Solve the system following of equations { ((x+y+z=2)),((2x+3y+z=1)),((x^2 +(y+2)^2 +(z−1)^2 =9)) :}


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 110087 by 1549442205PVT last updated on 27/Aug/20
$Solve the system following of equations { ((x+y+z=2)),((2x+3y+z=1)),((x^2 +(y+2)^2 +(z−1)^2 =9)) :}$
$Solve the system following of equations$ ${\begin{cases} x + y + z = 2 \\ 2 x + 3 y + z = 1 \\ x^{2} + {(y + 2)}^{2} + {(z - 1)}^{2} = 9 \end{cases}$
Commented by bemath last updated on 27/Aug/20
$[△((be)/(math))▽] { ((x+y+z = 2)),((2x+3y+z = 1)),((x^2 +(y+2)^2 +(z−1)^2 = 9)) :} ⇒z = z ⇒2−x−y=1−2x−3y ⇒ x+2y=−1 ⇒ (−1−2y)^2 +(y+2)^2 +(2x+3y)^2 =9 (1+2y)^2 +(y+2)^2 +(−2−y)^2 =9 2(y+2)^2 +(1+2y)^2 =9 2(y^2 +4y+4)+4y^2 +4y+1=9 6y^2 +12y=0→6y(y+2)=0 y=0 or y=−2 case(1) y=0→ { ((x=−1)),((z=3)) :} case(2) y=−2→ { ((x=3)),((z=1)) :} solution set is {(−1,0,3),(3,−2,1)}$
$[△ \frac{b e}{m a t h} ▽]$ ${\begin{cases} x + y + z = 2 \\ 2 x + 3 y + z = 1 \\ x^{2} + {(y + 2)}^{2} + {(z - 1)}^{2} = 9 \end{cases}$ $\Rightarrow z = z \Rightarrow 2 - x - y = 1 - 2 x - 3 y$ $\Rightarrow x + 2 y = - 1$ $\Rightarrow {(- 1 - 2 y)}^{2} + {(y + 2)}^{2} + {(2 x + 3 y)}^{2} = 9$ ${(1 + 2 y)}^{2} + {(y + 2)}^{2} + {(- 2 - y)}^{2} = 9$ $2 {(y + 2)}^{2} + {(1 + 2 y)}^{2} = 9$ $2 (y^{2} + 4 y + 4) + 4 y^{2} + 4 y + 1 = 9$ $6 y^{2} + 12 y = 0 \to 6 y (y + 2) = 0$ $y = 0 o r y = - 2$ $c a s e (1) y = 0 \to {\begin{cases} x = - 1 \\ z = 3 \end{cases}$ $c a s e (2) y = - 2 \to {\begin{cases} x = 3 \\ z = 1 \end{cases}$ $s o l u t i o n s e t i s {(- 1, 0, 3), (3, - 2, 1)}$
Commented by john santu last updated on 27/Aug/20

$n i c e$
Commented by 1549442205PVT last updated on 28/Aug/20

$Thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum