Question

question no 3


Solve on copy ​

Answer 1

Hey mate,.

Let be bat be -x

Let be ball be -y

7x+6y=3800---->1

3x+5y=1750---->2

Multiple equation1 with 5and equation 2 with 6

Now,

35x+30y=19000----->1

18x+30y=10500----->2

-----------------------------------

17x=8500

X=8500

-----------

17

X=500

To find the cost of one ball replace the value of x in any equation.

18(500)+30y=10500

9000+30y=10500

30y=10500-9000

30y=1500

y=50

Hope it will help you

Hope it will help you

Answer 2

HELLO DEAR FRIEND

SOLUTION IN THE ATTACHMENT.

Let the cost of bat be x

cost of ball be y

ATQ,

7x + 6y = 3800 .....(1)

3x + 5y = 1750 ....(2)

(1) × 3

(2) × 7

21x + 18y = 11400 .....(3)

21x + 35y = 12250 .....(4)

$\large \boxed{(3) - (4)}$

21x + 18y - (21x + 35y) =11400 - 12250

21x + 18y - 21x - 35y = -850

18y - 35y = -850

-17y = -850

$\large{y = \frac{ \large{ - 850}}{ \large{ - 17}} }$

$\large \boxed{y = 50}$

putting the value of y on (2)

3x + 5y = 1750

3x + 5(50) = 1750

3x + 250 = 1750

3x = 1750 - 250

3x = 1500

$\large{x = \frac{ \large{ 1500}}{ \large{ 3}} }$

$\large \boxed{x = 500}$

so, cost of bat =

$\large \boxed{ 500}$

cost of ball =

$\large \boxed{ 50}$

THANKS