Answer:
Solution:
Given that,
x⁴y⁵ + x⁵y⁴ = 810
➝ x⁴y⁴(x+y) = 810 ---(1)
[tex] \: [/tex]
Further given that,
x⁶y³ + x³y⁶ = 945
x³y³(x³y³) = 945
[tex] \: [/tex]
We know,
[tex] { \boxed{{x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}) }}[/tex]
So, using this identity, we get,
➝ x³y³( x + y )( x² - xy + y²) = 945
---(2)
[tex] \: [/tex]
On dividing equation (1) and (2), we get,
[tex] \longrightarrow\frac{ {x}^{2} - xy + {y}^{2} }{xy} = \frac{945}{810} [/tex]
[tex] \longrightarrow \: \frac{ {x}^{2} + {y}^{2} }{xy} - 1 = \frac{7}{6} [/tex]
[tex] \longrightarrow \: \frac{ {x}^{2} + {y}^{2} }{xy} = \frac{7}{6} + 1[/tex]
[tex] \longrightarrow \: \frac{ {x}^{2} + {y}^{2} }{xy} = \frac{7 + 6}{6} [/tex]
[tex] \longrightarrow \: \frac{ {x}^{2} + {y}^{2} }{xy} = \frac{13}{6} [/tex]
[tex] \: [/tex]
➝ 6x² + 6y² = 13xy
➝ 6x² - 13xy + 6y² = 0
➝ 6x² - 4xy - 9xy + 6y² = 0
➝ 2x( 3x - 2y ) - 3y( 3x - 2y ) = 0
➝ ( 3x - 2y ) ( 2x - 3y ) = 0
[tex] \longrightarrow \:{ \bold{ x = \frac{3y}{2} \: \: \: or \: \: x = \frac{2y}{3} }}[/tex]
[tex] \: [/tex]
[tex] \: [/tex]
[tex]{ \underline {\underline {\red{Case:- 1 }}}}[/tex]
[tex]{ \bold{When \: x = \frac{3y}{2}}} [/tex]
In Substating the value of x in equation (1) we get,
[tex] \frac{81 {y}^{4} }{16} \times {y}^{4} \times ( \frac{3y}{2} + y) = 810[/tex]
[tex] \frac{ {y}^{8} }{16} \times ( \frac{5y}{2} ) = 10[/tex]
[tex] {y}^{9} = 64[/tex]
[tex] {y}^{9} = {2}^{6} [/tex]
[tex] \: [/tex]
[tex] \longrightarrow \: { \bold{y = {(2)}^{ \frac{2}{3} } }}[/tex]
[tex] \longrightarrow \: { \bold{x = \frac{3}{2} {(2)}^{ \frac{2}{3} } }}[/tex]
[tex] \: [/tex]
[tex] \: [/tex]
[tex]{ \underline {\underline {\red{Case:- 2}}}}[/tex]
[tex]{ \bold{When \: x \: = \frac{2y}{3} \rightarrow \: \: y = \frac{3x}{2} }}[/tex]
On Substituting the value of y in equation (2), we get,
[tex] \frac{81 {x}^{4} }{16} \times {x}^{4} \times ( \frac{3x}{2} + x) = 810[/tex]
[tex] \frac{ {x}^{8} }{16} \times ( \frac{5x}{2} ) = 10[/tex]
[tex] {x}^{9} = 64[/tex]
[tex] {x}^{9} = {2}^{6} [/tex]
[tex] \: [/tex]
[tex] \longrightarrow \: { \bold{x \: = {(2)}^{ \frac{2}{3} } }}[/tex]
[tex] \longrightarrow \: { \bold{y = \frac{3}{2} {(2)}^{ \frac{2}{3} } }}[/tex]
