10类NCERT解决方案–第3章两个变量的线性方程对–练习3.3

问题1.用代换法求解下面的线性方程对

(i)x + y = 14且x – y = 4

解决方案：

x + y = 14 ……….. (1)

x – y = 4 ………….. (2)

x = 14 – y

Substitute x in (2)

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Transposing 2

y = 10/2

y = 5

x = 14 – y

x = 9

Therefore, x = 9 and y = 5.

(ii)s – t = 3且(s / 3)+(t / 2)= 26

解决方案：

s – t = 3 …….. (1)

(s/3) + (t/2) = 6 …………. (2)

s = 3 + t

Now, substitute the value of s in (2)

(3 + t) / 3 + (t/2) = 6

Taking 6 as LCM

(2(3 + t) + 3t) / 6 = 6

(6 + 2t + 3t) / 6 = 6

(6 + 5t) = 36

5t = 30

t = 6

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii)3x – y = 3和9x – 3y = 9

解决方案：

3x – y = 3 ……….. (1)

9x – 3y = 9 ……….(2)

From (1)

x = (3 + y) / 3

Substitute x in (2)

9(3 + y) / 3 – 3y = 9

9 + 3y – 3y = 9

0 = 0

Therefore, y has infinite values and x = (3 + y)/3 also has infinite values.

(iv)0.2x + 0.3y = 1.3和0.4x + 0.5y = 2.3

解决方案：

0.2x + 0.3y = 1.3 ……… (1)

 0.4x + 0.5y = 2.3 …….. (2)

From (1)

x = (1.3 – 0.3y) / 0.2 

Putting x in (2)

0.4(1.3 – 0.3y) / 0.2 + 0.5y = 2.3

2(1.3 – 0.3y) + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

2.6 – 0.1 y = 2.3

0.1 y = 0.3

y = 3

Substitute y in (1)

x = (1.3 – 0.3(3)) / 0.2 = (1.3 – 0.9) / 0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

(v)√2x+√3y= 0和√3x–√8y= 0

解决方案：

√2 x + √3 y = 0 …………… (1)

√3 x – √8 y = 0 …………..  (2)

From (1)

x = – (√3/√2)y 

Putting x in (2)

√3(-√3/√2)y – √8y = 0 

(-3/√2)y –  √8y = 0

-3y – 4y = 0

-7y = 0

y = 0

Therefore

x = 0

Therefore, x = 0 and y = 0.

(vi)(3x / 2)–(5y / 3)= -2和(x / 3)+(y / 2)=(13/6)

解决方案：

(3x/2) – (5y/3) = -2 ……………. (1)

 (x/3) + (y/2) = 13/6 ………. (2)

From (1)

(3/2)x = -2 + (5y/3)

(3/2)x = (-6 + 5y) / 3

x = ((-6 + 5y) / 3) * 2/3

⇒ x = 2(-6 + 5y) / 9 = (-12 + 10y) / 9 

Putting x in (2)

((-12 +10y)/9)/3 + y/2 = 13/6

(-12 + 10y)/27 + y/2 = 13/6

Taking 54 as LCM

-24 + 20y + 27y = 117

47y = 117 + 24

47y = 141

y = 3

x = (-12 + 30) / 9

x = 18/9

x = 2

Therefore, x = 2 and y = 3.

问题2。求解2x + 3y = 11和2x – 4y = – 24，从而找到y = mx + 3的’m’值。

解决方案：

2x + 3y = 11…………………………..(1)

2x – 4y = -24………………………… (2)

From (1)

x = (11 – 3y) / 2

Substituting x in equation (2)

2(11 – 3y) / 2 – 4y =- 24

11 – 3y – 4y = -24

-7y = -24 – 11

-7y = -35

y = 5

Putting y in (1)

x = (11 – 3 × 5) / 2 = -4/2 = -2

x = -2, y = 5

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Therefore, the value of m is -1.

问题3.形成以下问题的线性方程对，并通过替代方法找到它们的解决方案。

(i)两个数字之差为26，而一个数字为另一个的三倍。找到他们。

解决方案：

Let the two numbers be x and y 

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of y

3x – x = 26

2x = 26

x = 13 

y = 39

Therefore, the numbers are 13 and 39.

(ii)两个补充角中的较大者超过较小者18度。找到他们。

解决方案：

Let the larger angle by x^o and smaller angle be y^o.

Sum of two supplementary pair of angles is 180^o.

x + y = 180^o……………. (1)

x – y = 18^o ……………..(2)

From (1)

x = 180 – y

Substituting in (2)

180 – y – y = 18

-2y = -162

162 = 2y

y = 81^o

x = 180 – y

x = 180 – 81

= 99

Therefore, the angles are 99^o and 81^o.

(iii)板球队的教练以3800卢比的价格购买了7球拍和6球。后来，她以1750卢比的价格购买了3蝙蝠和5球。查找每个球棒和每个球的成本。

解决方案：

Let the cost of a bat be Rs. x and cost of a ball be Rs. y.

7x + 6y = 3800 ………………. (i)

3x + 5y = 1750 ………………. (ii)

From (i)

y = (3800 – 7x) / 6………………..(iii)

Substituting (iii) in (ii)

3x + 5(3800 – 7x) / 6 =1750

Taking 6 as LCM

18x + 19000 – 7x = 10500

11x = 10500 – 19000

⇒ -17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800 – 7 × 500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

(iv)城市出租车的费用包括固定费用和所覆盖距离的费用。对于10公里的距离，应支付的费用为105卢比，对于15公里的路程，应支付的费用为155卢比。固定费用和每公里的费用是多少?一个人要行驶25公里，需要支付多少费用?

解决方案：

Let the fixed charge be Rs x and per km charge be Rs y.

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1)

x = 105 – 10y ………………. (3)

Substituting the value of x in (2)

105 – 10y + 15y = 155

5y = 50

y = 10

Putting the value of y in (3)

x = 105 – 10 × 10

x = 105 – 100

x = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

(v)如果分子和分母都加2，则分数变为9/11。如果在分子和分母上都加上3，则它变成5/6。找到分数。

解决方案：

Let the fraction be x/y.

(x + 2)/(y + 2) = 9/11

By cross multiplication

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

By cross multiplication

(x + 3)/(y + 3) = 5/6

6x + 18 = 5y + 15

6x – 5y = -3 ………………. (2)

From (1) 

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2)

6(-4 + 9y)/11 -5y = -3

Taking 11 as the LCM

-24 + 54y – 55y = -33

-24 – y = -33

-y = -33 + 24

-y = -9

y = 9

Substituting the value of y in (3)

x = (-4 + 9 × 9)/11 

x = (-4 + 81)/11

x = 77/11

x = 7

Hence, the fraction is 7/9.

(vi)因此，雅各的五岁将是他儿子的三倍。五年前，雅各布的年龄是儿子的七倍。他们现在的年龄是几岁?

解决方案：

Let the age of Jacob = x years and that of son be y years.

(x + 5) = 3(y + 5) ………………… (i)

(x – 5) = 7(y – 5) ………………….. (ii)

From (i)

x + 5 = 3y + 15

x – 3y = 10……………. (iii)

From (ii)

x – 5 = 7y – 35

x – 7y = -30……………..(iv)

Subtracting (iv) from (iii)

-3y + 7y = 40

4y = 40

Transposing 4

y = 40/4

y = 10

Putting y = 10 in (iii)

x – 30 = 10

x = 40

Therefore, present age of Jacob is 40 years and that of son is 10 years