Question

Solve by 'CROSS MULTIPLICATION' method

$7. \: find \: \alpha \: and \: \beta \: \: in \: \: terms \: \: \\ of \: x \: , \: \: y \: if$

$\alpha x {}^{3} - \beta y = x - y$

and

$x \alpha + \beta y {}^{2} = xy {}^{2}$

Ans:

$( \frac{y(xy + x - y)}{x(x {}^{2} y + 1)}, \frac{x {}^{ {3}} y {}^{2} - x + y}{y(x {}^{2} y + 1)} )$


Class 10

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Answer 1

$\bold{Answer :}$

Let us learn the rule first.

Two equations are -

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

Then the solution be presented as

x/(b₁c₂ - b₂c₁) = y/(a₂c₁ - a₁c₂) = 1/(a₁b₂ - a₂b₁)

Given equations are

x³α - yβ - (x - y) = 0 ...(i)

xα + y²β - xy² = 0 ...(ii)

By cross multiplication, from (i) and (ii), we get

α/{(- y) (- xy²) + y² (x - y)}

= β/{- x (x - y) + xy² (x³)}

= 1/{x³y² - x (- y)}

or, α/(xy³ + xy² - y³)

= β/(- x² + xy + x⁴y²)

= 1/(x³y² + xy)

Then,

α = (xy³ + xy² - y³)/(x³y² + xy)

= [{y² (xy + x - y)}/{xy (x²y + 1)}]

= {y (xy + x - y)}/{x (x²y + 1)}

and

β = (- x² + xy + x⁴y²)/(x³y² + xy)

= [{x (x³y² - x + y)}/{xy (x²y + 1)}]

= (x³y² - x + y)/{y (x²y + 1)}

∴ the required solution be

α = {y (xy + x - y)}/{x (x²y + 1)},

β = (x³y² - x + y)/{y (x²y + 1)}

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