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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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  • $\begingroup$ wow.... thanks...!!!! $\endgroup$ – linux developer Mar 17 '13 at 4:49
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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

$$x^2+(6-x)^2=20\;.$$

After you multiply out the lefthand side, you have

$$x^2+36-12x+x^2=20\;,$$

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

$$2x^2-12x+16=0\;.$$

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

$$x^2-6x+8=0\;.$$

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

$$(x-2)(x-4)=0\;.$$

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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Hints:

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

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

Thus:

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

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  • $\begingroup$ Thank you for the suggestion $\endgroup$ – linux developer Mar 17 '13 at 4:49
  • $\begingroup$ @linuxdeveloper No problem. welcome to math.SE $\endgroup$ – Rustyn Mar 17 '13 at 4:50
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$(x+y)^2=36$

$x^2+y^2+2xy=36$

$x^2+y^2=20$

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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