + 2(3) = 12

a = 6/3 

a = 2

So, Now As

a = 1/x = 2

x = 1/2 

b = 1/y = 3

y = 1/3 

(ii) $\frac{2}{√x} + \frac{3}{√y} = 2$

$\frac{4}{√x} - \frac{9}{√y} = -1$

解决方案：

Lets, take 2/√x = a and 3/√y = b

Here, the two given equation will be as follows:

a + b = 2            -(1)

and,

2a – 3b =-1            -(2)

Now, by using Elimination method,

Multiply eq(1) by 3, and then add them

5a = 5

a = 1

Now putting a = 1 in eq(1), we get

1 + b = 2

b = 1

So, Now As

a = 2/√x = 1

√x = 2

x = 4

b = 3/√y = 1

√x = 3

y = 9

(iii) $\frac{4}{x}$ + 3年= 14

$\frac{3}{x}$ – 4y = 23

解决方案：

Lets, take 1/x = a

Here, the two given equation will be as follows:

4a + 3y = 14            -(1)

and,

3a – 4y = 23             -(2)

Now, by using Elimination method,

Multiply eq(1) by 3 and multiply eq(2) by 4, and then subtract them

-25y = 50

y = -2

Now putting y = -2 in eq(1), we get

4a + 3(-2) = 14

4a = 20

a = 5

So, Now As

a = 1/x  = 5

x = 1/5 

y = -2

(iv) $\frac{5}{x-1} + \frac{1}{y-2} = 2$

$\frac{6}{x-1} - \frac{3}{y-2} = 1$

解决方案：

Lets, take $\frac{1}{x-1}$ = a and, $\frac{1}{y-2}$ = b

Here, the two given equation will be as follows:

5a + b = 2            -(1)

and,

6a – 3b = 1             -(2)

Now, by using Elimination method,

Multiply eq(1) by 3, and then add them

21a = 7

a = 1/3 

Now putting a = 1/3  in eq(1), we get

5(1/3) + b = 2

b = 2 – 5/3

b = 1/3 

So, Now As

a = $\frac{1}{x-1} = \frac{1}{3}$

x – 1 = 3

x = 4

b = $\frac{1}{y-2} = \frac{1}{3}$

y – 2 = 3

y = 5

(v) $\frac{7x-2y}{xy} = 5$

$\frac{8x+7y}{xy}=15$

解决方案：

$\frac{7}{y} - \frac{2}{x}$ = 5

$\frac{8}{y} + \frac{7}{x}$ = 15

Lets, take 1/x = a and 1/y = b

Here, the two given equation will be as follows:

7b – 2a = 5             -(1)

and,

8b + 7a = 15             -(2)

Now, by using Elimination method,

Multiply eq(1) by 7, multiply eq(2) by 2 and then add them

65b = 65

b = 1

Now putting b = 1 in eq(1), we get

7(1) – 2a = 5

2a = 7 – 5

a = 1

So, Now As

a = 1/x = 1

x = 1

b = 1/y = 1

y = 1

(vi)6x + 3y = 6xy

2x + 4y = 5xy

解决方案：

Divide both the equations by xy, we get

$\frac{6}{y} + \frac{3}{x}$ = 6

$\frac{2}{y} + \frac{4}{x}$ = 5

Lets, take 1/x = a and, 1/y = b

Here, the two given equation will be as follows:

6b + 3a = 6 

Divide the above equation by 2,

2b + a = 2             -(1)

and,

2b + 4a = 5             -(2)

Now, by using Elimination method,

Subtract eq(1) from eq(2), we get

3a = 3

a = 1

Now putting a = 1 in eq(1), we get

2b + 1 = 2

b = 1/2 

So, Now As

a = 1/x = 1

x = 1

b = 1/y = 1/2 

y = 2

(vii) $\frac{10}{x+y} + \frac{2}{x-y}=4$

$\frac{15}{x+y} - \frac{5}{x-y}=-2$

解决方案：

Lets, take $\frac{1}{x+y}$ = a and $\frac{1}{x-y}$ = b

Here, the two given equation will be as follows:

10a + 2b = 4

Divide the above equation by 2,

5a + b = 2             -(1)

and,

15a – 5b = -2             -(2)

Now, by using Elimination method,

Multiply eq(1) by 3 and subtract them,

8b = 8

b = 1

Now putting b = 1 in eq(1), we get

5a + 1 = 2

a = 1/5 

So, Now As

a = $\frac{1}{x+y}$ = $\frac{1}{5}$

x + y = 5             -(3)

b = $\frac{1}{x-y}$ = 1

x – y = 1             -(4)

By adding eq(3) and (4), we get

2x = 6

x = 3 and y = 2

(viii) $\frac{1}{3x+y} + \frac{1}{3x-y}= \frac{3}{4}$

$\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = \frac{-1}{8}$

解决方案：

Lets, take $\frac{1}{3x+y}$ = a

and, $\frac{1}{3x-y}$ = b

Here, the two given equation will be as follows:

a + b = 3/4           -(1)

and,

$\frac{a}{2} - \frac{b}{2} = \frac{-1}{8}$

Multiply it by 2, we get

a – b = -1/4           -(2)

Now, by using Elimination method,

Add eq(1) and eq(1), we get

2a = 1/2 

a = 1/4 

Now putting a = 1/4 in eq(1), we get

$\frac{1}{4}$ + b = $\frac{3}{4}$

b = 1/2 

So, Now As

a =$\frac{1}{3x+y} = \frac{1}{4}$

3x + y = 4           -(3)

b = $\frac{1}{3x-y} = \frac{1}{2}$

3x – y = 2            -(4)

By adding eq(3) and eq(4), we get

6x = 6

 x = 1 and y = 1

问题2.将以下问题表达为一对方程，从而找到它们的解决方案：

(i)Ritu可以在2小时内向下游划20公里，在2小时内向上游划4公里。找到她在静水中划船的速度和水流的速度。

解决方案：

Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = (x + y) km/h

Upstream = (x – y) km/h

As Speed = $\frac{Distance}{Time}$

According to the given question,

x + y = 20/2 

x + y = 10           -(1)

and, 

x – y = 4/2 

x – y = 2            -(2)

Add eq(1) and eq(2), we get

2x = 12

x = 6 and y = 4

Hence, speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii)2名女性和5名男性可以在4天内一起完成刺绣工作，而3名女性和6名男性可以在3天内完成刺绣工作。找出一位女性独自完成工作所花费的时间，以及一位男性独自所花费的时间。

解决方案：

Let’s take, 

The total number of days taken by women to finish the work = x

The total number of days taken by men to finish the work = y

Work done by women in one day will be = 1/x 

Work done by women in one day will be = 1/y

So, according to the question

4($\frac{2}{x} + \frac{5}{y}$) = 1

And, 3($\frac{3}{x} + \frac{6}{y}$) = 1

Lets, take 1/x = a and,  1/y = b

Here, the two given equation will be as follows:

4(2a + 5b) = 1 

8a + 20b = 1            -(1)

and,

3(3a + 6b) = 1 

9a + 18b = 1            -(2)

Now, by using Cross multiplication method,

$\frac{a}{20-18} = \frac{b}{9-8} = \frac{1}{180-144}$

$\frac{a}{2} = \frac{b}{1} = \frac{1}{36}$

a = $\frac{2}{36} = \frac{1}{18}$

b = 1/36

So, Now As

a =$\frac{1}{x} = \frac{1}{18}$

x = 18

b = $\frac{1}{y} = \frac{1}{36}$

y = 36

Hence, number of days taken by women to finish the work = 18 days

Number of days taken by men to finish the work = 36 days.

(iii)鲁希(Roohi)部分地乘火车和部分巴士前往她家300公里。如果乘火车旅行60公里，然后乘公共汽车旅行，则需要4个小时。如果她乘火车旅行100公里，其余乘公共汽车旅行，则需要花费10分钟以上的时间。分别找到火车和公共汽车的速度。

解决方案：

Lets, take

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

$\frac{60}{x} + \frac{240}{y}$ = 4

$\frac{100}{x} + \frac{200}{y} = \frac{25}{6}$

Lets, take 1/x = a and 1/y = b

Here, the two given equation will be as follows:

60a + 240b = 4

Divide it by 4, we get

15a + 60b = 1           -(1)

and,

100a + 200b = 25/6 

Divide it by 25/6, we get

24a + 48b = 1           -(2)

Now, by using Cross multiplication method,

$\frac{a}{60-48} = \frac{b}{24-15} = \frac{1}{1440-720}$

$\frac{a}{12} = \frac{b}{9} = \frac{1}{720}$

a = $\frac{12}{720}$ = $\frac{1}{60}$

b = $\frac{9}{720}$ = $\frac{1}{80}$

So, Now As

a = $\frac{1}{x}$ = $\frac{1}{60}$

x = 60

b = $\frac{1}{y}$ = $\frac{1}{80}$

y = 80

Hence, speed of the train = 60 km/h

Speed of the bus = 80 km/h