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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineMario's Pizza just received two big orders from customers throwing parties. The first customer, Melissa, bought 44 regular pizzas and 33 deluxe pizzas and paid $111\$111. The second customer, Pablo, ordered 33 regular pizzas and 1010 deluxe pizzas, paying a total of $246\$246. What is the price of each pizza?\newlineEach regular pizza costs $\$_____, and each deluxe pizza costs $\$_____.

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Q. Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineMario's Pizza just received two big orders from customers throwing parties. The first customer, Melissa, bought 44 regular pizzas and 33 deluxe pizzas and paid $111\$111. The second customer, Pablo, ordered 33 regular pizzas and 1010 deluxe pizzas, paying a total of $246\$246. What is the price of each pizza?\newlineEach regular pizza costs $\$_____, and each deluxe pizza costs $\$_____.
  1. Define Prices: Let's denote the price of a regular pizza as rr and the price of a deluxe pizza as dd. Melissa's order gives us the first equation: 4r+3d=1114r + 3d = 111.
  2. Form Equations: Pablo's order gives us the second equation: 3r+10d=2463r + 10d = 246.
  3. Elimination Method: We now have a system of equations to solve:\newline4r+3d=1114r + 3d = 111\newline3r+10d=2463r + 10d = 246\newlineWe can use either substitution or elimination to solve this system. Let's use the elimination method.
  4. Eliminate Variable: To eliminate one of the variables, we can multiply the first equation by 1010 and the second equation by 33 to make the coefficients of dd the same:\newline(4r+3d)×10=111×10(4r + 3d) \times 10 = 111 \times 10\newline(3r+10d)×3=246×3(3r + 10d) \times 3 = 246 \times 3\newlineThis gives us:\newline40r+30d=111040r + 30d = 1110\newline9r+30d=7389r + 30d = 738
  5. Solve for rr: Now we subtract the second equation from the first to eliminate dd:
    (40r+30d)−(9r+30d)=1110−738(40r + 30d) - (9r + 30d) = 1110 - 738
    40r−9r=1110−73840r - 9r = 1110 - 738
    31r=37231r = 372
  6. Substitute and Solve: Divide both sides by 3131 to solve for rr:31r31=37231\frac{31r}{31} = \frac{372}{31}r=12r = 12
  7. Find Value of d: 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 first equation:\newline4(12)+3d=1114(12) + 3d = 111\newline48+3d=11148 + 3d = 111
  8. Find Value of d: 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 first equation:\newline4(12)+3d=1114(12) + 3d = 111\newline48+3d=11148 + 3d = 111Subtract 4848 from both sides to solve for dd:\newline3d=111−483d = 111 - 48\newline3d=633d = 63
  9. Find Value of d: 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 first equation:\newline4(12)+3d=1114(12) + 3d = 111\newline48+3d=11148 + 3d = 111 Subtract 4848 from both sides to solve for dd:\newline3d=111−483d = 111 - 48\newline3d=633d = 63 Divide both sides by 33 to find the value of dd:\newline3d3=633\frac{3d}{3} = \frac{63}{3}\newlined=21d = 21

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