What are the values of a and b for which the following pair of linear equations has infinite number of solutions 2x 3y 7 2ax A by 28?

What are the values of a and b for which the following pair of linear equations has infinite number of solutions 2x 3y 7 2ax A by 28?

What are the values of a and b for which the following pair of linear equations has infinite number of solutions 2x 3y 7 2ax A by 28?
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Question 32 Pair of Linear Equations in Two Variables - Exercise 3.3

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What are the values of a and b for which the following pair of linear equations has infinite number of solutions 2x 3y 7 2ax A by 28?

Answer:

Given pair of linear equations is

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

On comparing with ax + by + c = 0,

We get,

Here, a1 = 2, b1 = 3, c1 = – 7;

And a2 = 2p, b2 = (p + q), c2 = – 28;

a1/a2 = 2/2p

b1/b2 = 3/ (p+q)

c1/c2 = ¼

Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

a1/a2 = b1/b2 = c1/c2

1/p = 3/(p+q) = ¼

Taking first and third parts, we get

p = 4

Again, taking last two parts, we get

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

Here, we see that the values of p = 4 and q = 8 satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

What are the values of a and b for which the following pair of linear equations has infinite number of solutions 2x 3y 7 2ax A by 28?
What are the values of a and b for which the following pair of linear equations has infinite number of solutions 2x 3y 7 2ax A by 28?

The given system of equations can be written as2x + 3y - 7 = 0                         ….(i)2ax + (a + b)y – 28 = 0             ….(ii)This system is of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`where, `a_1 = 2, b_1= 3, c_1= -7 and a_2 = 2a, b_2 = a + b, c_2= – 28`

For the given system of linear equations to have an infinite number of solutions, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)``⇒2/(2a) = 3/(a+b) = (−7)/(−28)``⇒ 2/(2a) =( −7)/(−28 )= 1/4 and 3/(a+b) = (−7)/(−28) = 1/4`⇒ a = 4 and a + b = 12Substituting a = 4 in a + b = 12, we get4 + b = 12 ⇒ b = 12 – 4 = 8

Hence, a = 4 and b = 8.


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The given system of equations:8x + 5y = 98x + 5y - 9 = 0                   ….(i)kx + 10y = 15kx + 10y - 15 = 0                  ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 8, b_1= 5, c_1= -9 and a_2 = k, b_2 = 10, c_2= – 15`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`` i.e.,  8/k = 5/10 ≠( −9)/(−15)``i.e. , 8/k = 1/2 ≠ 3/5``8/k = 1/2 and 8/k ≠ 3/5``⇒ k = 16 and k ≠ 40/3`

Hence, the given system of equations has no solutions when k is equal to 16.


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The given system of equations:kx + 3y = 3kx + 3y - 3 = 0                    ….(i)12x + ky = 612x + ky - 6 = 0                    ….(ii)

These equations are of the following form:

`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = k, b_1= 3, c_1= -3 and a_2 = 12, b_2 = k, c_2= –6`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``i .e.,  k/12 = 3/k ≠ (−3)/(−6)``k/12 = 3/k and 3/k ≠ 1/2``⇒ k^2 = 36 and k ≠ 6``⇒ k = ±6 and k ≠ 6`

Hence, the given system of equations has no solution when k is equal to -6.


Page 4

The given system of equations:3x - y - 5 = 0                      ….(i)And, 6x - 2y + k = 0             ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where,`a_1 = 3, b_1= -1, c_1= -5 and a_2 = 6, b_2= -2, c_2 = k`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`` i.e., 3/6 = (−1)/(−2) ≠ −5/k``⇒(−1)/(−2) ≠ (−5)/k ⇒ k ≠ -10`

Hence, equations (i) and (ii) will have no solution if k ≠ -10.


Page 5

The given system of equations can be written askx + 3y + 3 - k = 0                       ….(i)12x + ky - k = 0                           ….(ii)This system of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`

where, `a_1 = k, b_1= 3, c_1 = 3 - k and a_2 = 12, b_2 = k, c_2= –k`

For the given system of linear equations to have no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ k/12 = 3/k ≠ (3−k)/(−k)``⇒k/12 = 3/k and 3/k ≠ (3−k)/(−k)``⇒ k^2 = 36 and -3 ≠ 3 - k`⇒ k = ±6 and k ≠ 6⇒k = -6

Hence, k = -6.


Page 6

The given system of equations:5x - 3y = 0                 ….(i)2x + ky = 0                 ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 5, b_1= -3, c_1 = 0 and a_2 = 2, b_2 = k, c_2 = 0`For a non-zero solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2)``⇒ 5/2 = (−3)/k``⇒5k = -6 ⇒ k = (−6)/5`

Hence, the required value of k is `(−6)/5`.