Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ A boutique in Winchester specializes in leather goods for men. Last month, the company sold $10$ wallets and $35$ belts, for a total of $\$1,450$ . This month, they sold $77$ wallets and $21$ belts, for a total of $\$3,710$ . How much does the boutique charge for each item? $\newline$ The boutique charges $\$\_\_\_\_$ for a wallet, and $\$\_\_\_\_$ for a belt.

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Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

A boutique in Winchester specializes in leather goods for men. Last month, the company sold

10

wallets and

35

belts, for a total of

\$1,450

. This month, they sold

77

wallets and

21

belts, for a total of

\$3,710

. How much does the boutique charge for each item?

\newline

The boutique charges

\$\_\_\_\_

for a wallet, and

\$\_\_\_\_

for a belt.

Equation Setup: Let's denote the price of each wallet as $w$ and the price of each belt as $b$ . The boutique sold $10$ wallets and $35$ belts for a total of $\$1,450$ . This gives us the equation $10w + 35b = 1450$ .
System of Equations: In the following month, they sold $77$ wallets and $21$ belts for a total of $\$3,710$ . This gives us the equation $77w + 21b = 3710$ .
Variable Elimination: We now have a system of two equations. We need to eliminate one of the variables, $w$ or $b$ . We choose to eliminate $b$ because its coefficients are not multiples of each other, and it will be easier to manipulate the equations to get integer coefficients for $b$ .
Coefficient Adjustment: To eliminate $b$ , we can multiply the first equation by $21$ (the coefficient of $b$ in the second equation) and the second equation by $35$ (the coefficient of $b$ in the first equation) to get the coefficients of $b$ to be the same. This gives us the new equations $210w + 735b = 30450$ and $2695w + 735b = 129850$ .
Variable Solving: We now subtract the first new equation from the second new equation to eliminate $b$ . This gives us $2485w = 99400$ . Solving for $w$ gives us $w = \frac{99400}{2485}$ , which simplifies to $w = 40$ .
Substitution and Final Solution: We substitute $w = 40$ into the first original equation $10w + 35b = 1450$ and solve for $b$ . This gives us $400 + 35b = 1450$ . Subtracting $400$ from both sides gives us $35b = 1050$ , and dividing both sides by $35$ gives us $b = 30$ .