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The equations 6x+5y=286x + 5y = 28 and 10x+14y=5810x + 14y = 58 represent the cost for lunch and dinner for a family eating out on vacation. If xx is the number of adults and yy is the number of children, how many adults are in the family?

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Full solution

Q. The equations 6x+5y=286x + 5y = 28 and 10x+14y=5810x + 14y = 58 represent the cost for lunch and dinner for a family eating out on vacation. If xx is the number of adults and yy is the number of children, how many adults are in the family?
  1. Write Equations: Write down the system of equations.\newlineWe have the following system of equations:\newline6x+5y=286x + 5y = 28 (11)\newline10x+14y=5810x + 14y = 58 (22)\newlineHere, xx represents the number of adults and yy represents the number of children.
  2. Multiply by 22: Multiply equation (11) by 22 to make the coefficients of yy in both equations the same.\newline2×(6x+5y)=2×282 \times (6x + 5y) = 2 \times 28\newline12x+10y=5612x + 10y = 56 (33)
  3. Subtract to Eliminate: Subtract equation (3)(3) from equation (2)(2) to eliminate yy.
    (10x+14y)−(12x+10y)=58−56(10x + 14y) - (12x + 10y) = 58 - 56
    10x+14y−12x−10y=210x + 14y - 12x - 10y = 2
    −2x+4y=2-2x + 4y = 2 (4)(4)
  4. Divide to Solve: Divide equation (44) by −2-2 to solve for xx.−2x−2=2−2\frac{-2x}{-2} = \frac{2}{-2}x=−1x = -1

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