solve for x and y using subsitution method : ax/b -by/a =a+b ; ax - by =2ab
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Hi...☺
Here is your answer...✌
Given equations :
![\frac{a}{b} x - \frac{b}{a} y = a + b \: \: \: \: \: \: \: \: \: .....(1)\\ \\ and \\ \\ ax - by = 2ab \\ \\ ax = 2ab + by \\ \\ x = \frac{2ab + by}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .....(2) \frac{a}{b} x - \frac{b}{a} y = a + b \: \: \: \: \: \: \: \: \: .....(1)\\ \\ and \\ \\ ax - by = 2ab \\ \\ ax = 2ab + by \\ \\ x = \frac{2ab + by}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .....(2)](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7Bb%7D+x+-+%5Cfrac%7Bb%7D%7Ba%7D+y+%3D+a+%2B+b+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+.....%281%29%5C%5C+%5C%5C+and+%5C%5C+%5C%5C+ax+-+by+%3D+2ab+%5C%5C+%5C%5C+ax+%3D+2ab+%2B+by+%5C%5C+%5C%5C+x+%3D+%5Cfrac%7B2ab+%2B+by%7D%7Ba%7D+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5C%3A+.....%282%29+)
Putting value of x in eq (1)
we get,
![\frac{a}{b} ( \frac{2ab + by}{a} ) - \frac{b}{a} y = a + b \\ \\ \frac{b(2a + y)}{b} - \frac{b}{a} y = a + b \\ \\ 2a + y - \frac{by}{a} = a + b \\ \\ y(1 - \frac{b}{a} ) = a + b - 2a \\ \\ y (\frac{a - b}{a} ) = - a + b \\ \\ y ( \frac{a - b}{a} ) = - (a - b) \\ \\ y = \frac{ - a(a - b)}{(a - b)} \\ \\ \implies y = - a \frac{a}{b} ( \frac{2ab + by}{a} ) - \frac{b}{a} y = a + b \\ \\ \frac{b(2a + y)}{b} - \frac{b}{a} y = a + b \\ \\ 2a + y - \frac{by}{a} = a + b \\ \\ y(1 - \frac{b}{a} ) = a + b - 2a \\ \\ y (\frac{a - b}{a} ) = - a + b \\ \\ y ( \frac{a - b}{a} ) = - (a - b) \\ \\ y = \frac{ - a(a - b)}{(a - b)} \\ \\ \implies y = - a](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%7D%7Bb%7D+%28+%5Cfrac%7B2ab+%2B+by%7D%7Ba%7D+%29+-+%5Cfrac%7Bb%7D%7Ba%7D+y+%3D+a+%2B+b+%5C%5C+%5C%5C+%5Cfrac%7Bb%282a+%2B+y%29%7D%7Bb%7D+-+%5Cfrac%7Bb%7D%7Ba%7D+y+%3D+a+%2B+b+%5C%5C+%5C%5C+2a+%2B+y+-+%5Cfrac%7Bby%7D%7Ba%7D+%3D+a+%2B+b+%5C%5C+%5C%5C+y%281+-+%5Cfrac%7Bb%7D%7Ba%7D+%29+%3D+a+%2B+b+-+2a+%5C%5C+%5C%5C+y+%28%5Cfrac%7Ba+-+b%7D%7Ba%7D+%29+%3D+-+a+%2B+b+%5C%5C+%5C%5C+y+%28+%5Cfrac%7Ba+-+b%7D%7Ba%7D+%29+%3D+-+%28a+-+b%29+%5C%5C+%5C%5C+y+%3D+%5Cfrac%7B+-+a%28a+-+b%29%7D%7B%28a+-+b%29%7D+%5C%5C+%5C%5C+%5Cimplies+y+%3D+-+a)
Putting y = -a in eq (2)
we get,
![x = \frac{2ab -ab }{a} \\ \\ x = \frac{ab}{a} \\ \\ \implies x = b x = \frac{2ab -ab }{a} \\ \\ x = \frac{ab}{a} \\ \\ \implies x = b](https://tex.z-dn.net/?f=x+%3D+%5Cfrac%7B2ab+-ab+%7D%7Ba%7D+%5C%5C+%5C%5C+x+%3D+%5Cfrac%7Bab%7D%7Ba%7D+%5C%5C+%5C%5C+%5Cimplies+x+%3D+b)
HENCE,
x = b and y = -a
Here is your answer...✌
Given equations :
Putting value of x in eq (1)
we get,
Putting y = -a in eq (2)
we get,
HENCE,
x = b and y = -a
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