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Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks.\newlineAllie is going to ship some gifts to family members, and she is considering two shipping companies. The first shipping company charges a fee of $16\$16 to ship a medium box, plus an additional $6\$6 per kilogram. A second shipping company charges $12\$12 for the same size of box, plus an additional $8\$8 per kilogram. At a certain weight, the two shipping methods will cost the same amount. How much will it cost? What is that weight?\newlineThe two shipping methods both cost $\$_____ at a weight of _____ kilograms.

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Q. Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks.\newlineAllie is going to ship some gifts to family members, and she is considering two shipping companies. The first shipping company charges a fee of $16\$16 to ship a medium box, plus an additional $6\$6 per kilogram. A second shipping company charges $12\$12 for the same size of box, plus an additional $8\$8 per kilogram. At a certain weight, the two shipping methods will cost the same amount. How much will it cost? What is that weight?\newlineThe two shipping methods both cost $\$_____ at a weight of _____ kilograms.
  1. Representation of Variables: Let xx represent the weight in kilograms and yy represent the total cost in dollars. \newlineFor the first shipping company: \newlineBase fee: $16\$16 \newlineCost per kilogram: $6\$6 \newlineEquation based on the given information: \newlineTotal cost = base fee + (cost per kilogram * weight) \newliney=6x+16y = 6x + 16 \newlineFor the first company, the equation is: y=6x+16y = 6x + 16
  2. First Shipping Company: For the second shipping company: \newlineBase fee: $12\$12 \newlineCost per kilogram: $8\$8 \newlineEquation based on the given information: \newlineTotal cost = base fee + (cost per kilogram ∗* weight) \newliney=8x+12y = 8x + 12 \newlineFor the second company, the equation is: y=8x+12y = 8x + 12
  3. Second Shipping Company: System of equations: \newliney=6x+16y = 6x + 16 \newliney=8x+12y = 8x + 12 \newlineTo find the weight at which the cost is the same, set the two equations equal to each other. \newline6x+16=8x+126x + 16 = 8x + 12
  4. System of Equations: Solve for xx: \newlineSubtract 6x6x from both sides of the equation: \newline6x+16−6x=8x+12−6x6x + 16 - 6x = 8x + 12 - 6x \newline16=2x+1216 = 2x + 12
  5. Solving for x: Subtract 1212 from both sides of the equation: \newline16−12=2x+12−1216 - 12 = 2x + 12 - 12 \newline4=2x4 = 2x
  6. Substituting xx into Equation: Divide both sides by 22 to solve for xx: \newline42=2x2\frac{4}{2} = \frac{2x}{2} \newline2=x2 = x \newlineSo, x=2x = 2
  7. Final Result: We found x=2x = 2. Now, find the value of yy by substituting xx into one of the original equations. \newlineSubstitute 22 for xx in y=6x+16y = 6x + 16: \newliney=6(2)+16y = 6(2) + 16 \newliney=12+16y = 12 + 16 \newliney=28y = 28 \newlineSo, y=28y = 28
  8. Final Result: We found x=2x = 2. Now, find the value of yy by substituting xx into one of the original equations. \newlineSubstitute 22 for xx in y=6x+16y = 6x + 16: \newliney=6(2)+16y = 6(2) + 16 \newliney=12+16y = 12 + 16 \newliney=28y = 28 \newlineSo, y=28y = 28 We found: \newlinex=2x = 2 \newliney=28y = 28 \newlineThe weight at which both shipping methods cost the same is 22 kilograms, and the cost at that weight is yy33.

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