help me solve this system of linear equation.

**JoanneMac** · December 4th 2008, 12:30 PM

in 3 variables, please. I'm tearing my hair out!!

3x^2+4y+z=7
2y+z=3
-5x^2+3y+8z=-31

Sorry to be such a pest, but math isn't my best subject.

**masters** · December 4th 2008, 01:21 PM

Originally Posted by JoanneMac

in 3 variables, please. I'm tearing my hair out!!

3x^2+4y+z=7
2y+z=3
-5x^2+3y+8z=-31

Sorry to be such a pest, but math isn't my best subject.

Hello Joanne,

Tearing your hair out will not help.

(1) $3x^2+4y+z=7$

(2) $2y+z=3$

(3) $-5x^2+3y+8z=-31$

Use (1) and (3) to eliminate x

$15x^2+20y+5z=35$

$-15x^2+9y+24z=-93$
----------------------------------
(4) $29y+29z=-58$

Solve (2) for z and substitute in (4)

(2) $z=3-2y$

(4) $29y+29(3-2y)=-58$

$29y+87-58y=-58$

$-29y=-145$

$\boxed{y=5}$

Substitute y=5 into (2) to find z

$z=3-2(5)$

$\boxed{z=-7}$

Substitute y=5 and z=-7 into (1)

(1) $3x^2+4(5)+(-7)=7$

$3x^2+20=14$

$3x^2=-6$

$x^2=-2$

$\boxed{x=\pm i\sqrt{2}}$

**Soroban** · December 4th 2008, 01:49 PM

Hello, JoanneMac!

I must assume that complex roots are expected.

$\begin{array}{cccc}3x^2 + 4y + z &=& 7 & {\color{blue}[1]} \\ \qquad\quad<br /> 2y + z &=& 3 & {\color{blue}[2]} \\ \text{-}5x^2 + 3y + 8z &=& \text{-}31 & {\color{blue}[3]} \end{array}$

From [2], we have: . $z \:=\:3 - 2y$

$\begin{array}{ccccccccc}\text{Substitute into {\color{blue}[1]}:} & 3x^2 + 4y + (3-2y) &=& 7 & \Rightarrow & 3x^2 + 2y &=& 4 & {\color{blue}[4]} \\<br /> \text{Substitute into {\color{blue}[3]}:} & \text{-}5x^2 + 3y + 8(3-2y) &=& \text{-}31 & \Rightarrow & \text{-}5x^2 - 13y &=& \text{-}55 & {\color{blue}[5]} \end{array}$

$\begin{array}{cccccccc}\text{Multiply {\color{blue}[4]} by 13:} & 39x^2 + 26y &=& 52 \\ \text{Multiply {\color{blue}[5]} by 2:} & \text{-}10x^2 - 26y &=& \text{-}110 \end{array}$

Add: . $29x^2 \:=\:-58 \quad\Rightarrow\quad x^2 \:=\:-2\quad\Rightarrow\quad\boxed{ x \:=\:\pm i\sqrt{2}}$

Substitute into [4]: . $3(\pm i\sqrt{2})^2 + 2y \:=\:4 \quad\Rightarrow\quad -6 +2y \:=\:4 \quad\Rightarrow\quad 2y \:=\:10 \quad\Rightarrow\quad\boxed{ y \:=\:5}$

Substitute into [2]: . $2(5) + z \:=\:3 \quad\Rightarrow\quad\boxed{ z \:=\:-7}$

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