Problem: 4 chairs and 3 tables cost Rs 2100 and 5 chairs and 2 tables cost Rs 1750. Find the cost of a chair and a table separately.

Solution:

Let the cost of a chair be 'x' and the cost of a table be 'y'.

We are given two equations:

4x + 3y = 2100 -----(1)
5x + 2y = 1750 -----(2)

We need to solve these equations to find the values of x and y.

Method 1: Elimination Method

We can eliminate one variable by adding or subtracting the two equations.

Multiplying equation (1) by 2 and subtracting it from equation (2) multiplied by 3, we get:

(3*5x + 3*2y) - (2*4x + 2*3y) = 15x + 6y - 8x - 6y = 7x = 525

Therefore, x = 75.

Substituting the value of x in equation (1), we get:

4(75) + 3y = 2100
300 + 3y = 2100
3y = 1800
y = 600

Therefore, the cost of a chair is Rs 75 and the cost of a table is Rs 600.

Method 2: Substitution Method

We can solve for one variable in terms of the other in one equation and substitute it in the other equation.

From equation (1), we have:

4x + 3y = 2100
y = (2100 - 4x)/3

Substituting this in equation (2), we get:

5x + 2((2100 - 4x)/3) = 1750

Multiplying both sides by 3, we get:

15x + 2(2100 - 4x) = 5250
15x + 4200 - 8x = 5250
7x = 1050
x = 150

Substituting this value in the equation for y, we get:

y = (2100 - 4(150))/3 = 600

Therefore, the cost of a chair is Rs 150 and the cost of a table is Rs 600.

Conclusion:

Using either method, we get that the cost of a chair is Rs 75 or Rs 150 and the cost of a table is Rs 600.

Let the cost of chair be x and cost of table be y
so 4x+3y=2100 (i)
5x+2y=1750 (ii)
multiply EQ.(i) by 5 and EQ. (ii) by 4
20x+15y=10500
20x+8y=7000

7y=3500, y 3500/7
y=500

4x+3y=2100
4x+3*500=2100
4x+1500=2100
4x=2100-1500
4x=600
x=600/4
x=150

so the cost of chair is 150
and cost of table is 500