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\( \frac { \left\{ \begin{array} { l } { 9 x \cdot y ^ { + \frac { 1 } { 2 } } = y ^ { 2 } } \\ { ( 3 y ) ^ { 3 } \cdot ( 27 x ) ^ { - \frac { 3 } { 2 } } = x ^ { 3 } } \end{array} \right. } { 3 } \)

Ask by King Reid. in Israel
Feb 03,2025

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Answer

La solution du système est \( x = 3 \) et \( y = 9 \).

Solution

Résolvons le système d'équations suivant : \[ \begin{cases} 9x \cdot y^{\frac{1}{2}} = y^{2} \\ (3y)^{3} \cdot (27x)^{-\frac{3}{2}} = x^{3} \end{cases} \] ### Étape 1 : Simplifier la première équation \[ 9x \cdot y^{\frac{1}{2}} = y^{2} \] Divisons les deux côtés par \( y^{\frac{1}{2}} \) (en supposant que \( y > 0 \)) : \[ 9x = y^{\frac{3}{2}} \quad \Rightarrow \quad x = \frac{y^{\frac{3}{2}}}{9} \] ### Étape 2 : Simplifier la deuxième équation \[ (3y)^{3} \cdot (27x)^{-\frac{3}{2}} = x^{3} \] Développons chaque terme : \[ (3y)^{3} = 27y^{3} \] \[ (27x)^{-\frac{3}{2}} = \frac{1}{(27x)^{\frac{3}{2}}} = \frac{1}{27^{\frac{3}{2}} \cdot x^{\frac{3}{2}}} = \frac{1}{3^{\frac{9}{2}} \cdot x^{\frac{3}{2}}} \] Ainsi, la deuxième équation devient : \[ 27y^{3} \cdot \frac{1}{3^{\frac{9}{2}} \cdot x^{\frac{3}{2}}} = x^{3} \] Simplifions : \[ \frac{27y^{3}}{3^{\frac{9}{2}} \cdot x^{\frac{3}{2}}} = x^{3} \] Sachant que \( 27 = 3^{3} \), nous avons : \[ \frac{3^{3} y^{3}}{3^{\frac{9}{2}} \cdot x^{\frac{3}{2}}} = x^{3} \quad \Rightarrow \quad \frac{y^{3}}{3^{\frac{3}{2}} \cdot x^{\frac{3}{2}}} = x^{3} \] Multiplions les deux côtés par \( 3^{\frac{3}{2}} \cdot x^{\frac{3}{2}} \) : \[ y^{3} = 3^{\frac{3}{2}} \cdot x^{\frac{9}{2}} \] ### Étape 3 : Substituer \( x \) dans la deuxième équation Nous avons déjà exprimé \( x \) en fonction de \( y \) : \[ x = \frac{y^{\frac{3}{2}}}{9} \] Substituons cette expression dans l'équation obtenue à l'étape précédente : \[ y^{3} = 3^{\frac{3}{2}} \cdot \left( \frac{y^{\frac{3}{2}}}{9} \right)^{\frac{9}{2}} \] Simplifions l'expression : \[ y^{3} = 3^{\frac{3}{2}} \cdot \frac{y^{\frac{27}{4}}}{9^{\frac{9}{2}}} = 3^{\frac{3}{2}} \cdot \frac{y^{\frac{27}{4}}}{3^{9}} = \frac{y^{\frac{27}{4}}}{3^{\frac{15}{2}}} \] Multipliant les deux côtés par \( 3^{\frac{15}{2}} \) : \[ 3^{\frac{15}{2}} \cdot y^{3} = y^{\frac{27}{4}} \] Divisons par \( y^{3} \) (en supposant \( y \neq 0 \)) : \[ 3^{\frac{15}{2}} = y^{\frac{27}{4} - 3} = y^{\frac{15}{4}} \] Résolvons pour \( y \) : \[ y^{\frac{15}{4}} = 3^{\frac{15}{2}} \quad \Rightarrow \quad y = \left(3^{\frac{15}{2}}\right)^{\frac{4}{15}} = 3^{2} = 9 \] ### Étape 4 : Trouver \( x \) Utilisons la première équation pour trouver \( x \) : \[ x = \frac{y^{\frac{3}{2}}}{9} = \frac{9^{\frac{3}{2}}}{9} = \frac{27}{9} = 3 \] ### Conclusion La solution du système est : \[ x = 3 \quad \text{et} \quad y = 9 \]

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Bonus Knowledge

Let's dive into the world of systems of equations in two variables! These types of equations allow us to find points of intersection, which are incredibly useful in fields like economics, physics, and engineering. Imagine using these calculations to optimize a production process or balance chemical equations—it's where math meets the real world in truly exciting ways! On another note, when you're working with systems like this, keep a sharp eye out for common pitfalls! One frequent mistake is misinterpreting the exponents, especially with fractions or negative values. It’s easy to accidentally flip them or mix up their signs. Double-check your work, and don’t hesitate to substitute back into the original equations to ensure your solutions are correct. A little caution can save you a lot of time!

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