Question

find the value of x and y by elimination method ​

Answer 1

Solution

Given :-

• a²x - b²y = a + b ----------(1)
• a³x - b³y = a²b² -----------(2)

Find :-

• Value of x & y

Explanation

Elimination Method

Multiply by a in equ(1) & 1 in equ(2)

• a³x - b²ay = a²+ba
• a³x - b³y = a²b²

_________________Sub. it's(Eliminate X)

➩ -b²ay + b³y = a²+ba - a²b²

➩ y(b³-b²a) = (a² + ba - a²b²)

➩ y = (a² + ba - a²b²)/(b³ - b²a)

Keep Value of y in equ(1)

➩ a²x - b² * [ (a² + ba - a²b²)/(b³ - b²a)] = (a+b)

➩a²x - b² * [ a² + ab - a²b²)/b²(b-a)] = (a+b)

➩ a²x = (a+b) + (a² + ab - a²b²)/(b-a)

➩ a²x = [(a²-b²) + (a² + ab - a²b²)]/(b-a)

➩ x = [(2a² - b² + ab - a²b²)/a²(b-a)]

Hence

• Value of x = [(2a² - b² + ab - a²b²)/a²(b-a)]
• Value of y = (a² + ba - a²b²)/(b³ - b²a)

_________________

Answer 2

* Answer↓

• $\large x= 【(2a^2-b^{2}+ab-a^{2}b^{2}) / (a^{2}(b-a) 】$

• $\large y= 【(a^{2}+ba-a^{2}b^{2}/(b^{3}-b^{2}a)】$

$\rule{300}2$

* G¡ven *

$a^{2}x-b^{2}y=a+b$-------eq(1)

$a^3x-b^2y=a^2b^2$-------eq(2)

◇ Solution ◇

$\rule{300}2$

→ Consider both of the given equation (1) and →(2)Multiply by 'a' in equation (1)

So,

→by equation (1)×equation(2). 【 For Eliminating x 】

we get,

$\rightarrow -b^{2}ay-b^{2}+b^{3}y=a^{2}+ba-a^{2}b^{2} \\ \rightarrow y(b^3y^2a)=a^{2}+ba-a^{2}b^{2} \\ \rightarrow y=\frac {a^{2}+ba-a^{2}b^{2}} {b^3-b^2a}$

By equation (1)

→ putting the value of y in equation (1)

$\rightarrow a^{2}x-b^{2}×[(a^2+ba-a^2b^2)/(b^3-b^2a)]=(a+b) \\\rightarrow a^{2}x-b^{2}×[(a^2+ba-a^2b^2)/(b^{2}(b-a)=(a+b) \\ \rightarrow^{2}x=[(a^2-b^2)+(a^2+ab-a^2 b^2) \\ \rightarrow x= 【(2a^2-b^{2}+ab-a^{2}b^{2}) / (a^{2}(b-a) 】$

$\rule{300}2$