How to solve simultaneous two variable system 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
 A: 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 $(x,y)=(4,2)$.
A: 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).
A: Hints:
$$x+y=6\Longrightarrow y=6-x$$
And now substitute in second equation:
$$x^2+(6-x)^2=20\Longrightarrow  x^2-6x+8=0\iff (x-4)(x-2)=0\ldots$$
A: $$
x+y = 6 \Longrightarrow x=6-y
$$
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?
A: $(x+y)^2=36$
$x^2+y^2+2xy=36$
$x^2+y^2=20$
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.
