Класс 10 Решения RD Sharma – Глава 3 Пара линейных уравнений с двумя переменными – Упражнение 3.5 | Набор 3
Вопрос 27. При каком значении а не имеет решения следующая система уравнений:
топор + 3у = а - 3
12х + есть = а
Решение:
Given that,
ax + 3y = a − 3 …(1)
12x + ay = a …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = a, b1 = 3, c1 = – (a − 3)
a2 = 12, b2 = a, c2 = − a
For unique solution, we have
a1/a2 = b1/b2 ≠ c1/c2
a/12 = 3/a ≠ -(a – 3)/-a
a-3 ≠ 3
a ≠ 6
And,
a/12 = 3/a
a2 = 36
a = + 6 or – 6?
a ≠ 6, a = – 6
Hence, when a = -6 the given set of equations will have no solution.
Вопрос 28. Найдите значение a, при котором имеет место следующая система уравнений
(i) Уникальное решение
(ii) Нет решения
кх + 2у = 5
3х + у = 1
Решение:
Given that,
kx + 2y − 5 = 0 …(1)
3x + y − 1 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = k, b1 = 2, c1 = −5
a2 = 3, b2 = 1, c2 = −1
(i) For unique solution, we have
a1/a2 ≠ b1/b2
k/3 ≠ 2
k ≠ 6
Hence, when k ≠ 6 the given set of equations will have unique solution.
(ii) For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
k/3 = 2/1 ≠ -5/-1
k/3 = 2/1
k = 6
Hence, when k = 6 the given set of equations will have no solution.
Вопрос 29. При каком значении c следующая система уравнений имеет бесконечно много решений (где c ≠ 0 c ≠ 0):
6х + 3у = с - 3
12x + су = с
Решение:
Given that,
6x + 3y − (c − 3) = 0 …(1)
12x + cy − c = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 6, b1 = 3, c1 = −(c − 3)
a2 = 12, b2 = c, c2 = – c
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
6/12 = 3/c = -(c + 3)/-c
6/12 = 3/c and 3/c = -(c + 3)/-c
c = 6 and c – 3 = 3
c = 6 and c = 6
Hence, when c = 6 the given set of equations will have infinitely many solution.
Вопрос 30. Найдите значение k, при котором имеет место следующая система уравнений
(i) Уникальное решение
(ii) Нет решения
(iii) Бесконечно много решений
2x + ку = 1
3х - 5у = 7
Решение:
Given that,
2x + ky = 1 …(1)
3x − 5y = 7 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = k, c1 = −1
a2 = 3, b2 = −5, c2 = −7
(i) For unique solution, we have
a1/a2 ≠ b1/b2
2/3 ≠ -k/-5 ≠ -10/3
Hence, when k ≠ -10/3 the given set of equations will have unique solution.
(ii) For no solution, we have
a1/a2 = b1/b2 ≠ c1/c2
2/3 = k/-5 ≠ -1/-7
2/3 = k/-5 and k/-5 ≠ -1/-7
k = -10/3 and k ≠ -5/7
k = -10/3
Hence, when k = -10/3 the given set of equations will have no solution.
(iii) For the given system to have infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/3 = k/-5 = -1/-7
Clearly a1/a2 ≠ c1/c2
Hence. there is no value of k for which the given set of equation has infinitely many solution.
Вопрос 31. При каком значении k следующая система уравнений будет представлять совпадающие прямые:
х + 2у + 7 = 0
2x + ку + 14 = 0
Решение:
Given that,
x + 2y + 7 = 0 …(1)
2x + ky + 14 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = 2, c1 = 7
a2 = 2, b2 = k, c2 = 14
The given system of equation will represent the coincident
lines if they have infinitely many solutions
So,
a1/a2 = b1/b2 = c1/c2
1/2 = 2/k = 7/14
1/2 = 2/k = 7/14
k = 4
Hence, when k = 4 the given set of equations will have infinitely many solution.
Вопрос 32. Найдите значение k, при котором следующая система уравнений имеет единственное решение:
топор + бай = с,
лк + мой = п
Решение:
Given that,
ax + by − c = 0 …(1)
lx + my − n = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = a, b1 = b, c1 = − c
a2 = l, b2 = m, c2 = − n
For unique solution, we have
a1/a2 ≠ b1/b2
a/l ≠ b/m
am ≠ bl
Hence, when am ≠ bl the given set of equations will have unique solution.
Вопрос 33. Найдите такие значения a и b, что следующая система линейных уравнений имеет бесконечно много решений:
(2а - 1)х + 3у - 5 = 0,
3х + (б - 1)у - 2 = 0
Решение:
Given that,
(2a − 1)x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a2 = 3, b2 = b − 1, c2 = −2
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
(2a – 1)/3 = 3/(b – 1) = -5/-2
(2a – 1)/3 = 3/(b – 1) and 3/(b – 1) = -5/-2
2(2a − 1) = 15 and 6 = 5(b − 1)
4a − 2 = 15 and 6 = 5b − 5
4a = 17 and 5b = 11
So, a = 17/4 and b = 11/5
Вопрос 34. Найдите такие значения a и b, что следующая система линейных уравнений имеет бесконечно много решений:
2х - 3у = 7,
(а + б) х - (а + б - 3) у = 4а + б
Решение:
Given that,
2x − 3y − 7 = 0 …(1)
(a + b)x − (a + b − 3)y − (4a + b) = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = −3, c1 = −7
a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a + b) = -3/-(a + b – 3) = -7/-(4a + b)
2/(a + b) = -3/-(a + b – 3) and -3/-(a + b – 3) = -7/-(4a + b)
2(a + b − 3) = 3(a + b) and 3(4a + b) = 7(a + b − 3)
2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21
a + b = −6 and 5a − 4b = −21
a = − 6 − b
Substituting the value of a in 5a − 4b = −21, and we will get,
5( -b – 6) – 4b = -21
− 5b − 30 − 4b = − 21
9b = − 9 ⇒ b = −1
As a = – 6 – b
a = − 6 + 1 = − 5
So, a = – 5 and b = –1.
Вопрос 35. Найдите такие значения p и q, что следующая система линейных уравнений имеет бесконечно много решений:
2х - 3у = 9,
(p + q)x + (2p − q)y = 3(p + q + 1)
Решение:
Given that,
2x − 3y − 9 = 0 …(1)
(p + q)x + (2p − q)y − 3(p + q + 1) = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −9
a2 = (p + q), b2 = (2p − q), c2 = -3(p + q + 1)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(p + q) = 3/(2p – q) = -9/-3(p + q + 1)
2/(p + q) = 3/(2p – q) and 3/(2p – q) = -9/-3(p + q + 1)
2(2p – q) = 3(p + q) and (p + q + 1) = 2p – q
4p – 2q = 3p + 3q and -p + 2q = -1
p = 5q and p – 2q = 1
Substituting the value of p in p – 2q = 1, we have
3q = 1
q = 1/3
Substituting the value of p in p = 5q we have
p = 5/3
So, p = 5/3 and q = 1/3.
Вопрос 36. Найдите значения a и b, при которых следующая система уравнений имеет бесконечно много решений:
(i) (2а - 1)х + 3у = 5,
3х + (б - 2)у = 3
(ii) 2х - (2а + 5)у = 5,
(2б + 1)х - 9у = 15
(iii) (a − 1)x + 3y = 2,
6х + (1 - 2b)у = 6
(iv) 3x + 4y = 12,
(а + б) х + 2 (а - б) у = 5а - 1
(v) 2x + 3y = 7,
(а - 1)х + (а + 1)у = 3а - 1
(vi) 2x + 3y = 7,
(а - 1)х + (а + 2)у = 3а
(vii) 2x + 3y = 7,
(а – б)х + (а + б)у = 3а + б + 2
(viii) х + 2у = 1,
(а - б) х + (а + б) у = а + б - 2
(ix) 2x + 3y = 7,
2ах + ау = 28 – по
Решение:
(i) Given that,
(2a − 1)x + 3y − 5 = 0 …(1)
3x + (b − 2)y − 3 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2a − 1, b1 = 3, c1 = −5
a2 = 3, b2 = b − 2, c2 = -3(p + q + 1)
For infinitely many solution, we have
a1/a2 = b1/b2 = c1/c2
(2a – 1)/3 = -3/(b – 2) = -5/-3
(2a – 1)/3 = -3/(b – 2) and -3/(b – 2) = -5/-3
2a – 1 = 5 and – 9 = 5(b – 2)
a = 3 and -9 = 5b – 10
a = 3 and b = 1/5
So, a = 3 and b = 1/5.
(ii) Given that,
2x − (2a + 5)y = 5 …(1)
(2b + 1)x − 9y = 15 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = – (2a + 5), c1 = −5
a2 = (2b + 1), b2 = −9, c2 = −15
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(2b + 1) = (-2a + 5)/-9 = -5/-15
2/(2b + 1) = (-2a + 5)/-9 and (-2a + 5)/-9 = -5/-15
6 = 2b + 1 and 2a + 5
3 = b = 5/2 and a = -1
So, a = – 1 and b = 5/2.
(iii) Given that,
(a − 1)x + 3y = 2 …(1)
6x + (1 − 2b)y = 6 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = a – 1, b1 = 3, c1 = −2
a2 = 6, b2 = 1 − 2b, c2 = −6
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
(a – 1)/6 = 3/(1 – 2b) = 2/6
(a – 1)/6 = 3/(1 – 2b) and 3/(1 – 2b) = 2/6
a – 1 = 2 and 1 – 2b = 9
a – 1 = 2 and 1 – 2b = 9
a = 3 and b = -4
a = 3 and b = -4
So, a = 3 and b = −4.
(iv) Given that,
3x + 4y − 12 = 0 …(1)
(a + b)x + 2(a − b)y − (5a − 1) = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 3, b1 = 4, c1 = −12
a2 = (a + b), b2 = 2(a − b), c2 = – (5a − 1)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
3/(a + b) = 4/2(a – b) = 12/(5a – 1)
3/(a + b) = 4/2(a – b) and 4/2(a – b) = 12/(5a – 1)
3(a – b) = 2a + 2b and 2(5a – 1) = 12(a – b)
a = 5b and -2a = -12b + 2
On substituting a = 5b in -2a = -12b + 2, we have
-2(5b) = -12b + 2
−10b = −12b + 2 ⇒ b = 1
Thus a = 5
So, a = 5 and b = 1.
(v) Given that,
2x + 3y − 7 = 0 …(1)
(a − 1)x + (a + 1)y − (3a − 1) = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −7
a2 = (a − 1), b2 = (a + 1), c2 = – (3a − 1)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a – b) = 3/(a + 1) = -7/(3a – 1)
2/(a – b) = 3/(a + 1) and 3/(a + 1) = -7/(3a – 1)
2(a + 1) = 3(a – 1) and 3(3a – 1) = 7(a + 1)
2a – 3a = -3 – 2 and 9a – 3 = 7a + 7
a = 5 and a = 5
So, a = 5 and b = 1.
(vi) Given that,
2x + 3y − 7 = 0 …(1)
(a − 1)x + (a + 2)y − 3a = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −7
a2 = (a − 1), b2 = (a + 2), c2 = −3a
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a – b) = 3/(a + 2) = -7/-3a
2/(a – b) = 3/(a + 2) and 3/(a + 2) = -7/-3a
2(a + 2) = 3(a – 1) and 3(3a) = 7(a + 2)
2a + 4 = 3a – 3 and 9a = 7a + 14
a = 7 and a = 7
So, a = 7 and b = 1.
(vii) 2x + 3y – 7 = 0, …(1)
(a – b)x + (a + b)y – 3a – b + 2 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −7
a2 = (a − b), b2 = (a + b), c2 = −(3a + b – 2)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/(a – b) = 3/(a + b) = -7/−(3a + b – 2)
2/(a – b) = 3/(a + b) and 3/(a + b) = -7/−(3a + b – 2)
2(a + b) = 3(a – b) and 3(3a + b – 2) = 7(a + b)
2a + 2b = 3a – 3b and 9a + 3b – 6 = 7a + 7b
So, a = 5 and b = 1.
(viii) x + 2y – 1 = 0 …(1)
(a − b)x + (a + b)y – a – b + 2 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = 2, c1 = −1
a2 = (a − b), b2 = (a + b), c2 = −(a + b – 2)
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
1/(a – b) = 2/(a + b) = -1/−(a + b – 2)
1/(a – b) = 2/(a + b) and 2/(a + b) = -1/−(a + b – 2)
(a + b) = 2(a – b) and 2(a + b – 2) = (a + b)
a + b = 2a – 2b and 2a + 2b – 4 = a + b
So, a = 3 and b = 1.
(ix) 2x + 3y – 7 = 0 …(1)
2ax + ay – 28 + by = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −7
a2 = 2a, b2 = (a + b), c2 = −28
For infinitely many solutions, we have
a1/a2 = b1/b2 = c1/c2
2/2a = 3/(a + b) = -7/−28
2/2a = 3/(a + b) and 3/(a + b) = 7/28
2(a + b) = 6a and 84 = 7(a + b)
So, a = 4 and b = 8.
Вопрос 37. При каких значениях λ пара линейных уравнений λx + y = λ 2 и x + λy = 1 имеет
(i) нет решения?
(ii) бесконечно много решений?
(iii) единственное решение?
Решение:
Given that,
λx + y = λ2 …(1)
x + λy = 1 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = λ, b1 = 1, c1 = -λ2
a2 = 1, b2 = λ, c2 = -λ2
(i) For no solution,
a1/a2 = b1/b2 ≠ c1/c2
λ = 1/λ ≠ -λ2/-1
λ2 – 1 = 0
λ = 1, -1
Here we will take only λ = -1 because at λ = 1 the linear
equation will have infinite many solutions.
(ii) For infinite many solutions,
a1/a2 = b1/b2 = c1/c2
λ = 1/λ = λ2 /1
λ(λ – 1) = 0
When λ ≠ 0 then λ = 1
(iii) For Unique solution,
a1/a2 ≠ b1/b2
λ ≠ 1/λ