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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineA boutique in Winchester specializes in leather goods for men. Last month, the company sold 1010 wallets and 3535 belts, for a total of $1,450\$1,450. This month, they sold 7777 wallets and 2121 belts, for a total of $3,710\$3,710. 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 elimination, and fill in the blanks.\newlineA boutique in Winchester specializes in leather goods for men. Last month, the company sold 1010 wallets and 3535 belts, for a total of $1,450\$1,450. This month, they sold 7777 wallets and 2121 belts, for a total of $3,710\$3,710. How much does the boutique charge for each item?\newlineThe boutique charges $____\$\_\_\_\_ for a wallet, and $____\$\_\_\_\_ for a belt.
  1. Equation Setup: Let's denote the price of each wallet as ww and the price of each belt as bb. The boutique sold 1010 wallets and 3535 belts for a total of $1,450\$1,450. This gives us the equation 10w+35b=145010w + 35b = 1450.
  2. System of Equations: In the following month, they sold 7777 wallets and 2121 belts for a total of $3,710\$3,710. This gives us the equation 77w+21b=371077w + 21b = 3710.
  3. Variable Elimination: We now have a system of two equations. We need to eliminate one of the variables, ww or bb. We choose to eliminate bb because its coefficients are not multiples of each other, and it will be easier to manipulate the equations to get integer coefficients for bb.
  4. Coefficient Adjustment: To eliminate bb, we can multiply the first equation by 2121 (the coefficient of bb in the second equation) and the second equation by 3535 (the coefficient of bb in the first equation) to get the coefficients of bb to be the same. This gives us the new equations 210w+735b=30450210w + 735b = 30450 and 2695w+735b=1298502695w + 735b = 129850.
  5. Variable Solving: We now subtract the first new equation from the second new equation to eliminate bb. This gives us 2485w=994002485w = 99400. Solving for ww gives us w=994002485w = \frac{99400}{2485}, which simplifies to w=40w = 40.
  6. Substitution and Final Solution: We substitute w=40w = 40 into the first original equation 10w+35b=145010w + 35b = 1450 and solve for bb. This gives us 400+35b=1450400 + 35b = 1450. Subtracting 400400 from both sides gives us 35b=105035b = 1050, and dividing both sides by 3535 gives us b=30b = 30.

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