Question

Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$.

Answer 1

Answer:

13.31 and - 3.31

Step-by-step explanation:

(x + y)^2 = x^2 + y^2 + 2xy ; x + y = 10; x^2 + y^2 = 56

This means ;

10^2 = 56 + xy;

100-56 = XY

44 = xy---------(3)

From x + y = 10

x = 10- y

Substitute that in eqn 3;

We have: 44 = (10-y) y

44 =10y - y^2

Y^2-10y -44 = 0;

From formula method of quadratic equation;

Y =[ - (-10) +_ √ (-10)^2 + 4 × 1 x (-44)] / 2 × 1

y = [ 10 +_√ (100 + 176) ]/ 2]

y = [ 10 +_ √ 276]/ 2

y = [ 10 + √ 276]/ 2 or [ 10 _ √ 276]/ 2

y = 13.31 or - 3.31

Similarly x = 10 - 13.31 or 10 - (-3.31)

x = - 3.31 or 13.31

Answer 2

The required ordered pairs are (5+√3, 5-√3) and  (5-√3, 5+√3).

Given equations are:

$x+y=10$......(1)

$x^{2} +y^{2} =56$.......(2)

What is an equation?

An equation is a statement that equates to two statements.

On squaring equation (1) we get

$x^{2} +y^{2} +2xy=100$

From equation(2) $x^{2} +y^{2} =56$

So, $2xy=44$

$xy=22$

$y=\frac{22}{x}$

Put $y=\frac{22}{x}$ in equation (1)

We get $x+\frac{22}{x} =10$

$x^{2} -10x+22=0$

$x=5+\sqrt{3}$

$x=5-\sqrt{3}$

If  $x=5+\sqrt{3}$,

$y=10-x$

$y=5-\sqrt{3}$

So the ordered pair is $(5+\sqrt{3}, 5-\sqrt{3} )$

If  $x=5-\sqrt{3}$,

$y=10-x$

$y=5+\sqrt{3}$

So the ordered pair is $(5-\sqrt{3}, 5+\sqrt{3} )$

Hence, the required ordered pairs are $(5+\sqrt{3}, 5-\sqrt{3} )$ and $(5-\sqrt{3}, 5+\sqrt{3} )$.

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