Question 32 Pair of Linear Equations in Two Variables - Exercise 3.3
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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.
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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.
Page 2
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.
Page 3
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`.
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