Question

7 chairs and 5 tables cost 3950 and 8 chairs and 11 tables cost 6100. Find the cost of one chair and one table separately.

Answer 1

Answer:

$\huge{\red{\boxed{\red{\boxed{\green{\underline{\sf{x=350\:and\: y=300-}}}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\red{\boxed{\blue{\underline{\sf{\red{SOLUTION-}}}}}}}}}$

$chairs = x \\ tables = y \\ according \: to \: question \\ 7x + 5y = 3950 \\ we \: use \: substitution \: method \: to \: solve \\ 7x = 3950 - 5y \\ x = \frac{3950 - 5y}{7} - - - - - (1) \\ 8x + 11y = 6100 - - - - - (2) \\ putting \: value \: of \: x\:in \: (2) \\ 8 \times \frac{(3950 - 5y)}{7} + 11y = 6100 \\ = ) \frac{31600 - 40y}{7} + \frac{11y}{1} = 6100 \\ = ) \frac{31600 - 40y + 77y}{7} = 6100 \\ = )31600 + 37y = 42700 \\ = )37y = 42700 - 31600 \\ = )37y = 11100 \\ = )y = \frac{11100}{37} \\ = )y = 300 \\ putting \: value \: of \: y \: in \: (1) \\ = ) x = \frac{3950 - 5 \times 300}{7} \\ = ) x = \frac{3950 - 1500}{7} \\ = )x = \frac{2450}{7} \\ = )x = 350$

>> COST OF CHAIR = 350

>> COST OF TABLE = 300

$\huge{\red{\boxed{\red{\boxed{\green{\underline{\sf{x=350\:and\: y=300-}}}}}}}}$