3 years ago

One positive integer is 5 times another positive integer and their product is 320. What are the positive integers?

1 answer:

7 0

Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
x = 5y
xy = 320

Substitute the first equation into the second equation
(5y)(y) = 320
5y^2 = 320
y^2 = 64
y = 8 (y must be positive)

The integers are (x, y) = (40, 8).

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4 years ago

When p2 – 4p is subtracted from p2 + p – 6, the result is 5p-6 To get p – 9, subtract

jeka57 [31]

$(p^{2} + p - 6) - (x) = p - 9

$ is what is asked.

So all you need to do is a value that will result to p - 9.

x = $(p^{2} + p - 6) - p + 9$ To isolate x we just transpose the values from the other side of the equation and do the opposite operation.

Combine like terms and just copy the terms that do not have similar expressions:

p will cancel out because p - p = 0
-6 and + 9 will come together because they are like terms: -6 + 9 = 3
$p^{2}$ does not have a pair so it will remain as is. you will be left with:

$p^{2} + 3$

Let us check:

$p^{2} + p - 6$
- $p^{2} + 0 + 3$
----------------------------
$0 + p - 9$ or $p -9$