Question-201722

Question Number 201722 by Calculusboy last updated on 11/Dec/23

Commented by Frix last updated on 11/Dec/23

There also should be complex solutions

Answered by Sutrisno last updated on 11/Dec/23

x^{2} + y^{2} + x y = 3

{(x + y)}^{2} - x y = 3

x y = {(x + y)}^{2} - 3 \dots (1)

((x + y) + 1) (4 + {(x + y)}^{2} - 3 + 2 (x + y)) = 27

((x + y) + 1) ({(x + y)}^{2} + 2 (x + y) + 1) = 27

((x + y) + 1) {((x + y) + 1)}^{2} = 27

{((x + y) + 1)}^{3} = 27

x + y = 2 \dots (2)

x y = 2^{2} - 3 \to x y = 1

x (2 - x) = 1

x^{2} - 2 x + 1 = 0

x = 1, y = 1

Commented by Calculusboy last updated on 11/Dec/23

t h a n k s s i r

Answered by esmaeil last updated on 12/Dec/23

x + y = s \to S^{2} - p = 3

x y = p \to (s + 1) (4 + p + 2 s) = 27 \to

(s + 1) ((s^{2} + 1 + 2 s) = 27 \to

(s + 1) = 3 \to s = 2 \to p = 1

a^{2} - 2 a + 1 = 0 \to x = y = 1

Answered by Rasheed.Sindhi last updated on 12/Dec/23

S l i g h t l y d i f f e r e n t w a y

{\begin{cases} x^{2} + y^{2} + x y = 3 \dots \dots . (i) \\ (x + y + 1) (4 + x y + 2 x + 2 y) = 27 \dots (i i) \end{cases}

(i i) \Rightarrow (x + y + 1) (1 + 3 + x y + 2 x + 2 y) = 27

(x + y + 1) (1 + x^{2} + y^{2} + x y + x y + 2 x + 2 y) = 27

(x + y + 1) (1 + x^{2} + y^{2} + 2 x y + 2 x + 2 y) = 27

(x + y + 1) ({(x + y)}^{2} + 2 (x + y) + 1) = 27

(x + y + 1) {(x + y + 1)}^{2} = 27

{(x + y + 1)}^{3} = 27

x + y + 1 = 3

x + y = 2 \dots . . (i i i)

x^{2} + y^{2} + 2 x y = 4

x^{2} + y^{2} + x y + x y = 4

3 + x y = 4

x y = 1 \dots \dots . (i)

(i i i) & (i v) : x = 1, y = 1

Commented by Calculusboy last updated on 15/Dec/23

t h a n k s s i r

Answered by Rasheed.Sindhi last updated on 12/Dec/23

A way leading \begin{matrix} ALL \end{matrix} the solutions :

{\begin{cases} x^{2} + y^{2} + x y = 3 \dots . (i) \\ (x + y + 1) (4 + x y + 2 x + 2 y) = 27 \dots (i i) \end{cases}

(i i) \Rightarrow (x + y + 1) (1 + 3 + x y + 2 x + 2 y) = {(3)}^{3}

(x + y + 1) (1 + x^{2} + y^{2} + x y + x y + 2 x + 2 y) = {(3)}^{3}

(x + y + 1) ({(x + y)}^{2} + 2 (x + y) + 1) = {(3)}^{3}

(x + y + 1) {(x + y + 1)}^{2} = {(3)}^{3}

{(x + y + 1)}^{3} = {(3)}^{3}

{(x + y + 1)}^{3} - {(3)}^{3} = 0

{(x + y + 1) - 3} {{(x + y + 1)}^{2} + 3 (x + y + 1) + 9} = 0^{★ ★}

Missing \left or extra \right

x + y = 2 \Rightarrow y = 2 - x

(i) \Rightarrow x^{2} + {(2 - x)}^{2} + x (2 - x) = 3

x^{2} - 2 x + 1 = 0

{(x - 1)}^{2} = 0

x = 1^{✓} \Rightarrow y = 2 - x = 2 - 1 = 1^{✓}

^{★} x + y + 1 = \frac{- 3 \pm \sqrt{9 - 4 (9) (1)}}{2} = \frac{- 3 \pm 3 \sqrt{3}}{2}

x + y = \frac{- 3 \pm 3 \sqrt{3}}{2} - 1 = \frac{- 5 \pm 3 \sqrt{3}}{2}

y = \frac{- 5 \pm 3 \sqrt{3}}{2} - x

(i) \Rightarrow {(x + y)}^{2} - x y = 3 \Rightarrow {(\frac{- 5 \pm 3 \sqrt{3}}{2})}^{2} - x (\frac{- 5 \pm 3 \sqrt{3}}{2} - x) = 3

l e t \frac{- 5 \pm 3 \sqrt{3}}{2} = a

a^{2} - x (a - x) - 3 = 0

x^{2} - a x + a^{2} - 3 = 0

x = \frac{a \pm \sqrt{a^{2} - 4 (a^{2} - 3)}}{2} = \frac{a \pm \sqrt{12 - 3 a^{2}}}{2}

y = a - x = a - \frac{a \pm \sqrt{12 - 3 a^{2}}}{2} = \frac{a \mp \sqrt{12 - 3 a^{2}}}{2}

w h e r e a = \frac{- 5 \pm 3 \sqrt{3}}{2}

Commented by Rasheed.Sindhi last updated on 12/Dec/23

^{★ ★} x + y + 1 - 3 = 0 ∣ {(x + y + 1)}^{2} + 3 (x + y + 1) + 9 = 0

x + y + 1 = 3, 3 ω, 3 ω^{2}

x + y = 2, 3 ω - 1, 3 ω^{2} - 1

(i) \Rightarrow {(x + y)}^{2} - x y = 3 \Rightarrow x y = {(x + y)}^{2} - 3

x y = 4 - 3, {(3 ω - 1)}^{2} - 3, {(3 ω^{2} - 1)}^{2} - 3

\dots \dots

\dots .

Commented by Calculusboy last updated on 15/Dec/23

n i c e s o l u t i o n s i r

Answered by ajfour last updated on 12/Dec/23

x + y = s, x y = m

s^{2} = 3 + m

(s + 1) (2 s + m + 4) = 27

(s + 1) (s^{2} + 2 s + 1) = 27

s = - 1 + 3, - 1 + 3 ω, - 1 + 3 ω^{2}

x, y = \frac{s \pm \sqrt{3 (4 - s^{2})}}{2}

s a y i f s = 2

x, y = \frac{s}{2} = 1

Commented by Calculusboy last updated on 15/Dec/23

t h a n k s s i r