Question 30 Pair of Linear Equations in Two Variables - Exercise 3.3
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Answer:
The given pair of linear equations is
kx + 3y = k – 3 …(i)
12x + ky = k …(ii)
On comparing the equations (i) and (ii) with ax + by = c = 0,
We get,
a1 = k, b1 = 3, c1 = -(k – 3)
a2 = 12, b2 = k, c2 = – k
Then,
a1 /a2 = k/12
b1 /b2 = 3/k
c1 /c2 = (k-3)/k
For no solution of the pair of linear equations,
a1/a2 = b1/b2≠ c1/c2
k/12 = 3/k ≠ (k-3)/k
Taking first two parts, we get
k/12 = 3/k
k2 = 36
k = + 6
Taking last two parts, we get
3/k ≠ (k-3)/k
3k ≠ k(k – 3)
k2 – 6k ≠ 0
so, k ≠ 0,6
Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.
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Text Solution
Solution : as we know that `a_1x+b_1y+c_1 = 0` <br> `a_2x+b_2y+c_2=0` <br> has infinity solutions only when <br> `a_1/a_2 = b_1/b_2 = c_1/c_2` <br> so here, `k/12= 3/k = -(k-3)/(-k)` <br> `k/12= 3/k` <br> `k^2 = 36` <br> `k= +-6` <br> case 1 : `6/12=3/6=-(6-3)/(-6)` <br> `1/2=1/2=1/2` <br> case 2: `k=-6` <br> `-6/12=3/-6=-(-6-3)/-(-6)` <br> `-1/2=-1/2 = -(-9)/(-(-6)=3/2` <br> `=-1/2=-1/2=3/2` false statement<br> `:.` k = 6 answer
Text Solution
Solution : Comparing the given equations with <br> `a_(1)x + b_(1)y + c_(1) = 0` <br> and `a_(2)x + b_(2)y + c_(2) = 0` <br> We have <br> `(a_(1))/(a_(2)) = (k)/(12), (b_(1))/(b_(2)) = (3)/(k)` and `(c_(1))/(c_(2)) = (-(k + 3))/(k)` <br> Now, for coincident lines <br> `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))` <br> implies `(k)/(12) = (3)/(k) = - ((k+3))/(k)` <br> `{:("Taking first two expressions, we get"),((k)/(12) = (3)/(k)" ...(1)"),(implies" "k^(2) = 36),(therefore " "k = pm 6 ):}:|{:("Taking last two expressions, we get"),((3)/(k) = - ((k + 3))/(k)" "...(2)),(implies " "3k = - k^(2) - 3k),(implies " "k^(2)+ 6k = 0 ),(implies " " k(k + 6) = 0),(therefore " k = 0 or k = - 6 "):}` <br> But k = - 6 is a value which satisfies both the equations (1) and (2). <br> `therefore` k = - 6
For what value of ' K ' will the following pair of linear equations have infinitely many solutions Kx +3 y = k 3 ; 12 x + ky = k[ or k x+3 y k+3=0 ; 12 x+k y k=0]
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