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Albert is going to hire a makeup artist for a fashion show and is comparing prices. Bonnie charges $20 as a booking fee and an additional $40 per hour. Linda charges $45 per hour, plus a booking fee of $10. Depending on the length of the show, the cost could end up being the same for either artist. How long would the show be?

Albert is going to hire a makeup artist for a fashion show and is comparing prices. Bonnie charges $20 \$ 20 as a booking fee and an additional $40 \$ 40 per hour. Linda charges $45 \$ 45 per hour, plus a booking fee of $10 \$ 10 . Depending on the length of the show, the cost could end up being the same for either artist. How long would the show be?

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Q. Albert is going to hire a makeup artist for a fashion show and is comparing prices. Bonnie charges $20 \$ 20 as a booking fee and an additional $40 \$ 40 per hour. Linda charges $45 \$ 45 per hour, plus a booking fee of $10 \$ 10 . Depending on the length of the show, the cost could end up being the same for either artist. How long would the show be?
  1. Define Show Length: Let's denote the length of the show in hours as hh. We will set up an equation for the total cost for each makeup artist and then equate them to find the value of hh where the costs are the same.\newlineBonnie's total cost = Booking fee + (Hourly rate ×\times Number of hours)\newlineBonnie's total cost = $20+($40×h)\$20 + (\$40 \times h)
  2. Calculate Bonnie's Cost: Similarly, we calculate Linda's total cost.\newlineLinda's total cost = Booking fee + (Hourly rate ×\times Number of hours)\newlineLinda's total cost = $10+($45×h)\$10 + (\$45 \times h)
  3. Calculate Linda's Cost: Now we equate Bonnie's total cost to Linda's total cost to find the point where they are the same.\newline$20+($40×h)=$10+($45×h)\$20 + (\$40 \times h) = \$10 + (\$45 \times h)
  4. Equate Total Costs: We will solve for 'h' by first subtracting $10\$10 from both sides of the equation to isolate the terms with 'h'.\newline$20−$10+($40×h)=$10−$10+($45×h)\$20 - \$10 + (\$40 \times h) = \$10 - \$10 + (\$45 \times h)\newline$10+($40×h)=$45×h\$10 + (\$40 \times h) = \$45 \times h
  5. Isolate 'h' Terms: Next, we will subtract $40×h\$40 \times h from both sides to get 'h' by itself.$10+($40×h)−($40×h)=($45×h)−($40×h)\$10 + (\$40 \times h) - (\$40 \times h) = (\$45 \times h) - (\$40 \times h)$10=$5×h\$10 = \$5 \times h
  6. Solve for 'h: Finally, we divide both sides by $5\$5 to solve for 'h'.\newline105=(5×h)5\frac{10}{5} = \frac{(5 \times h)}{5}\newlineh = 22

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