Solve the following equation :
$x + y = z, 3 x - 2 y + 17 z = 0, x^{3} + 3 y^{3} + 2 x^{3} = 167.$

Open in App

Solution

$x + y - z = 0...... (i) 3 x - 2 y + 17 z = 0...... (i i) x^{3} + {3 y}^{3} + 2 z^{3} = 167.... (i i i)$

Applying cross multiplication on $(i)$ and $(i i)$

$\frac{x}{17 - 2} = \frac{- y}{17 + 3} = \frac{z}{- 2 - 3} \frac{x}{- 3} = \frac{y}{4} = \frac{z}{1} = k \Rightarrow x = - 3 k, y = 4 k, z = k$

Substituting $x, y$ and $z$ in $(i i i)$

${(- 3 k)}^{3} + {3 (4 k)}^{3} + 2 {(k)}^{3} = 167 167 k^{3} = 167 k = 1 \Rightarrow x = - 3, y = 4, z = 1$

Suggest Corrections

Similar questions

x + y - z = 0

x - 2 y + z = 0

3 x + 6 y - 5 z = 0

x + 2 y + z = 1

3 x + y + z = 6

x + 2 y = 0

x + y = 3.3 \frac{0.6}{3 x - 2 y} = - 1; 3 x - 2 y \neq 0

Z = 3 x + 2 y

x + 2 y \leq 10, 3 x + y \leq 15, x, y \geq 0

x + y = 5 x y

3 x + 2 y = 13 x y,

x \neq 0, y \neq 0

Join BYJU'S Learning Program