{ ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :} solve for x &y in term a


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Question Number 155991 by cortano last updated on 07/Oct/21
${ ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :} solve for x &y in term a$
${\begin{cases} a (x + 2) + y = 3 a \\ a + {2 x}^{3} = y^{3} + (a + 2) x^{3} \end{cases}$ $solve for x & y in term a$
Commented byRasheed.Sindhi last updated on 07/Oct/21
$Mr cortano, if you want answer in terms of a , that is x,y depepend upon a the following answer is right the answer you are looking for: x=((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 )) y=a−a(((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 ))) Fix value of a,calculate x & y there is the solution which is satisfied by both given equations for that fixed value of a. For easy cases substitute a=±13 x=((−1−2(±13)^2 ±(√(12(±13)^2 −3)))/(2−2(±13)^2 )) =((−339±45)/(−336))=(7/8),(8/7) y_1 =13−13((7/8))=((13)/8) y_2 =13−13((8/7))=−((13)/7) (x,y)=((7/8),((13)/8)),((8/7),−((13)/7)) for a=13 VERIFICATION: { ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :} { ((13((7/8)+2)+((13)/8)=3(13))),((13+2((7/8))^3 =(((13)/8))^3 +(13+2)((7/8))^3 )) :} { ((13(((7+16)/8))+((13)/8)=3(13))),((((13.8^3 +2.7^3 )/8^3 )=((13^3 +15.7^3 )/8^3 ))) :} { ((13(((23+1)/8))=3(13)⇒13(3)=3(13))),((((7342)/8^3 )=((7342)/8^3 ))) :} V E R I F I E ^( ) D$
$M r c o r t a n o, i f y o u w a n t a n s w e r$ $i n t e r m s o f a, t h a t i s x, y d e p e p e n d$ $u p o n a t h e f o l l o w i n g a n s w e r i s$ $r i g h t t h e a n s w e r y o u a r e l o o k i n g$ $f o r :$ $x = \frac{- 1 - {2 a}^{2} \pm \sqrt{{12 a}^{2} - 3}}{2 - {2 a}^{2}}$ $y = a - a (\frac{- 1 - {2 a}^{2} \pm \sqrt{{12 a}^{2} - 3}}{2 - {2 a}^{2}})$ $F i x v a l u e o f a, c a l c u l a t e x & y$ $t h e r e i s t h e s o l u t i o n w h i c h i s$ $s a t i s f i e d b y b o t h g i v e n e q u a t i o n s$ $f o r t h a t f i x e d v a l u e o f a .$ $F o r e a s y c a s e s s u b s t i t u t e a = \pm 13$ $x = \frac{- 1 - 2 {(\pm 13)}^{2} \pm \sqrt{12 {(\pm 13)}^{2} - 3}}{2 - 2 {(\pm 13)}^{2}}$ $= \frac{- 339 \pm 45}{- 336} = \frac{7}{8}, \frac{8}{7}$ $y_{1} = 13 - 13 (\frac{7}{8}) = \frac{13}{8}$ $y_{2} = 13 - 13 (\frac{8}{7}) = - \frac{13}{7}$ $(x, y) = (\frac{7}{8}, \frac{13}{8}), (\frac{8}{7}, - \frac{13}{7}) f o r a = 13$ $VERIFICATION :$ ${\begin{cases} a (x + 2) + y = 3 a \\ a + {2 x}^{3} = y^{3} + (a + 2) x^{3} \end{cases}$ ${\begin{cases} 13 (\frac{7}{8} + 2) + \frac{13}{8} = 3 (13) \\ 13 + 2 {(\frac{7}{8})}^{3} = {(\frac{13}{8})}^{3} + (13 + 2) {(\frac{7}{8})}^{3} \end{cases}$ ${\begin{cases} 13 (\frac{7 + 16}{8}) + \frac{13}{8} = 3 (13) \\ \frac{13 . 8^{3} + 2 . 7^{3}}{8^{3}} = \frac{13^{3} + 15 . 7^{3}}{8^{3}} \end{cases}$ ${\begin{cases} 13 (\frac{23 + 1}{8}) = 3 (13) \Rightarrow 13 (3) = 3 (13) \\ \frac{7342}{8^{3}} = \frac{7342}{8^{3}} \end{cases}$ $V \overset{}{E R I F I E} D$
	Answered by Rasheed.Sindhi last updated on 07/Oct/21
	${ ((a(x+2)+y=3a.......(i))),((a+2x^3 =y^3 +(a+2)x^3 ......(ii))) :} solve for x &y in term a (i):y=3a−a(x+2)=a(1−x)...(iii) (ii):y^3 =a+2x^3 −(a+2)x^3 =a+x^3 (2−a−2) =a(1−x^3 )..........(iv) (iv)&(iii):a(1−x^3 )=a^3 (1−x)^3 1−x^3 =a^2 (1−x)^3 [a≠0] 1−x^3 −a^2 (1−x)^3 =0 (1−x)(1+x+x^2 )−a^2 (1−x)^3 =0 (1+x+x^2 )−a^2 (1−x)^2 =0 [x≠1] x^2 +x+1−a^2 (x^2 −2x+1)=0 x^2 −a^2 x^2 +x+2xa^2 +1−a^2 =0 (1−a^2 )x^2 +(1+2a^2 )x+(1−a^2 )=0 x=((−(1+2a^2 )±(√((1+2a^2 )^2 −4(1−a^2 )^2 )))/(2(1−a^2 ))) △=1+4a^4 +4a^2 −4(1−2a^2 +a^4 ) =1+4a^4 +4a^2 −4+8a^2 −4a^4 =12a^2 −3 x=((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 )) (iii):y=a(1−x) y=a−a(((−1−2a^2 ±(√(12a^2 −3)))/(2−2a^2 ))) (to be simplified)$
	${\begin{cases} a (x + 2) + y = 3 a . . . . . . . (i) \\ a + {2 x}^{3} = y^{3} + (a + 2) x^{3} . . . . . . (i i) \end{cases}$ $solve for x & y in term a$ $(i) : y = 3 a - a (x + 2) = a (1 - x) . . . (iii)$ $(ii) : y^{3} = a + {2 x}^{3} - (a + 2) x^{3} = a + x^{3} (2 - a - 2)$ $= a (1 - x^{3}) . . . . . . . . . . (iv)$ $(iv) & (iii) : a (1 - x^{3}) = a^{3} {(1 - x)}^{3}$ $1 - x^{3} = a^{2} {(1 - x)}^{3} [a \neq 0]$ $1 - x^{3} - a^{2} {(1 - x)}^{3} = 0$ $(1 - x) (1 + x + x^{2}) - a^{2} {(1 - x)}^{3} = 0$ $(1 + x + x^{2}) - a^{2} {(1 - x)}^{2} = 0 [x \neq 1]$ $x^{2} + x + 1 - a^{2} (x^{2} - 2 x + 1) = 0$ $x^{2} - a^{2} x^{2} + x + {2 xa}^{2} + 1 - a^{2} = 0$ $(1 - a^{2}) x^{2} + (1 + {2 a}^{2}) x + (1 - a^{2}) = 0$ $x = \frac{- (1 + {2 a}^{2}) \pm \sqrt{{(1 + {2 a}^{2})}^{2} - 4 {(1 - a^{2})}^{2}}}{2 (1 - a^{2})}$ $△ = 1 + {4 a}^{4} + {4 a}^{2} - 4 (1 - {2 a}^{2} + a^{4})$ $= 1 + {4 a}^{4} + {4 a}^{2} - 4 + {8 a}^{2} - {4 a}^{4}$ $= {12 a}^{2} - 3$ $x = \frac{- 1 - {2 a}^{2} \pm \sqrt{{12 a}^{2} - 3}}{2 - {2 a}^{2}}$ $(iii) : y = a (1 - x)$ $y = a - a (\frac{- 1 - {2 a}^{2} \pm \sqrt{{12 a}^{2} - 3}}{2 - {2 a}^{2}})$ $(to be simplified)$
	Commented bycortano last updated on 07/Oct/21

	$x = 1 solution sir \Rightarrow y = 0$
	Answered by Rasheed.Sindhi last updated on 07/Oct/21
	${ ((a(x+2)+y=3a)),((a+2x^3 =y^3 +(a+2)x^3 )) :} solve for x &y in term a { ((y=3a−a(x+2)=a(1−x)...A)),((y^3 =a+2x^3 −(a+2)x^3 =a(1−x^3 )...B)) :} A & B : a^3 (1−x)^3 =a(1−x^3 ) a^2 (1−x)^3 −(1−x^3 )=0 (1−x){a^2 (1−x)−(1+x+x^2 )}=0 x=1 ∣ a^2 −a^2 x−1−x−x^2 =0^★ If x=1 a(x+2)+y=3a⇒a(1+2)+y=3a ⇒3a+y=3a⇒y=0 x=1,y=0 (Answer free of a) ^★ Will produce results in terms of a$
	${\begin{cases} a (x + 2) + y = 3 a \\ a + {2 x}^{3} = y^{3} + (a + 2) x^{3} \end{cases}$ $solve for x & y in term a$ ${\begin{cases} y = 3 a - a (x + 2) = a (1 - x) . . . A \\ y^{3} = a + {2 x}^{3} - (a + 2) x^{3} = a (1 - x^{3}) . . . B \end{cases}$ $A & B : a^{3} {(1 - x)}^{3} = a (1 - x^{3})$ $a^{2} {(1 - x)}^{3} - (1 - x^{3}) = 0$ $(1 - x) {a^{2} (1 - x) - (1 + x + x^{2})} = 0$ $x = 1 ∣ a^{2} - a^{2} x - 1 - x - x^{2} = 0^{★}$ $If x = 1$ $a (x + 2) + y = 3 a \Rightarrow a (1 + 2) + y = 3 a$ $\Rightarrow 3 a + y = 3 a \Rightarrow y = 0$ $x = 1, y = 0 (Answer free of a)$ $^{★} W i l l p r o d u c e r e s u l t s i n t e r m s o f a$
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