What is the value of K so that the system of equations 3x y 5 0 and 6x 2y k 0 have infinitely many solutions?

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.

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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.

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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`.

The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and graphically.

Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y

Case I. Cost of 3 bats = 3x

Cost of 6 balls = 6y

According to question,

3x + 6y = 3900

Case II. Cost of I bat = x

Cost of 3 more balls = 3y

According to question,

x + 3y = 1300

So, algebraically representation be

3x + 6y = 3900

x + 3y = 1300

Graphical representation :

We have,    3x + 6y = 3900

⇒    3(x + 2y) = 3900

⇒    x + 2y = 1300

⇒    a = 1300 - 2y

Thus, we have following table :

We have,    x + 3y = 1300

⇒    x = 1300 - 3y

Thus, we have following table :

When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations.

Solution

Short Answer

Here, a1 = 3, b1 = -1 and c1 = -5
a2 = 6, b2 = -2 and c2 = k

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