# Algebra I

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The sum of three numbers is 147. The second number is 4 more than two times the first number. The third number is five less than three times the first number. Find the three numbers.

Can you please solve it? I have no idea how to do this. Thanks a lot for any help!

• Algebra I -

Let the first number be x

"The second number is 4 more than two times the first number"
the second number is 2x + 4

"The third number is five less than three times the first number"
the third number is 3x - 5

your first sentence "The Sum of three numbers is 147"
x + 2x+4 + 3x-5 = 147
6x = 148
x = 74/3
then second number is 2(74/3) + 4 = 160/3
the third number is 3(74/3) - 5 = 39

check: 74/3 + 160/3 + 69 = 147

• Algebra I -

Call the numbers A, B and C

We know
A + B + C = 147

We are told that:
B = 2 A + 4 (two times the first number and add 4)
and
C = 3 A - 5 (five less than three times the first number)

This is called substituting. Now we can substitute (2 A + 4) for B whenever we see it.

Using this, we can rewrite

A + B + C = 147

as

A + (2 A + 4) + (3 A - 5) = 147

since we know B and C in terms of A.
Make sure you understand this step before continuing.

Gathering the As and the numbers together, we have:

A + 2A + 3A + 4 - 5 = 147
6 A + 4 - 5 = 147.
6 A - 1 = 147.
6 A = 148.
A = 148 / 6.

A = 24 2/3

B = 2 A + 4 = 49 + 4 + 1/3 = 53 1/3

C = 74 - 5 = 69.

• Algebra I -

i think the answeris letter c

• Algebra I -

eat my ass out a straw

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