Mẹo Hướng dẫn For what value of k the following system of equations have infinite solutions 2x-3y=7 2022
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Nội dung chính Show- Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.5For what value of k will the following equations have infinite solutions 2x 3y 7?For what value of k the system has infinite solutions?How do you find the solution to a system of infinite solutions?For what value of k will the following equations have infinitely many solutions KX 3y
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Buy NowSolution : The given equations are
` 2x - 3y - 7 =0`,
` ( k + 1 ) x + ( 1 - 2k ) y + ( 4 - 5k ) = 0 `
These equations are of the form
` a_ 1 x + b_ 1 y + c_ 1 = 0 and a_ 2 x + b_ 2 y + c_ 2 = 0 `
where ` a_ 1 = 2, b_ 1 = - 3, c_ 1 = - 7 `
and `a_ 2 = ( k + 1 ), b_ 2 = (1 - 2k ), c_ 2 = ( 4- 5k )`
Let the given system of equations have infinitely many solutions.
Then, `(a_ 1 ) /(a_ 2 ) = (b_ 1 )
/( b_ 2 ) = (c _ 1 ) /(c_2)`
`rArr ( 2)/(( k+ 1 )) = (-3)/((1- 2k )) = ( -7)/(( 4- 5k ))`
`rArr (2)/(( k + 1 ) ) = (3)/( ( 2k - 1 )) = ( 7)/(( 5k - 4))`
` rArr ( 2)/(( k + 1 )) = ( 3)/(( 2k - 1 )) = ( 7)/(( 5k - 4))`
`rArr ( 2)/((k + 1 )) = (3)/(( 2k - 1 )) and (3)/((2k -1 )) = ( 7)/((5 k - 4))`
`rArr 4k - 2 = 3k + 3 and 15k - 12 = 14k - 7`
` rArr k = 5 and k = 5`
Hence, ` k = 5`
A given system of equations may or may not have a solution. Sometimes it can also have infinitely many solutions. All these conditions for solvability are studied in this exercise. The RD Sharma Solutions Class 10 prepared by experts BYJU’S can help students get a strong conceptual knowledge on the subject. Also, the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.5 PDF given below is available for students for further clarifications.
Chapter 3 Pair Of Linear…
Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.5
In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, find it from 1 to 4:
1. x − 3y = 3
3x − 9y = 2
Solution:
The given system of equations is:
x − 3y – 3 = 0
3x − 9y − 2 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 1, b1 = −3, c1 = −3
a2 = 3, b2 = −9, c2 = −2
So according to the question, we get
a1 / a2 = 1/3
b1 / b2 = −3/ −9 = 1/3 and,
c1 / c2 = −3/ −2 = 3/2
⇒ a1 / a2 = b1/ b2 ≠ c1 / c2
Hence, we can conclude that the given system of equation has no solution.
2. 2x + y = 5
4x + 2y = 10
Solution:
The given system of equations is:
2x + y – 5 = 0
4x + 2y – 10 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 2, b1 = 1, c1 = −5
a2 = 4, b2 = 2, c2 = −10
So according to the question, we get
a1 / a2 = 2/4 = 1/2
b1 / b2 = 1/ 2 and,
c1 / c2 = −5/ −10 = 1/2
⇒ a1 / a2 = b1/ b2 = c1 / c2
Hence, we can conclude that the given system of equation has infinity many solutions.
3. 3x – 5y = 20
6x – 10y = 40
Solution:
The given system of equations is:
3x – 5y – 20 = 0
6x – 10y – 40 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 3, b1 = -5, c1 = −20
a2 = 6, b2 = -10, c2 = −40
So according to the question, we get
a1 / a2 = 3/6 = 1/2
b1 / b2 = -5/ -10 = 1/2 and,
c1 / c2 = -20/ −40 = 1/2
⇒ a1 / a2 = b1/ b2 = c1 / c2
Hence, we can conclude that the given system of equation has infinity many solutions.
4. x – 2y = 8
5x – 10y = 10
Solution:
The given system of equations is:
x – 2y – 8 = 0
5x – 10y – 10 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 1, b1 = -2, c1 = −8
a2 = 5, b2 = -10, c2 = -10
So according to the question, we get
a1 / a2 = 1/5
b1 / b2 = -2/ -10 = 1/5 and,
c1 / c2 = -8/ −10 = 4/5
⇒ a1 / a2 = b1/ b2 ≠ c1 / c2
Hence, we can conclude that the given system of equation has no solution.
Find the value of k for which the following system of equations has a unique solution: (5-8)
5. kx + 2y = 5
3x + y = 1
Solution:
The given system of equations is:
kx + 2y – 5 = 0
3x + y – 1 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = k, b1 = 2, c1 = −5
a2 = 3, b2 = 1, c2 = -1
So according to the question,
For unique solution, the condition is
a1 / a2 ≠ b1 / b2
k/3 ≠ 2/1
⇒ k ≠ 6
Hence, the given system of equations will have unique solution for all real values of k other than 6.
6. 4x + ky + 8 = 0
2x + 2y + 2 = 0
Solution:
The given system of equations is:
4x + ky + 8 = 0
2x + 2y + 2 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 4, b1 = k, c1 = 8
a2 = 2, b2 = 2, c2 = 2
So according to the question,
For unique solution, the condition is
a1 / a2 ≠ b1 / b2
4/2 ≠ k/2
⇒ k ≠ 4
Hence, the given system of equations will have unique solution for all real values of k other than 4.
7. 4x – 5y = k
2x – 3y = 12
Solution
The given system of equations is:
4x – 5y – k = 0
2x – 3y – 12 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 4, b1 = 5, c1 = -k
a2 = 2, b2 = 3, c2 = 12
So according to the question,
For unique solution, the condition is
a1 / a2 ≠ b1 / b2
4/2 ≠ 5/3
⇒k can have any real values.
Hence, the given system of equations will have unique solution for all real values of k.
8. x + 2y = 3
5x + ky + 7 = 0
Solution:
The given system of equations is:
x + 2y – 3 = 0
5x + ky + 7 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 1, b1 = 2, c1 = -3
a2 = 5, b2 = k, c2 = 7
So according to the question,
For unique solution, the condition is
a1 / a2 ≠ b1 / b2
1/5 ≠ 2/k
⇒ k ≠ 10
Hence, the given system of equations will have unique solution for all real values of k other than 10.
Find the value of k for which each of the following system of equations having infinitely many solution: (9-19)
9. 2x + 3y – 5 = 0
6x + ky – 15 = 0
Solution:
The given system of equations is:
2x + 3y – 5 = 0
6x + ky – 15 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 2, b1 = 3, c1 = -5
a2 = 6, b2 = k, c2 = -15
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
2/6 = 3/k
⇒ k = 9
Hence, the given system of equations will have infinitely many solutions, if k = 9.
10. 4x + 5y = 3
kx + 15y = 9
Solution:
The given system of equations is:
4x + 5y – 3= 0
kx + 15y – 9 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 4, b1 = 5, c1 = -3
a2 = k, b2 = 15, c2 = -9
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
4/ k = 5/ 15 = −3/ −9
4/ k = 1/ 3
⇒k = 12
Hence, the given system of equations will have infinitely many solutions, if k = 12.
11. kx – 2y + 6 = 0
4x – 3y + 9 = 0
Solution:
The given system of equations is:
kx – 2y + 6 = 0
4x – 3y + 9 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = k, b1 = -2, c1 = 6
a2 = 4, b2 = -3, c2 = 9
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
k/ 4 = −2/ −3 = 2/ 3
⇒k = 8/ 3
Hence, the given system of equations will have infinitely many solutions, if k = 8/3.
12. 8x + 5y = 9
kx + 10y = 18
Solution:
The given system of equations is:
8x + 5y – 9 = 0
kx + 10y – 18 = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 8, b1 = 5, c1 = -9
a2 = k, b2 = 10, c2 = -18
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
8/k = 5/10 = −9/ −18 = 1/2
⇒k=16
Hence, the given system of equations will have infinitely many solutions, if k = 16.
13. 2x – 3y = 7
(k+2)x – (2k+1)y = 3(2k-1)
Solution:
The given system of equations is:
2x – 3y – 7 = 0
(k+2)x – (2k+1)y – 3(2k-1) = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 2, b1 = -3, c1 = -7
a2 = (k+2), b2 = -(2k+1), c2 = -3(2k-1)
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
2/ (k+2) = −3/ −(2k+1) = −7/ −3(2k−1)
2/(k+2) = −3/ −(2k+1)and −3/ −(2k+1)= −7/ −3(2k−1
⇒2(2k+1) = 3(k+2) and 3×3(2k−1) = 7(2k+1)
⇒4k+2 = 3k+6 and 18k – 9 = 14k + 7
⇒k=4 and 4k = 16 ⇒k=4
Hence, the given system of equations will have infinitely many solutions, if k = 4.
14. 2x + 3y = 2
(k+2)x + (2k+1)y = 2(k-1)
Solution:
The given system of equations is:
2x + 3y – 2= 0
(k+2)x + (2k+1)y – 2(k-1) = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 2, b1 = 3, c1 = -5
a2 = (k+2), b2 = (2k+1), c2 = -2(k-1)
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
2/ (k+2) = 3/ (2k+1) = −2/ −2(k−1)
2/ (k+2) = 3/ (2k+1) and 3/(2k+1) = 2/2(k−1)
⇒2(2k+1) = 3(k+2) and 3(k−1) = (2k+1)
⇒4k+2 = 3k+6 and 3k−3 = 2k+1
⇒k = 4 and k = 4
Hence, the given system of equations will have infinitely many solutions, if k = 4.
15. x + (k+1)y = 4
(k+1)x + 9y = 5k + 2
Solution:
The given system of equations is:
x + (k+1)y – 4= 0
(k+1)x + 9y – (5k + 2) = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 1, b1 = (k+1), c1 = -4
a2 = (k+1), b2 = 9, c2 = -(5k+2)
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
1/ k+1 = (k+1)/ 9 = −4/ −(5k+2)
1/ k+1 = k+1/ 9 and k+1/ 9 = 4/ 5k+2
⇒9 = (k+1)2 and (k+1)(5k+2) = 36
⇒9 = k2 + 2k + 1 and 5k2+2k+5k+2 = 36
⇒k2+2k−8 = 0 and 5k2+7k−34 = 0
⇒k2+4k−2k−8 = 0 and 5k2+17k−10k−34 = 0
⇒k(k+4)−2(k+4) = 0 and (5k+17)−2(5k+17) = 0
⇒(k+4)(k−2) = 0 and (5k+17) (k−2) = 0
⇒k = −4 or k = 2 and k = −17/5 or k = 2
Its seen that k=2 satisfies both the condition.
Hence, the given system of equations will have infinitely many solutions, if k = 9.
16. kx + 3y = 2k + 1
2(k+1)x + 9y = 7k + 1
Solution:
The given system of equations is:
kx + 3y – (2k + 1) = 0
2(k+1)x + 9y – (7k + 1) = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = k, b1 = 3, c1 = – (2k+1)
a2 = 2(k+1), b2 = 9, c2 = – (7k+1)
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
k/ 2(k+1) = 3/ 9 and 3/ 9 = -(2k+1)/ -(7k+1)
3k = 2k +2 and 7k+1 = 3(2k+1) = 6k + 3
k = 2 and k = 2
Hence, the given system of equations will have infinitely many solutions, if k = 2.
17. 2x + (k-2)y = k
6x + (2k-1)y = 2k + 5
Solution:
The given system of equations is:
2x + (k-2)y – k = 0
6x + (2k-1)y – (2k+5) = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 2, b1 = k-2, c1 = – k
a2 = 6, b2 = 2k-1, c2 = -2k-5
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
2/6 = (k-2)/ (2k-1) and (k-2)/ (2k-1) = – k/ -2k-5
4k -2 = 6k -12 and (k-2)(2k+5) = k(2k-1)
2k = 10 and 2k2 – 4k + 5k – 10 = 2k2 – k
⇒ k = 5 and 2k = 10 ⇒ k = 5
Hence, the given system of equations will have infinitely many solutions, if k = 5.
18. 2x + 3y = 7
(k+1)x + (2k-1)y = 4k+1
Solution:
The given system of equations is:
2x + 3y – 7= 0
(k+1)x + (2k-1)y – (4k+1) = 0
The above equations are of the form
a1 x + b1 y − c1 = 0
a2 x + b2 y − c2 = 0
Here, a1 = 2, b1 = 3, c1 = – 7
a2 = (k+1), b2 = 2k-1, c2 = – (4k+1)
So according to the question,
For unique solution, the condition is
a1 / a2 = b1 / b2 = c1 / c2
2/ (k+1) = 3/ (2k−1) = −7/ −(4k+1)
2/ (k+1) = 3/(2k−1) and 3/ (2k−1) = 7/(4k+1)
2(2k−1) = 3(k+1) and 3(4k+1) = 7(2k−1)
⇒4k−2 = 3k+3 and 12k + 3 = 14k − 7
⇒k = 5 and 2k = 10
⇒k = 5 and k = 5
Hence, the given system of equations will have infinitely many solutions, if k = 5.
Hence, the value of k is 7.
Summary: (i) The values of a and b for which the equations 2x + 3y = 7 and (a - b) x + (a + b) y = 3a + b - 2 will have infinitely many solutions will be a = 5 and b = 1.
Therefore, the given system of equations will have infinitely many solutions, if k=7.
Hence, the given system of equations will have infinitely many solutions, if k=2.
For what value of k will the following equations have infinite solutions 2x 3y 7?
Therefore, the given system of equations will have infinitely many solutions, if k=7.For what value of k the system has infinite solutions?
So, for k=2 the given system of the linear equations has infinite solutions.How do you find the solution to a system of infinite solutions?
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For what values of k will the following pair of linear equations have infinitely many solutions? kx + 3y - (k – 3) = 0. 12x + ky - k = 0. The value of k which satisfies both the equations is 6. Tải thêm tài liệu liên quan đến nội dung bài viết For what value of k the following system of equations have infinite solutions 2x-3y=7