The sum of 3 numbers is 61. The second number is 5 times the first. While the 3rd is 2 less then the first. How do you find the numbers?

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The sum of 3 numbers is 61. The second number is 5 times the first. While the 3rd is 2 less then the first. How do you find the numbers?

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Answer 1 · 2015-10-18T22:42:37

$9$ , $45$ , and $7$

Explanation:

Let's say that your numbers are $x$ , $y$ , and $z$ .

The first thing to write down is that their sum is equal to $61$

$x + y + z = 61" " " "color(purple)((1))$

Next, you know that the second number, $y$ , is five times bigger than the first, $x$ . This is equivalent to saying that you need to multiply $x$ by $5$ to get $y$

$y = 5 * x" " " "color(purple)((2))$

Finally, you know that if you subtract $2$ from $x$ you get the $z$

$z = x - 2" " " "color(purple)((3))$

Use equations $color(purple)((2))$ and $color(purple)((3))$ to find the value of $x$ in equation $color(purple)((1))$

$x + 5x + x - 2 =61$

$7x = 63 implies x = 63/7 = color(green)(9)$

Now take this value of $x$ and find $y$ and $z$

$y = 5 * (9) = color(green)(45)$

$z = (9) - 2 = color(green)(7)$

The three numbers are $9$ , $45$ , and $7$ .

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The sum of 3 numbers is 61. The second number is 5 times the first. While the 3rd is 2 less then the first. How do you find the numbers?

1 Answer

Explanation:

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