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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineThe drama club is selling gift baskets to raise money for new costumes. During the fall play, they sold a combined 2222 regular gift baskets and 77 deluxe gift baskets, earning a total of $627\$627. During the spring musical, they sold 1414 regular gift baskets and 22 deluxe gift baskets, earning a total of $318\$318. How much are they charging for the different sized gift baskets?\newlineThe drama club is charging $____\$\_\_\_\_ for a regular gift basket and $____\$\_\_\_\_ for a deluxe gift basket.

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Q. Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineThe drama club is selling gift baskets to raise money for new costumes. During the fall play, they sold a combined 2222 regular gift baskets and 77 deluxe gift baskets, earning a total of $627\$627. During the spring musical, they sold 1414 regular gift baskets and 22 deluxe gift baskets, earning a total of $318\$318. How much are they charging for the different sized gift baskets?\newlineThe drama club is charging $____\$\_\_\_\_ for a regular gift basket and $____\$\_\_\_\_ for a deluxe gift basket.
  1. Define Prices: Let's denote the price of a regular gift basket as rr and the price of a deluxe gift basket as dd.
  2. Equation for Fall Play: According to the information given for the fall play, the equation representing the total earnings from the gift baskets is:\newline22r+7d=62722r + 7d = 627
  3. Equation for Spring Musical: Similarly, for the spring musical, the equation representing the total earnings is: 14r+2d=31814r + 2d = 318
  4. System of Equations: We now have a system of two equations with two variables:\newline22r+7d=62722r + 7d = 627\newline14r+2d=31814r + 2d = 318
  5. Elimination Method: To solve the system, we can use the method of substitution or elimination. Let's use the elimination method. We will multiply the second equation by 77 to match the coefficient of dd in the first equation:\newline(14r+2d)×7=318×7(14r + 2d) \times 7 = 318 \times 7\newline98r+14d=222698r + 14d = 2226
  6. Subtract Equations: Now we subtract the first equation from the modified second equation to eliminate dd:$98r+14d\$98r + 14d - 22r+7d22r + 7d = 22262226 - 627627\)98r+14d−22r−7d=2226−62798r + 14d - 22r - 7d = 2226 - 62776r+7d−7d=159976r + 7d - 7d = 159976r=159976r = 1599
  7. Solve for r: Divide both sides by 7676 to solve for r:\newline76r76=159976\frac{76r}{76} = \frac{1599}{76}\newliner=21r = 21
  8. Substitute r: Now that we have the value for rr, we can substitute it back into one of the original equations to solve for dd. Let's use the second equation:\newline14r+2d=31814r + 2d = 318\newline14(21)+2d=31814(21) + 2d = 318\newline294+2d=318294 + 2d = 318
  9. Solve for d: Subtract 294294 from both sides to solve for d:\newline294+2d−294=318−294294 + 2d - 294 = 318 - 294\newline2d=242d = 24
  10. Solve for d: Subtract 294294 from both sides to solve for dd: \newline294+2d−294=318−294294 + 2d - 294 = 318 - 294\newline2d=242d = 24Divide both sides by 22 to find the value of dd:\newline2d2=242\frac{2d}{2} = \frac{24}{2}\newlined=12d = 12

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