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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineA boutique in Booneville specializes in leather goods for men. Last month, the company sold 6060 wallets and 2222 belts, for a total of $3,422\$3,422. This month, they sold 8181 wallets and 2222 belts, for a total of $4,073\$4,073. How much does the boutique charge for each item?\newlineThe boutique charges $\$_____ for a wallet, and $\$_____ for a belt.

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Q. Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineA boutique in Booneville specializes in leather goods for men. Last month, the company sold 6060 wallets and 2222 belts, for a total of $3,422\$3,422. This month, they sold 8181 wallets and 2222 belts, for a total of $4,073\$4,073. How much does the boutique charge for each item?\newlineThe boutique charges $\$_____ for a wallet, and $\$_____ for a belt.
  1. Define Variables: Let's denote the price of a wallet as w w and the price of a belt as b b . We are given that 6060 wallets and 2222 belts were sold for a total of \(3\),\(422\) last month. This can be represented by the equation:\(\newline\)\[ 60w + 22b = 3422 \]
  2. Given Sales Data: We are also given that this month, they sold \(81\) wallets and \(22\) belts for a total of 44,073073. This can be represented by the equation:\newline81w+22b=4073 81w + 22b = 4073
  3. System of Equations: We now have a system of equations to solve:\newline{60w+22b=342281w+22b=4073 \begin{cases} 60w + 22b = 3422 \\ 81w + 22b = 4073 \end{cases} \newlineTo solve this system, we can use the method of elimination or substitution. Let's use elimination to solve for one of the variables.
  4. Elimination Method: To eliminate b b , we can multiply the first equation by 1-1 and add it to the second equation:\newline1×(60w+22b)=1×3422 -1 \times (60w + 22b) = -1 \times 3422 \newline60w22b=3422 -60w - 22b = -3422 \newline81w+22b=4073 81w + 22b = 4073 \newlineAdding the two equations:\newline(81w+22b)+(60w22b)=4073+(3422) (81w + 22b) + (-60w - 22b) = 4073 + (-3422) \newline21w=651 21w = 651
  5. Solve for Wallet Price: Now we can solve for w w :\newline21w=651 21w = 651 \newlinew=65121 w = \frac{651}{21} \newlinew=31 w = 31 \newlineSo, the boutique charges \(31\) for a wallet.
  6. Substitute Wallet Price: Now that we have the price of a wallet, we can substitute \( w = 31 \) into one of the original equations to find \( b \). Let's use the first equation:\(\newline\)\[ 60(31) + 22b = 3422 \]\(\newline\)\[ 1860 + 22b = 3422 \]\(\newline\)\[ 22b = 3422 - 1860 \]\(\newline\)\[ 22b = 1562 \]
  7. Solve for Belt Price: Now we can solve for \( b \):\(\newline\)\[ 22b = 1562 \]\(\newline\)\[ b = \frac{1562}{22} \]\(\newline\)\[ b = 71 \]\(\newline\)So, the boutique charges 7171 for a belt.

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