SOLUTION: The cost of two chairs and three tables for an office is $1,800.00. After a month, the cost of each chair and each table increased by 20%. The office again bought 6 chairs and 2 ta

Let x be the cost of a chair previous month; y be the cost of a table previous month.

Then the purchase equation for the previous month is

    2x + 3y = 1800  dollars.    (1)


Next month the cost of a chair is 1.2x; the cost of a table is 1.2y.

Next month purchase equation is

    6*(1.2x) + 2*(1.2y) = 4800  dollars.    (2)


Transform equation (2) by dividing all its terms by 1.2. You will get

    6x + 2y = 4000.    (3)


Thus we have a system of two equations

    2x + 3y = 1800     (1')
    6x + 2y = 4000     (3')


To solve, multiply equation (1') by 3; keep equation (3') as is

    6x + 9y = 5400     (1'')
    6x + 2y = 4000     (3'')


From equation (1'') subtract equation (3'').  You will get

         9y - 2y = 5400 - 4000

            7y   =    1400

             y   =    1400/7 = 200.


Thus one table cost was 200 dollars previous month;  
so next month new price is 1.2*200 = 240 dollars per table. 


Now from equation (1),

    2x + 3*200 = 1800,  2x = 1800 - 3*200 = 1200,  x = 1200/2 = 600 dollars.


Thus one chair cost was 600 dollars previous month;
so next month new price is 1.2*600 = 720 dollars per chair.


ANSWER.  New prices are 240 dollars per table and 720 dollars per chair.