An algebra problem by Fidel Simanjuntak

Algebra Level 3

{ x 2 + y 2 = x y x + y = 12 \large \begin {cases} x^2+y^2=xy \\ x+y=12\end {cases} If the above system of equations holds true for some real numbers x x and y y , find 1 x + 1 y \dfrac{1}{x}+\dfrac{1}{y} .

1 3 \frac{1}{3} 1 4 \frac{1}{4} 1 12 \frac{1}{12} 1 6 \frac{1}{6}

Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Rishabh Jain
Jul 2, 2016

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

⟹ 3 x y = x 2 + y 2 + 2 x y = ( x + y ) 2 = 144 \implies 3xy=x^2+y^2+2xy=(x+y)^2=144

⟹ x y = 144 3 = 48 \implies xy=\dfrac{144}{3}=48

Now,

1 x + 1 y = x + y x y = 12 48 = 1 4 \dfrac 1x+\dfrac 1y=\dfrac{x+y}{xy}=\dfrac{12}{48}=\boxed{\dfrac 14}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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

1 2 2 = 3 x y 12^2 = 3xy

x y = 144 3 xy = \frac{144}{3}

= 48 =48

1 x + 1 y = x + y x y \frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy}

12 48 = 1 4 \frac{12}{48} = \frac{1}{4}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...