Question

I need some help with this: to be calculated as a system: x+y=4 xy=3

Answer 1

it need to be calculated as R x R

Answer 2

x + y = 4 -------------------- (1)

xy = 3 -------------------- (2)

These two equations are given to you.

Answer 3

square the equation (1) both sides.

Answer 4

What do you get?

Answer 5

@paperboy please follow the above instruction and let me know what you got...

Answer 6

What about using substitution?

Answer 7

@mathslover my main language is not romanian so corect me if I'm wrong x^2 + y^2 = 16

Answer 8

y = 4 - x

x(4 - x) = 3

Answer 9

@paperboy - no , (x+y)^2 = x^2 + y^2 + 2xy

and yes @Hero 's method may be a lot easier.

Answer 10

How about solving for 'y' in the equation xy=3 ?
y = 3/x
then substituting this into the other equation?
x + y =4
x + 3/x = 4
Multiplying both sides by 'x'
x² +3 = 4x
x² -4x +3 =0
You should be able to finish this calculation.

Answer 11

That may work too @Hero

But the method I am using here is the one I love the most! :)

It will go like this :

x + y = 4

Squaring both sides :

(x+y)^2 = 16

x^2 + y^2 + 2xy = 16

x^2 + y^2 + 2(3) = 16 [as xy = 3]

x^2 + y^2 = 10

(x-y)^2 = x^2 + y^2 - 2xy

= 10 - 2(3) = 4

therefore , (x-y)^2 = 4

or x-y = 2

thus, x + y = 4 and x - y = 2

Thus, 2x = 6 , x = 3

and y = 1

Answer 12

You only do that if you want to intentionally make the steps longer. There's no need to square both sides.

Answer 13

i would have let mathslover finish explaining his method, then jump in and say, you can do it this way too.... interrupting in between will just lead to confusion.....just saying

Answer 14

@hartnn, no one interrupted anyone

Answer 15

Well, hero, I squared both sides to get the value of x^2 + y^2 which i can put in (x-y)^2 expression and solve for x-y.

Let us stop our discussion and concentrate on teaching @paperboy in proper way. Hero, your method is a bit easier, so you please teach him your method.

Answer 16

No, you can continue with your method. I'm just going to leave now since I'm interrupting. Thanks @hartnn

Answer 17

As you wish,

@paperboy :

First of all, (x+y)^2 = x^2 + y^2 + 2xy not x^2 + y^2

So, you get now : (x^2 + y^2 + 2xy) = 16

right?

Answer 18

yes

Answer 19

so now, can you substitute the value of xy there in the equation :

x^2 + y^2 + 2xy = 16

Answer 20

but i need to solve it as R x R ( R*R) and i think that @wolf1728 's method was what the exercise asked for

Answer 21

Oh, didn't notice that. Sorry, then use @Hero 's method... or @wolf1728 's method.

Answer 22

Hero may like to teach that method now!

Answer 23

No, I just came to say good job on your method @mathslover. It's a good alternative method.

Answer 24

Thanks Hero.

Paper boy,

you are given with xy = 3

and since x + y = 4 or y = 4 - x

therefore : xy = 3 becomes : x(4-x)= 3

Answer 25

can u solve it now?

Answer 26

I think I can, thanks

Answer 27

Great, good luck! Also,
As you are new to this site,
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Good Luck for your learning on this site. Do let me know if you get any problem while working on this site.

Answer 28

@mathslover can you help solve this as you last said? It seems like I'm not doing it right

"you are given with xy = 3

and since x + y = 4 or y = 4 - x

therefore : xy = 3 becomes : x(4-x)= 3"

Answer 29

Great to see that you tried it yourself first. Nice spirit.

See, x(4-x)=3

4x - x^2 = 3

right?

Answer 30

yes, now what i do next?

Answer 31

I can write it as : x^2 + 3 - 4x = 0

Answer 32

or x^2 - 4x + 3 = 0

or x^2 - 3x - x + 3 = 0

x(x-3) - (x-3) = 0

(x-1)(x-3) = 0

so, x = 1 or x = 3

Answer 33

getting it?

Answer 34

yes, thanks a lot