For what value of k will the following pair of linear equations have infinitely many solutions?

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.

Find the value of k for which the following pair of linear equations has infinitely many solutions.
2x + 3y = 7, (k +1) x+ (2k -1) y = 4k + 1

We have,

`2x + 3y = 7 ⇒ 2x + 3y - 7 = 0`
`(k + 1) x + (2k - 1)y = 4k + 1 ⇒ (k + 1)x (2k -1)y - (4k + 1) = 0`

For infinitely many solutions

`a_1/a_2 = b_1/b_2 = c_1/c_2`

⇒ `(2)/(k+1) = (3)/((2k -1)) = (-7)/-(4k +1)`

⇒ `(2)/(k+1) = (3)/(2k -1)`

⇒ `2(2k + 1) = 3 (k+1)`

⇒`4k - 2 = 3k + 3`

⇒`4k - 3k = 3 +2`

`k = 5`

⇒ `(2)/(k+1) = (-7)/-(4k +1)`