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**Hint**: We will consider the chair as x and table as y. Then we will form two equations from the data given. Then on solving these two equations we will get the value of x and y and that is the cost of chair and table.

**Complete step-by-step answer**:

Let the cost of one chair be x and the cost of one table be y.

Now from the first statement, Two chairs and three tables costs Rs.5650

The equation so formed is, \[2x + 3y = 5650\] …eq1

From second statement, three chairs and two tables cost Rs.7100

The equation so formed is, \[3x + 2y = 7100\] …eq2

Now to solve these equations we will use elimination method,

eq1+eq2,

\[2x + 3y + 3x + 2y = 5650 + 7100\]

On adding the same terms,

\[5x + 5y = 12750\]

Dividing both sides by 5,

\[x + y = 2550\] ….eq3

Now to eliminate any one variable we need it in opposite sign,

eq2-eq1,

\[3x + 2y - \left( {2x + 3y} \right) = 7100 - 5650\]

on solving we get,

\[3x + 2y - 2x - 3y = 1450\]

subtracting the same terms,

\[x - y = 1450\] ….eq4

now to solve the equations eq3+-eq4,

\[x + y + x - y = 2550 + 1450\]

Now adding same terms,

\[2x = 4000\]

Dividing both sides by 2,

\[x = 2000\]

This is the cost of one chair.

Now using eq3 we get,

\[y = x - 1450\]

Putting the value of x,

\[y = 2000 - 1450\]

On subtracting we get,

\[y = 550\]

This is the cost of one table.

Thus the cost of one chair is Rs.2000 and that of one table is Rs.550.

**Note**: Note that substitution method can also be used to solve the equations. But here the elimination method is easier to solve the equations. But it is important to notice the equations that are formed do not get wrong in any sign. Because that is the only mistake generally students make.

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