Question

6 pans + 7 pencils = 58 and 4 pans + 3 pencils = 32 . you buy 3 pen and 3 pencil so How meny money have you need .​

Answer 1

$\sf\large\underline{Given:}$

$\tt{\implies Cost\:_{(6\:pens)}+Cost\:_{(7\:pencils)}=58}$

$\tt{\implies Cost\:_{(4\:pens)}+Cost\:_{(3\:pencils)}=32}$

$\sf\large\underline{To\: Find:}$

$\tt{\implies Cost\:_{(3\:pens)}+Cost\:_{(3\:pencils)}=?}$

$\sf\large\underline{Solution:}$

$\tt{\implies Let,\:The\:cost\:a\:pen\:be\:x}$

$\tt{\implies Let,\:The\:cost\:a\:pencil\:be\:y}$

• Here solving this question by equation process]

$\tt{\implies 6x+7y=58-----(i)}$

$\tt{\implies 4x+3y=32-----(ii)}$

• In eq 1st multiplying by 4 and eq 2nd multiplying by 6 than subtracting each other]

$\tt{\implies 24x+28y=232}$

$\tt{\implies 24x+18y=192}$

• By solving equation we get, here]

$\tt{\implies 10y=40}$

$\tt{\implies y=4}$

• Putting the value of y=4 in eq 1]

$\tt{\implies 6x+7y=58}$

$\tt{\implies 6x+7(4)=58}$

$\tt{\implies 6x+28=58}$

$\tt{\implies 6x=58-28}$

$\tt{\implies 6x=30}$

$\tt{\implies x=5}$

$\sf\large{Hence,}$

$\tt{\implies Cost\:_{(3\:pens)}+Cost\:_{(3\:pencils)}}$

$\tt{\implies Cost\:_{(3x)}+Cost\:_{(3y)}}$

• Here, putting the value of X and y]

$\tt{\implies 3(5)+3(4)}$

$\tt{\implies 15+12}$

$\tt{\implies Rs.27}$

Therefore, I need Rs.27 to buy 3 pens and 3 pencils.