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part 2: two pizza orders (harder!) - 4) $3(2x + 5) + 4(x + 1)$ - 5) $2(…

Question

part 2: two pizza orders (harder!)

  • 4) $3(2x + 5) + 4(x + 1)$
  • 5) $2(3y + 4x + 6) + 5(x + 2)$
  • 6) $4(x + 3y + 2) + 3(2x + y)$

part 3: ultimate challenge

  • 7) $5(2x + 3y + 4) + 2(3x + y + 6)$
  • 8) $3(4x + 2y + 5) + 4(x + 3y + 2)$
  • 9) $2(5x + 4y + 3) + 3(2x + y + 7)$

Explanation:

Response

Let's solve each problem one by one using the distributive property and combining like terms.

Problem 4: \( 3(2x + 5) + 4(x + 1) \)

Step 1: Distribute the coefficients

Using the distributive property \( a(b + c) = ab + ac \), we get:
\( 3(2x) + 3(5) + 4(x) + 4(1) \)
\( = 6x + 15 + 4x + 4 \)

Step 2: Combine like terms

Combine the \( x \)-terms and the constant terms:
\( (6x + 4x) + (15 + 4) \)
\( = 10x + 19 \)

Step 1: Distribute the coefficients

Using the distributive property:
\( 2(3y) + 2(4x) + 2(6) + 5(x) + 5(2) \)
\( = 6y + 8x + 12 + 5x + 10 \)

Step 2: Combine like terms

Combine the \( x \)-terms and the constant terms:
\( 8x + 5x + 6y + (12 + 10) \)
\( = 13x + 6y + 22 \)

Step 1: Distribute the coefficients

Using the distributive property:
\( 4(x) + 4(3y) + 4(2) + 3(2x) + 3(y) \)
\( = 4x + 12y + 8 + 6x + 3y \)

Step 2: Combine like terms

Combine the \( x \)-terms, \( y \)-terms, and the constant term:
\( (4x + 6x) + (12y + 3y) + 8 \)
\( = 10x + 15y + 8 \)

Answer:

\( 10x + 19 \)

Problem 5: \( 2(3y + 4x + 6) + 5(x + 2) \)