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Gr 8 question

The sum of two numbers is 36, and one of the numbers is three times the other. find the two numbers

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The two numbers are 9 and 27.

Solution:
Suppose the two numbers are x and y, then,

x + y = 36 ---------1
and
x = 3y -------------2

Solving equation 1 and 2...

3y+y=36 => 4y=36 => y = 9
x + 9 = 36 => x = 27

Please rate and comment if my answer is right.

Posted on Feb 01, 2010

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I'm trying to figure out a 3x3 magic square where the rows equal 21 18 10 and the columns equal 15 24 13 using the numbers 1-9


I don't think you can do it! Could you check your numbers.

If we are using the numbers 1-9 once and only once in a 3x3 magic square, the sum of the rows + the sum of the columns should be 90, since 1+2+3+4+5+6+7+8+9 in the rows adds up to 45 and 1+2+3+4+5+6+7+8+9 in the columns adds up to 45.

When trying to solve, the magic number seem to be the sums 24 and 10. To get 24, the only three numbers that add to 24 are 7 + 8 +9. Similarly to get 10, the lowest numbers re 1 and 2 and the smallest big number to use is 7. I then ran out of number trying to get 21, 13 and 15 sums.

Good luck,

Paul

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Find the smallest two consecutive even integers such that three times the smaller is at least 19 more than the sum of the two integers


Let x be the smallest number.
Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)
three times the smaller 3(x)
19 more -19
sum of the two integers - (x) +( x+2)

Pulling it together,
3x -19 = x + x +2
collect like terms
3x - 19 = 2x + 2
Put all the constants on one side by adding 19 to both sides.
3x - 19 +19= 2x + 2 + 19
3x = 2x +21
Subtract 2x from both sides to have all the x's on one side.
3x - 2x = 2x +21 - 2x
x = 21
The other number is 21 + 2, or 23
Check:three times the smaller = 3 x 21 = 63
sum of the two integers = 21 + 23 or 44
is 63 at least 19 more than 44






Feb 19, 2015 | Office Equipment & Supplies

1 Answer

Consecutive even integers


38.
The three numbers are 34, 36, and 38.

Jun 03, 2014 | Office Equipment & Supplies

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Find two numbers such that the sum of the first and three times the second is 5 and the sum of the second and 2 times the first is 8


Let the two numbers be X and Y:
X + 3Y = 5
2X + Y = 8

Rewrite the second equation to Y = 8 - 2X and substitute in first equation:
X + 3(8-2X) = 5
X + 24 - 6X = 5
-5X + 24 = 5
-5X = 5 - 24
-5X = -19
X = 19/5 = 3.8
Y = 8 - 2(3.8) = 8 - 7.6 = 0.4

Check: 3.8 + 3(0.4) = 3.8 + 1.2 = 5
2(3.8) + .4 = 7.6 + 0.4 = 8

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1 Answer

How to use numbers 1-9 to add up three sums using said numbers only once


1+2+3 is 6.
4+5+6 is 15.
7+8+9 is 24.
That's three sums, using each number only once.

There's more to the problem than this, isn't there? Perhaps making them add up to the SAME sum?

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1 Answer

Cindy is 5 years older than Alex and 7 years older than peter sum


Three unknowns with three equasions:
cindy is 5 years older than alex. write it as c = a + 5
cindy is 7 years older than peter. write it as c = p + 7
The sum of their ages is 21. write it as c + a + p = 21
first solve for peters age in terms of alex's age.
c = p + 7 or
(a + 5) = p + 7 or
p = (a + 5) - 7 or
p = a - 2
substitute in and solve the equasion:
(a + 5) + a + (a - 2) = 21 or
3a + 3 = 21 or
3a = 24 or
a = 8

c = a + 5 or
c = 8 + 5 = 13

c = p + 7 or
p = c - 7 or
p = 13 - 7 = 6

cindy is 13, alex is 8 and peter is 6
This is called simultaneous equasions and will always work if you have the same number of unknowns as you have equasions (or relationships).

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Algebra


Is this somebody's home work assignment or what??? You got the wrong forum dude.

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