10类RD Sharma解–第3章两个变量的线性方程对–练习3.4 |套装1

通过交叉乘法解决以下每个方程组。

问题1. x + 2y + 1 = 0和2x – 3y – 12 = 0

解决方案：

Given that,

x + 2y + 1 = 0

2x – 3y – 12 = 0

On comparing both the equation with the geneal form we get

 a₁ = 1, b₁ = 2, c₁ = 1, a₂ = 2, b₂ = −3, c₂ = -12

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(-24 – (-3)) = y/(-2 – (-12)) = 1/(-3 – 4)

x/(-24 + 32) = y/(2 + 12) = 1/ (-3 – 4)

x = -21/-7

So, x = 3

and 

y = 14/-7

y = -2

Hence, x = 3 and y = -2

问题2。3x + 2y + 25 = 0和2x + y + 10 = 0

解决方案：

Given that,

3x +2y + 250

2x + y + 10 = 0

On comparing both the equation with the geneal form we get

 a₁ = 3, b₁ = 2, c₁ = 25, a₂ = 2, b₂ = 1, c₂ = 10

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(20 – 25) = y/(50 – 30) = 1/(3 – 4)

x/-5 = y/20 = 1/-1

x = -5 × (-1) 

x = 5

x = -5 × (-1) = 5

y/20 = -1

y = 20 × (-1) = −20

Hence, x = 5y and y = −2

问题3. 2x + y = 35和3x + 4y = 65

解决方案：

Given that,

2x + y = 35

3x + 4y = 65

Or 2x + y – 35 = 0

3x + 4y – 65 = 0

On comparing both the equation with the geneal form we get

 a₁ = 2, b₁ = 1, c₁ = -35, a₂ = 3, b₂ = 4, c₂ = -65

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(-65 – (-140)) = y/(-105-(-65 × 2)) = 1/( 8 – 3)

x/(-65 + 140) = y/( -105 + 130) = 1/5

x = 75/5

x = 15

y = 25/5

y = 5

Hence, x = 15 and y = 5

问题4：2x – y = 6和x – y = 2

解决方案：

Given that,

2x – y = 6

x – y = 2

Or 2x – y – 6 = 0

x – y – 2 = 0

On comparing both the equation with the geneal form we get

 a₁ = 2, b₁ = -1, c₁ = -6, a₂ = 1, b₂ = -1, c₂ = -2

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(2 – 6) = y/(-6 + 4) = 1/(-2 + 1)

So, x = 4

and y = 2

Hence, x = 4 and y = 2

问题5.(x + y)/ xy = 2和(x-y)/ xy = 6

解决方案：

Given that,

(x + y)/xy = 2,

(x − y)/xy = 6

Or, 1/x + 1/y – 2 = 0

and,

1/x – 1/y – 6 = 0

On comparing both the equation with the geneal form we get

 a₁ = 1, b₁ = 1, c₁ = -2, a₂ = 1, b₂ = -1, c₂ = -6

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

(1/x)/(-6 + 2) = (1/y)/(2 + 6) = 1/(1 + 1)

So, x = -1/2

and y = 1/4

Hence, x = -1/2 and y = 1/4

问题6。ax + by = a – b和bx – ay = a + b

解决方案：

Given that,

ax + by = a – b

bx – ay = a + b

Or, ax + by – (a – b) = 0

bx – ay – (a + b) = 0

On comparing both the equation with the geneal form we get

 a₁ = a, b₁ = b, c₁ = -(a – b), a₂ = b, b₂ = -a, c₂ = (a + b)

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(−b(a + b) − (-a)(-(a − b))) = y/((-(a − b) × b) – (-(a + b) × a) = 1/ (-a² – b²)

x/(-ab – b² – a² + ab) = y/(-ab – b² + a² + ab) = 1/-(-a² + b²)

So, 

x/-(-a² + b²) = 1/-(-a² + b²) 

x = 1

and, y/(a² + b²) = 1/-(-a² + b²)

y = -1

Hence, x = 1 and y = -1

问题7. x + ay = b和ax – by = c

解决方案：

Given that,

x + ay = b

ax – by = c

Or, x + ay – b = 0

ax – by – c = 0

On comparing both the equation with the geneal form we get

a₁ = 1, b₁ = a, c₁ = – b, a₂ = a, b₂ = -b, c₂ = -c

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(-ac – b²) = y/(-ab + c) = 1/(-b – a²)

x/(- ac – b²) = 1/(-b – a²) 

x = (ac + b²)/(a² + b)

and,

y/(- ab + c) = 1/(-b – a²)

y =(ab – c)/(a² + b)

Hence, x = (ac + b²)/(a² +b) and y =(ab – c)/(a² + b)

问题8. ax + by = a ²和bx + ay = b ²

解决方案：

Given that,

ax + by = a²

bx + ay = b²

Or, ax + by – a² = 0

bx + ay – b² = 0

On comparing both the equation with the geneal form we get

a₁ = a, b₁ = b, c₁ = – a², a₂ = b, b₂ = a, c₂ = -b²

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/(-b³ + a³) = y/(-a²b + ab²) = 1/(a² – b²)

x/(a³ – b³) = y/(-ab(a – b)) = 1/(a² – b²)

x= (a³ – b³)/(a² – b²)

x = ((a – b)(a² + ab + b²))/(a + b)(a – b)

x = (a² + ab + b²)/a + b

and,

y/(-ab(a – b)) = 1/(a² – b²)

(a – b)(a² + ab + b²) (a + ba – b)

 y = -ab(a – b)/(a + b)(a – b)

y= -ab/(a + b)

Hence, x = (a² + ab + b²)/(a + b) and y = -ab/(a + b)

问题9.(5 /(x + y))–(2 /(x – y))= -1和(15 /(x + y))+(7 /(x – y))= 10

解决方案：

Given that,

(5/(x + y)) – (2/(x – y)) = -1

(15/(x + y)) + (7/(x – y)) = 10

Let, us assume

x + y = a, and x – y = b, then

(5/a) – (2/b) + 1 = 0

(15/a) – (7/b) + 10 = 0

On comparing both the equation with the geneal form we get

a₁ = 5, b₁ = -2, c₁ = – 1, a₂ = 15, b₂ = 7, c₂ = -10

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

((1/a)/(20 – 7)) = ((1/b)(15 + 50)) = 1/(35 + 30)

((1/a)/13) = ((1/b)/65) = 1/65

So,

1/a = 13/65

a = 5

and,

1/b = 65/65

b =1

= x + y = 5……       (1)

= x – y = 1…..      (2)

Now, by adding eq(1) and (2) then we get,

x = 3 and y = 2

Hence, x = 3 and y = 2

问题10(2 / x)+(3 / y)= 13和(5 / x)–(4 / y)= -2

解决方案：

Given that,

(2/x) +(3/y) = 13

(5/x) – (4/y) = -2

Or, (2/x) + (3/y) – 13 = 0

(5/x) – (4/y) + 2 = 0

On comparing both the equation with the geneal form we get

a₁ = 2, b₁ = 3, c₁ = – 13, a₂ = 5, b₂ = -4, c₂ = 2

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

 = ((1/x)/(6 – 52)) = ((1/y)(-65 – 4)) = 1/(-8 – 5)

= ((1/x)/-46) = ((1/y)/(-69)) = 1/(-23)

Now,

1/x = -46/(-23)

x = 1/2

and,

1/y = -69/-23

y = 1/3

Hence, x = 1/2 and y = 1/3

问题11((11x + y))+(6 /(x-y))= 5和(38 /(x + y))+(21(x-y))= 9

解决方案：

Given that,

(57/(x + y)) + (6/(x – y)) = 5

(38/(x + y)) + (21(x – y)) = 9

Let us considered

x + y = p

x – y = q

So,

(57/p) + (6/q) – 5 = 0

(38/p) + (21/q) – 9 = 0

On comparing both the equation with the geneal form we get

a₁ = 57, b₁ = 6, c₁ = – 5, a₂ = 38, b₂ = 21, c₂ = -9

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

((1/p)/(-54 + 105)) = ((1/q)/(-190 + 513)) = 1/(1197 – 228)

((1/p)/51) = ((1/q)/323) = 1/969

So,

((1/p)/51) = 1/969

1/p = 51/969

p = 19

and,

((1/q)/323) = 1/969

1/q = 323/969

q = 3

Now, adding eq x + y = 19 and x – y = 3 then we get:

x = 11 and y = 8

Hence, x = 11 and y = 8

问题12。(x / a)+(y / b)= 2并且ax – by = a ² – b ²

解决方案：

Given that,

(x/a) + (y/b) = 2

ax – by = a² – b²

(x/a) + (y/b) – 2 = 0

ax – by – (a² – b²) = 0

On comparing both the equation with the geneal form we get

a₁ = 1/a, b₁ = 1/b, c₁ = – 2, a₂ = a, b₂ = -b, c₂ = -(a² – b²)

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/((-1/b)(a² + b²)-2b) = y/((-2a) + (1/a)(a² – b²)) = 1/((-b/a) – (a/b))

x/((-a²/b) – b) = y/(-a – (b²/a)) = 1/-(a² + b²)/ab

x/(-(a² + b²)/b) = y/(-(a² + b²)/a)) = 1/-(a² + b²)/ab

So,

 x/(-(a² + b²)/b) = 1/-( a² + b²)/ab

x = -(a² + b²)ab/-b(a² + b²) 

x = a

and,

 y/(-(a² + b²)/a)) = 1/-(a² + b²)/ab

y = -(a² + b²)ab/-a(a² + b²) 

y = b

Hence, x = a and y = b

问题13.(x / a)+(y / b)= a + b和(x / a ² )+(y / b ² )= 2

解决方案：

Given that,

(x/a) + (y/b) = a + b

(x/a²) + (y/b²) = 2

(x/a) + (y/b) – a + b = 0

(x/a²) + (y/b²) – 2 = 0

On comparing both the equation with the geneal form we get

a₁ = 1/a, b₁ = 1/b, c₁ = -(a + b), a₂ = 1/ a², b₂ = 1/ b², c₂ = -2

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

(x/ (-2/b) + (a/b²) + (1/b)) = (-y/(-2/a) + (1/a) + (b/ a²)) = (1/(-1/ ab²) – (-1/ a²b))

Now,

x/((a – b)b²) = 1/(-1/ab²) – (-1/a²b)

x = a²

and,

-y/((-a – b)/a²) + (1/a) + (b/b²) = 1/(-1/ab²) – (-1/a²b)

y = b²

Hence, x = a² and y = b²

问题14. x / a = y / b和ax + by = a ² + b ²

解决方案：

Given that,

x/a = y/b

ax + by = a² + b²

On comparing both the equation with the geneal form we get

a₁ = 1/a, b₁ = 1/b, c₁ = 0, a₂ = a, b₂ = b, c₂ = -(a² + b²)

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

x/((a² + b²)/b) = y/((a² + b²)/a) = 1/((a/b) + (b/a))

Now,

x/((a² + b²)/b) = 1/((a/b) + (b/a))

x = a

and,

 y/((a² + b²)/a) = 1/((a/b) + (b/a))

y = b

Hence, x = a and y = b