The sum and product of 2 numbers are 9 and 20 respectively. find the numbers. steps please...?

Question

no direct answers please

Answer 1

x + y = 9
x y = 20

x + 20/x = 9
x² + 20 = 9x
x² - 9x + 20 = 0
(x - 5)(x - 4) = 0
x = 4 , x = 5

Answer 2

First write out what you know:
Let's call the two numbers x and y. Sum means addition. Product means multiplication.

x + y = 9
xy = 20

From the second equation we can see that:
y = 20/x

Let's put this into the first equation:

x + 20/x = 9

To simplify, I'll multiply each term by x giving:

x2(that's x squared) + 20 = 9x

or x2 - 9x + 20 = 0.

Now comes the tricky part. You have to factorise. Trial and error is often a good way but there is also an equation you can use. Look this up yourself. Anyway what you get is:

(x-5)(x-4) = 0

Multiply out the brackets to reasure yourself that this is the same as the previous expression.
If two thing multiplied together make zero, then one or the other (or both) must be zero.
So:
(x-5) = 0 or (x-4) = 0

so x = 5 or x = 4.

If x is 5 then y is 4 and vice-versa. Either way, you have your answer.

Answer 3

there are only 4 possible combinations of whole numbers that add to 9 (1&8, 2&7, 3&6, 4&5). Work out which one of those combinations multiplies to 20.

Answer 4

Let numbers be x and y
x + y =9 so y=9-x
xy =20
x(9-x) =20
x^2 -9x +20 =0
(x -4)(x-5)=0
so x =4 or 5
(y=5 or 4)

Answer 5

the numbers are 5 & 4.

Answer 6

Suppose the numbers are x and y.

Sum of the numbers,
ie, x+y =9
ie, y = 9-x

product of the numbers.
xy = 20

ie, x (9-x) = 20
ie, 9x -x^2 = 20
ie, x^2 -9x = - 20
ie, x^2 -9x + 20 = 0
ie, x^2 -5x - 4x + 20 = 0
ie, x(x-5) -4 (x-5) =0
ie, (x-5)(x-4)=0

herefore x= 4 or x=5

if x = 4, then y = 9-x = 5
if x=5, , then y = 9-x = 4
So the two numbers are 4 and 5

Answer 7

Let x be the first number.
the second number will be 20/x.
(20/x)+x=9
20+x^2=9x
x^2-9x+20=0
(x-5)(x-4)=0
The numbers are 5 and 4

Answer 8

x+y=9 equ(1)
xy=20 equ(2)
equ(2) x=20/y
substitute the value of xin equ(1)
20/y+y=9
take LCM and take y to the other side, then
20+y(square)=9y
rearrange
y(square) - 9y +20=0
factorise
y(square) - 5y -4y+20=0
take y common from the first two terms and 4 from other two terms
y(y-5) - 4(y-5)=0
(y-4)(y-5)=0
therefore, y=5 or 4
now substitute both the value of y in equ(2) to find the value of x
x=20/y
first put y=4
then x=5
put y=5
then x=4
so the points are (5,4) or (4,50
anything can be substituted, both are right.