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The cost of two chairs and three tables for an office is $1800.
After a month, the cost of each chair and each table increased by 20%.
The office again bought 6 chairs and 2 tables at $4800.
Calculate the new cost of a chair and a table.
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It was solved at this forum one or two months ago. See my solution under this link
https://www.algebra.com/algebra/homework/percentage/percentage.faq.question.1203149.html
For your convenience, I copy-paste it from there to here again.
Solution
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.
Solved.