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In summary, the sum of three numbers is 95, with the second number being 5 more than the first, and the third number being 3 times the second. The numbers are x=15, y=20, and z=60.

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Let \( x + y + z = 95 \), \( y = x + 5 \), and \( z = 3y \).

You can now use Elimination, Substitution, or Matrices to solve. I will use substitution by taking the 2nd and 3rd equations and getting the 1st equation in terms of \( y \).

\( y = x+ 5 \) subtract \( 5 \) from both sides.

\( x = y - 5 \)

Substitute \( x = y - 5 \) and \( z = 3y \) into \( x + y + z = 95 \):

\( ( y -5 ) + y + (3y) = 95 \) combine like terms

\( 5y - 5 = 95 \) add \( 5 \) to both sides

\( 5y = 100 \) divide by \( 5 \) on both sides

\( y = 20 \)

Substitute \( y = 20 \) into \( x = y - 5 \):

\( x = (20) - 5 \) simplify

\( x = 15 \)

Substitute \( y = 20 \) into \( z = 3y \):

\( z = 3(20) \) simplify

\( z = 60 \)

ANSWER: \( x = 15 \), \( y = 20 \), and \( z = 60 \)

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Prepare thyself for more questions.SquareOne said:

Let \( x + y + z = 95 \), \( y = x + 5 \), and \( z = 3y \).

You can now use Elimination, Substitution, or Matrices to solve. I will use substitution by taking the 2nd and 3rd equations and getting the 1st equation in terms of \( y \).

\( y = x+ 5 \) subtract \( 5 \) from both sides.

\( x = y - 5 \)

Substitute \( x = y - 5 \) and \( z = 3y \) into \( x + y + z = 95 \):

\( ( y -5 ) + y + (3y) = 95 \) combine like terms

\( 5y - 5 = 95 \) add \( 5 \) to both sides

\( 5y = 100 \) divide by \( 5 \) on both sides

\( y = 20 \)

Substitute \( y = 20 \) into \( x = y - 5 \):

\( x = (20) - 5 \) simplify

\( x = 15 \)

Substitute \( y = 20 \) into \( z = 3y \):

\( z = 3(20) \) simplify

\( z = 60 \)

ANSWER: \( x = 15 \), \( y = 20 \), and \( z = 60 \)

Spoon feeding can be very addictive.

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Spoon feeding can be very "additive." (Dance)jonah said:Prepare thyself for more questions.

Spoon feeding can be very addictive.

-Dan

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