Question

solve for x and y. ax+by=3ab a²x+b²y=a+b Guys here is the question again and plzzzz ans it​

Answer 1

$ax + by = 3ab...(i)$

${a}^{2} x + {b}^{2} y = a + b...(ii)$

$y = \frac{a(3b - x)}{b} ...from(i)$

$substitute \: the \: value \: of \: x \: in \: (ii) \: we \: get \:$

${a}^{2} x + {b}^{2} \frac{a(3b - x)}{b } = a + b$

${a}^{2} x + 3a {b}^{2} - abx = a + b$

$x = \frac{a + b - 3a {b}^{2} }{ {a}^{2 -}ab }$

$therefore \: y = \frac{a(3b - x)}{b}$

$= \frac{a(3b - \times \frac{a + b - 3a {b}^{2} }{ {a}^{2} - ab} )}{b}$

$= \frac{(3 {a}^{2} b - 3a {b}^{2} - a - b + 3a {b}^{2} )}{b (a - b)}$

$= \frac{3 {a}^{2} b - a - b}{b(a - b)}$

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Answer 2

Step-by-step explanation:

ax+by=3ab...(i)

{a}^{2} x + {b}^{2} y = a + b...(ii)a

2

x+b

2

y=a+b...(ii)

y = \frac{a(3b - x)}{b} ...from(i)y=

b

a(3b−x)

...from(i)

substitute \: the \: value \: of \: x \: in \: (ii) \: we \: get \:substitutethevalueofxin(ii)weget

{a}^{2} x + {b}^{2} \frac{a(3b - x)}{b } = a + ba

2

x+b

2

b

a(3b−x)

=a+b

{a}^{2} x + 3a {b}^{2} - abx = a + ba

2

x+3ab

2

−abx=a+b

x = \frac{a + b - 3a {b}^{2} }{ {a}^{2 -}ab }x=

a

2−

ab

a+b−3ab

2

therefore \: y = \frac{a(3b - x)}{b}thereforey=

b

a(3b−x)

= \frac{a(3b - \times \frac{a + b - 3a {b}^{2} }{ {a}^{2} - ab} )}{b}=

b

a(3b−×

a

2

−ab

a+b−3ab

2

)

= \frac{(3 {a}^{2} b - 3a {b}^{2} - a - b + 3a {b}^{2} )}{b (a - b)}=

b(a−b)

(3a

2

b−3ab

2

−a−b+3ab

2

)

= \frac{3 {a}^{2} b - a - b}{b(a - b)}=

b(a−b)

3a

2

b−a−b