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`
or
⇒ `(2)/(k+1) = (-7)/-(4k +1)`
⇒ `2(4k + 1) = 7 (k+1)`
⇒ `8k + 2 = 7k + 2`
⇒`8k - 7k = 7- 2`
`k = 5`
Hence, the value of k is 5 for which given equations have infinitely many solutions.
Concept: Pair of Linear Equations in Two Variables
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