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My maths knowledge is rusty and need some help in brushing it up. I tried to google around could not get what i am looking for

How to solve the below equation $x + y = 6$ and $x^2 + y^2 = 20$

Help is greatly appreciated

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The easiest way for you probably substitution.

Since $x+y=6$, $y=6-x$. Therefore, substitute it into the other equation, you get $$x^2+(6-x)^2=20$$ $$x^2+x^2-12x+36=20$$ $$x^2-6x+8=0$$ $$(x-2)(x-4)=0$$

Therefore, the two roots are $2$ and $4$.

Note that $x$ and $y$ are interchangable, you get two pairs of solutions

$(x,y)=(2,4)$ and $(x,y)=(4,2)$.

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wow.... thanks...!!!! – linux developer Mar 17 '13 at 4:49

Start with the simpler equation, $x+y=6$, and solve it for $y$ in terms of $x$; you get $y=6-x$. Now substitute that value of $y$ into the other equation to get


After you multiply out the lefthand side, you have


and collecting all terms on one side of the equation leaves you with


You might as well simplify by dividing through by $2$:


At this point you can either invoke the quadratic formula or factor the quadratic to get


A product of numbers is $0$ if and only if at least one of the factors is $0$, so the solutions to this equation are $x-2=0$ and $x-4=0$, i.e., $x=2$ and $x=4$. Recall that $y=6-x$, so the solutions to the original pair of equations are $x=2,y=4$ and $x=4,y=2$ (and you can of course check that both are correct by substituting them into the original equations).

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$$x+y=6\Longrightarrow y=6-x$$

And now substitute in second equation:

$$x^2+(6-x)^2=20\Longrightarrow x^2-6x+8=0\iff (x-4)(x-2)=0\ldots$$

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$$ x+y = 6 \Longrightarrow x=6-y $$


$$ (6-y)^2 + y^2 = 20 \Longrightarrow 16-12y+2y^2 = 0 \Longrightarrow\\ 8-6y + y^2 = 0 \Longrightarrow (y-4)(y-2)=0\\ $$ Can you take it from here?

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Thank you for the suggestion – linux developer Mar 17 '13 at 4:49
@linuxdeveloper No problem. welcome to math.SE – Rustyn Mar 17 '13 at 4:50




Therefore, $xy= 8$

There are four possible pairs $(4,2)(-4,-2)(-2,-4)(2,4)$ , Since $x+y=6$ Only the $+ve$ pairs are the solution.

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