10类RD Sharma解–第3章两个变量的线性方程对–练习3.4 |套装2

通过交叉乘法解决以下每个方程组。

问题15. 2ax + 3by = a + 2b和3ax + 2by = 2a + b

解决方案：

Given that,

2ax + 3by = a + 2b

3ax + 2by = 2a + b

On comparing both the equation with the geneal form we get

 a₁ = 2a, b₁ = 3b, c₁ = -(a + 2b), a₂ = 3a, b₂ = 2b, c₂ = -(2a + b)

Now by using cross multiplication we get

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

x/(-3b(2a + b) + 2b(a + 2b)) = y/(-3a(a + 2b) + 2a(2a + b)) = 1/(4ab – 9ab)

x/(b² – 4ab) = y/(a² – 4ab) = 1/-5ab

x/(-4ab + b²) = 1/-5ab

x/b(b – 4a) = 1/-5ab

x = (4a – b)/5a

and,

= -y/(-a² + 4ab) = 1/-5ab

= -y/a(-a + 4b) = 1/-5ab

 y = (4b – a)/5b

Hence, x = (4a – b)/5a and y = (4b – a)/5b

问题16. 5ax + 6by = 28和3ax + 4by = 18

解决方案：-

Given that,

5ax + 6by = 28

3ax + 4by = 18

Or, 5ax + 6by – 28 = 0

3ax + 4by – 18 = 0

Here, a₁ = 5a, b₁ = 6b, c₁=-28

a₂= 3a, b₂ = 4b, c₂ =-18

On comparing both the equation with the geneal form we get

 a₁ = 5a, b₁ = 6b, c₁ = -28, a₂ = 3a, b₂ = 4b, c₂ = -18

Now by using cross multiplication we get

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

x/4b = -y/-6a = 1/2ab

x/4b = 1/2b

x = 2/a

and,

-y/-6a = 1/2ab

y = 3/b

Hence, x = 2/a and y = 3/b

问题17.(a + 2b)x +(2a – b)y = 2和(a – 2b)x +(2a + b)y = 3

解决方案：

Given that,

(a + 2b)x + (2a – b)y = 2

(a – 2b)x + (2a + b)y = 3

Or 

(a + 2b) x + (2a – b)y – 2 = 0

(a – 2b) x + (2a + b)y – 3 = 0

On comparing both the equation with the geneal form we get

 a₁ = a + 2b, b₁ = 2a – b, c₁ = -2, a₂ = a – 2b, b₂ = 2a + b, c₂ = -3

Now by using cross multiplication we get

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

x/(-3(2a – b)) + (2(2a + b)) = y/(-2(a – 2b)) + (3(a + 2b)) = 1/((a + 2b)(2a + b) – (a – 2b)(2a – b))

x/(5b – 2a) = y/(a + 10b) = 1/(2a² + 5ab + 2b² – 2a² + 5ab – 2b²)

x/(5b – 2a) = y/(a + 10b) = 1/10ab

So,

x/(5b – 2a) = 1/10ab

x= (5b – 2a)/10ab

and,

y/(a + 10b) = 1/10ab

y = (a + 10b)/10ab

Hence, x = (5b – 2a)/10ab and y= (a + 10b)/10ab

问题18. x((a – b)+(ab /(a – b)))= y((a + b)–(ab /(a + b)))和x + y = 2a ²

解决方案：

Given that,

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

Or on solving we get

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

= x((a² + b² – 2ab + ab)/(a – b)) – y((a² + b² + 2ab – ab)/(a + b)) = 0

and,

x + y = 2a²

On comparing both the equation with the geneal form we get

 a₁ = (a² + b² – 2ab + ab)/(a – b), b₁ = (a² + b² + 2ab – ab)/(a + b), c₁ = 0, 

a₂ = 1, b₂ = 1, c₂ = 2a²

Now by using cross multiplication we get

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

⇒ x/(2a²((a² + b² – 2ab + ab)/(a – b)) – 0) = y/(0 + 2a²((a² + b² + 2ab – ab)/(a + b))) 

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

⇒ x/(2a²((a² + b² – 2ab + ab)/(a – b))) = y/(-2a²)((a² + b² + 2ab – ab)/(a + b)) = 1/(2a³/(a² – b²))

Now,

x/(2a²((a² + b² – 2ab + ab)/(a – b)) = 1/(2a³/(a² – b²))

x = (2a²(a² + ab + b²)(a² – b²)) / 2a³(a + b)

x = (a³ – b³)/a

and,

 y/(-2a²)((a² + b² + 2ab – ab)/(a + b)) = 1/(2a³/(a² – b²))

y = (2a²(a² – ab + b²)(a² – b²))/2a³(a – b)

y = a³ + b³/a

Hence, x = (a³ – b³)/a and y = a³+ b³/a

问题19。bx + cy = a + b和ax [(1 /(a – b))–(1 /(a + b))] + cy [(1 /(b – a))–(1 /( b + a))] = 2a /(a + b)

解决方案：

Given that,

bx + cy = a + b

ax[(1/(a – b)) – (1/(a + b))] + cy[(1/(b – a)) – (1/(b + a))] = 2a/(a + b)

Or

bx + cy -(a + b) = 0

ax((1/(a – b)) – (1/(a + b))) + cy(((1/(b – a)) – (1/(b + a))) – 2a/(a + b) = 0

ax(2b/(a² – b²))) + cy(2a/(b² – a²)) – 2a/(a + b) = 0

On comparing both the equation with the geneal form we get

 a₁ = b, b₁ = c, c₁ = -(a + b),

a₂ = 2b/(a² – b²), b₂ = 2a/(b² – a²), c₂ = 2a/(a + b) 

Now by using cross multiplication we get

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

⇒ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b² – a²)))) = y/((-(a + b)2ab)/(a² – b²)) + (2ab/(a + b)) 

= 1/((2abc/(b²– a²)) – (2abc/(a² – b²)))

⇒ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b² – a²)))) = y/((-(a + b)2ab)/(a² – b²)) + (2ab/(a + b)) 

= 1/(-4abc/(a² – b²))

⇒ x/(-2ac((1/(a + b)) + (1/(a – b))) = y/(2ab((-1/(a – b)) + (1/(a + b))) = 1/(-4abc/(a² – b²))

⇒ x/(-4a²c/(a² – b²)) = y/(4ab²/(a² – b²)) = 1/1/(-4abc/(a² – b²))

So,

x/(-4a²c/(a² – b²)) = 1/(-4abc/(a² – b²))

x = a/b

and,

y/(4ab²/(a² – b²)) = 1/(-4abc/(a² – b²))

y = b/c

Hence, x = a/b, y = b/c

问题20.(a – b)x +(a + b)y = 2a ² – 2b ²和(a + b)(x + y)= 4ab

解决方案：

Given that,

(a – b) x + (a + b) y = 2a² – 2b²

(a + b) (x + y) = 4ab

(a – b) x + (a + b) y – 2(a² – b²) = 0

(a + b)x +  (a + b)y – 4ab = 0

On comparing both the equation with the geneal form we get

 a₁ = a – b, b₁ = a + b, c₁ = -2,

a₂ = a + b, b₂ = a + b, c₂ = -4ab

Now by using cross multiplication we get

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

⇒ x/(-(a + b)4ab + 2(a + b) (a² – b²)) = y/(− 2(a² − b²)(a + b) + 4ab(a – b)) 

= 1/((a − b)(a + b) − (a + b)(a + b))

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

x = (2ab – a² + b²)/b

and,

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

y = (a – b)(a² + b²)/ b(a + b)

Hence, x = (2ab – a² + b²)/b and y = (a – b)(a² + b²)/ b(a + b)

问题21. a ² x + b ² y = c ²和b ² x + a ² y = d ²

解决方案：

Given that,

a²x + b²y = c²

b²x + a²y = d²

Or

a²x + b²y  – c² = 0

b²x + a²y – d² = 0

On comparing both the equation with the geneal form we get

 a₁ = a², b₁ = b², c₁ = -c²,

a₂ = b², b₂ = a², c₂ = -d²

Now by using cross multiplication we get

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

x/(-b²d² + a²c²) = y/(-c²b² + a²d²) = 1/(a⁴-b⁴)

 x/(a²c² – b²d²) = y/(a²d² – c²b²) = 1/(a⁴-b⁴)

Therefore,

 = x/(a²c² – b²d²) = 1/(a⁴ – b⁴)

x = (a²c² – b²d²)/(a⁴ – b⁴)

and,

= y/(a²d² – c²b²) = 1/(a⁴-b⁴)

y = (a²c² – b²d²) / (a⁴-b⁴)

Hence, x = (a²c² – b²d²)/(a⁴ – b⁴),  y = (a²c² – b²d²) / (a⁴ – b⁴)

问题22. ax + by =(a + b)/ 2和3x + 5y = 4

解决方案：

Given that,

ax + by = (a+b)/2

3x + 5y = 4

Or 

ax + by – (a + b)/2 = 0

3x + 5y – 4 = 0

On comparing both the equation with the geneal form we get

 a₁ = a, b₁ = b, c₁ = -(a + b)/2,

a₂ = 3, b₂ = 5, c₂ = -4

Now by using cross multiplication we get

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

= x/(-4b + 5((a + b)/2)) = y/(-3((a + b)/2) + 4a) = 1/(5a – 3b)

= x/((5a – 3b)/2) = y/((5a – 3b)/2) = 1/(5a – 3b)

Now,

x/((5a – 3b)/2) = 1/(5a – 3b)

x = (5a – 3b)/(2(5a – 3b))

x = 1/2

and,

y/((5a – 3b)/2) = 1/(5a – 3b)

y = (5a – 3b)/(2(5a – 3b))

y = 1/2

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

问题23. 2(ax – by)+ a + 4b = 0和2(bx + ay)+ b – 4a = 0

解决方案：

Given that,

2 (ax – by) + (a + 4b) = 0

2 (bx + ay) + (b – 4a) = 0

On comparing both the equation with the geneal form we get

 a₁ = 2a, b₁ = -2b, c₁ = a + 4b,

a₂ = 2b, b₂ = 2a, c₂ = b – 4a

Now by using cross multiplication we get

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

= x/((-2b(a + 4b)) – (2a(b – 4a ))) = y/((2b(a + 4b)) – (2a(b – 4a))) = 1/(4a² + 4b²)

= x/(-2b² + 8ab – 2ab + 8a²) = y/(2ab + 8b² – 2ab + 8a²) = 1/4(a² + b²)

= x/-2(a² + b²) = y/8(a² + b²) = 1/4(a² + b²)

So, 

= x/-2(a² + b²) = 1/4(a² + b²)

x = -1/2

and,

= y/8(a² + b²) = 1/4(a² + b²)

y = 2

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

问题24. 6(ax + by)= 3a + 2b和6(bx – ay)= 3b – 2a

解决方案：

given that,

6 (ax + by) = 3a + 2b

6 (bx – ay) = 3b – 2a 

6 (ax + by) -(3a + 2b)=0….       (1)

6 (bx – ay) -(3b – 2a) =0…..       (2)

On comparing both the equation with the geneal form we get

 a₁ = 6a, b₁ = 6b, c₁ = -(3a – 2b),

a₂ = 6b, b₂ = 66a, c₂ = -(3b – 2a)

Now by using cross multiplication we get

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

= x/(-6b(3b – 2a) – 6a(3a – 2b)) = y/(-6b(3a – 2b) + 6a(3b – 2a)) = 1/(-36a² – 36b²)

= x/(-18(a² + b²)) = y/(-12(a² + b²)) = 1/(-36(a² + b²))

Therefore,

x/(-18(a² + b²)) = 1/(-36(a² + b²))

x = 1/2

and,

y/(-12(a² + b²)) = 1/(-36(a² + b²))

y = 1/3

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

问题25.(a ² / x)−(b ² / y)= 0和(a ² b / x)−(b ² a / y)= a + b，x，y≠0

解决方案：

Given that,

(a²/x) − (b²/y) = 0

(a²b/x) − (b²a/y) = a + b 

Or

(a²b/x) − (b²a/y) – (a + b) = 0

On comparing both the equation with the geneal form we get

 a₁ = a², b₁ = -b², c₁ = 0,

a₂ = a²b, b₂ = b²a, c₂ = -(a + b)

Now by using cross multiplication we get

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

= (1/x)/(b²(a + b) – 0) = (1/y)/(0 + (a²(a + b))) = 1/(a³b² – a²b³)

= (1/x)/(b²(a + b)) = (1/y)/(a²(a + b)) = 1/a²b²(a + b)

So,

= (1/x)/(b²(a + b)) = 1/a²b²(a + b)

x = a²

and,

= (1/y)/(a²(a + b)) = 1/a²b²(a + b)

y = b²

Hence, x = a² and y = b²

问题26. mx – ny = m ² + n ²和x + y = 2m

解决方案：

Given that,

mx – ny = m² + n²

x + y = 2m

Or

mx – ny -(m² + n²) = 0

x + y – 2m = 0

On comparing both the equation with the geneal form we get

 a₁ = m, b₁ = -n, c₁ = -(m² + n²),

a₂ = 1, b₂ = 1, c₂ = -2m

Now by using cross multiplication we get

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

= x/(2mn + (m² + n²)) = y/(-(m² + n²) + 2m²) = 1/(m + n)

= x/(m + n)² = y/(m² – n²) = 1/(m + n)

Therefore,

 x/(m + n)² = 1/(m + n)

x = m + n

and,

y/(m² – n²) = 1/(m + n)

y = m – n

Hence, x = m + n, y = m – n

问题27.(ax / b)–(by / a)= a + b和ax – by = 2ab

解决方案：

Given that,

(ax/b) – (by/a) = a + b

ax – by = 2ab

Or

(ax/b) – (by/a) – (a + b) = 0

ax – by – 2ab = 0

On comparing both the equation with the geneal form we get

 a₁ = a/b, b₁ = -b/a, c₁ = -(a + b),

a₂ = a, b₂ = b, c₂ = -2ab

Now by using cross multiplication we get

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

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

So,

x/b(b – a) = 1/(b – a)

x = b

and,

-y/a(-a + b) = 1/(b – a)

y = -a

Hence, x = b, y = -a

问题28.(b / a)x +(a / b)y = a ² + b ²和x + y = 2ab

解决方案：

Given that,

(b/a)x + (a/b)y = a² + b²

x + y = 2ab

Or

(b/a)x + (a/b)y – (a² + b²) = 0

x + y – 2ab = 0

On comparing both the equation with the geneal form we get

 a₁ = b/a, b₁ = a/b, c₁ = -(a² + b²),

a₂ = 1, b₂ = 1, c₂ = -2ab

Now by using cross multiplication we get

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

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

Therefore,

 x/(b² – a²) = 1/((b² – a²)/ab)

x = ab

y/(-b² + a²) = 1/((b² – a²)/ab)

y = ab

Hence, x = ab, y = ab