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b) Emma's limo service charges $50 plus $2.10 per km. Sarah's limo service charges $42 plus $2.20 per km. When do they charge the same amount?

b) Emma's limo service charges $50 \$ 50 plus $2.10 \$ 2.10 per km. Sarah's limo service charges $42 \$ 42 plus $2.20 \$ 2.20 per km. When do they charge the same amount?

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Q. b) Emma's limo service charges $50 \$ 50 plus $2.10 \$ 2.10 per km. Sarah's limo service charges $42 \$ 42 plus $2.20 \$ 2.20 per km. When do they charge the same amount?
  1. Identify Cost Functions: Identify the cost functions for both Emma's and Sarah's limo services.\newlineEmma's cost function: E(km)=50+2.10×kmE(\text{km}) = 50 + 2.10 \times \text{km}\newlineSarah's cost function: S(km)=42+2.20×kmS(\text{km}) = 42 + 2.20 \times \text{km}
  2. Set Equations Equal: Set the two cost functions equal to each other to find the distance at which they charge the same amount. 50+2.10×km=42+2.20×km50 + 2.10 \times km = 42 + 2.20 \times km
  3. Isolate Variable kmkm: Begin to isolate the variable kmkm by moving the terms without kmkm to one side of the equation.\newline2.10×km−2.20×km=42−502.10 \times km - 2.20 \times km = 42 - 50
  4. Combine Like Terms: Combine like terms to simplify the equation.\newline−0.10×km=−8-0.10 \times km = -8
  5. Divide to Solve: Divide both sides of the equation by −0.10-0.10 to solve for kmkm.km=−8−0.10km = \frac{-8}{-0.10}
  6. Calculate kmkm Value: Calculate the value of kmkm.km=80km = 80

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