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AleksAgata [21]
4 years ago
6

Can someone help me plzzz !!!

Mathematics
1 answer:
rewona [7]4 years ago
6 0

To solve for y, you need to isolate/get the variable "y" by itself in the equation:

a(n + y) = 10y + 32      First distribute a into (n + y)

(a)n + (a)y = 10y + 32

an + ay = 10y + 32        Subtract an on both sides

an - an + ay = 10y + 32 - an

ay = 10y + 32 - an      Subtract 10y on both sides to get "y" on one side of the equation

ay - 10y = 10y - 10y + 32 - an

ay - 10y = 32 - an       Now take out the y in (ay - 10y)

y(a - 10) = 32 - an       Divide (a - 10) on both sides to get "y" by itself

\frac{y(a-10)}{(a-10)} =\frac{32-an}{(a-10)}

y=\frac{32-an}{a-10}

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  more than half are being used

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Perhaps easiest to understand is the most straightforward: figure the total number of seats and the number of seats being used. Check to see if that is more than half.

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Another way to look at this is to compare utilization to 1/2 in each carriage class.

  In first class 1/2 -3/8 = 1/8 of 64 seats = 8 seats fewer than 1/2 the seats are being used.

  In standard class, 7/13 -1/2 = 14/26 -13/26 = 1/26 of 78 seats = 3 seats more than 1/2 the seats are being used. There are 6 standard class carriages, so 6·3 = 18 more than half the seats in the standard-class carriages are being used.

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Combine the terms by multiplying into a single fraction.

\bf{\dfrac{5}{2}x-7=\dfrac{3}{4}x+14   }

Find the common denominator.

\bf{\dfrac{5x}{2}-7=\dfrac{3}{4}x+14   }

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\bf{\dfrac{5x(-7)}{2}=\dfrac{3}{4}x+14   }

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\bf{\dfrac{5x-14}{2}=\dfrac{3}{4}x+14   }

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\bf{\dfrac{5x-14}{2}=\dfrac{3x}{4}+14   }

Find the common denominator.

\bf{\dfrac{5x-14}{2}=\dfrac{3x}{4}+\dfrac{4\cdot14}{4}    }

Combine fractions with the lowest common denominator.

\bf{\dfrac{5x-14}{2}=\dfrac{3x+4\cdot14}{4}   }

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\bf{\dfrac{5x-14}{2}=\dfrac{3x+56}{4}   }

Eliminate the denominators of the fractions.

\bf{4\cdot\dfrac{5x-14}{2}=4\cdot\dfrac{3x+56}{4}   }

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\bf{10x-28=3x+56 }

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\bf{10x-28+28=3x+56+28 }

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\bf{10x-3x=3x+84-3x }

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\bf{7x=84 }

Divide both sides by the same factor.

\bf{x=\dfrac{84}{7} }

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\bf{x=12 \ \ \ === > \ \ \ Answer}

↓

<h3>Verification</h3>

Let x=12.

  1. \bf{\dfrac{5}{2}\times12-7=\dfrac{3}{4}\times12+14   }
  2. \bf{\dfrac{5\times12}{2}-7=\dfrac{3}{4}\times12+14   }
  3. \bf{\dfrac{60}{2}-7=\dfrac{3}{4}\times12+14   }
  4. \bf{30-7=\dfrac{3}{4}\times12+14   }
  5. \bf{30-7=\dfrac{3\times12}{4}+14   }
  6. \bf{30-7=\dfrac{36}{4}+14   }
  7. \bf{30-7=9+14   }
  8. \bf{23=9+14 }
  9. \bf{23=23}

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