Question

Solve by Elimination by equating Coefficients method

5. ax + by = c ;

bx + ay = 1 + c


Class 10

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Answer 1

Here is your solution :

⇒ ax + by = c -------- ( 1 )

⇒ bx + ay = 1 + c -------- ( 2 )

Multiply the eq'n( 1 ) by ' a ',

⇒ a( ax + by ) = ac

⇒ a²x + aby = ac ------ ( 3 )

Multiply the eq'n ( 2 ) by 'b',

⇒b( bx + ay ) = b( 1 + c )

⇒ b²x + aby = b + bc ------ ( 4 )

Subtract the eq'n ( 4 ) from eq'n ( 3 ),

⇒ a²x + aby - b²x - aby = ac - b - bc

⇒ a²x - b²x = ac - bc - b

⇒ x( a² - b² ) = ac - bc - b

•°• x = [ ac - bc - b ] / ( a² - b² )

Substitute the value of ' x ' in eq'n( 1 ),


⇒ ax + by = c

⇒ a[ ( ac - bc - b ) / ( a² - b² ) ] + by = c

⇒ [ ( a²c - abc - ab ) / ( a² - b² ) ] + by = c

⇒ by = c - [ ( a²c - abc - ab ) / ( a² - b² ) ]

⇒ by = [ a²c - b²c - a²c + abc + ab ] / ( a² - b² )

⇒ by = ( abc + ab - b²c ) / ( a² - b² )

⇒ by = b( ac + a - bc ) / ( a² - b² )

⇒ y = [ b( ac + a - bc ) ] / [ b( a² - b² ) ]

•°• y = ( ac + a - bc ) / ( a² - b² )

Hence,

⇒ x = ( ac - bc - b ) / ( a² - b² )

⇒ y = ( ac - bc + a ) / ( a² - b² )

Hope it helps !!

Answer 2

ax + by = c --1equation
bx + ay = ( 1 + c ) --2equation

$\mathbb{multiply \: \: by \: \: b \: \: on \: \: 1 equation \: \: \: and \: \: by \: \: a \: on \: \: 2equation}$

Hence, it will be :

abx + b²y = cb ---3equation
abx + a²y = a + ac ---4equation

$\textcolor{red}{subtract \: \: 4equation \: \: from \: \: 3equation}$

abx + b²y = cb
abx + a²y = ( a + ac )
-__(-)____(-)_______
b²y - a²y = cb - a - ac
________________

abx will cancel, we get

=> y( b² - a² ) = c( b - a ) - a

=> y( b - a ) ( b + a ) = cb - ca - a

$=> y = \frac{cb - ca - a}{(b - a)(b + a)}$

Putting the value of y in 1equation,

=> ax + by = c

$=> ax + b( \frac{cb - ca - a}{(b - a)(b + a)} ) = c \\ \\ \\ => ax = \frac{c( {b}^{2} - {a}^{2} ) - c {b}^{2} + abc + ab }{(b - a)(b + a)} \\ \\ \\ => ax = \frac{c {b}^{2} - c {a}^{2} - c {b}^{2} + abc + ab }{(b - a)(b + a)} \\ \\ \\ => x = \frac{-c {a}^{2} +abc + ab }{(b - a)(b + a)a}$

x = [ - ca² + abc + ab ] / [ a( b - a ) ( b + a ) ]

x = [ a( - ca+ bc + b ) ] / [ a( b - a ) ( b + a )]

x = ( bc + b - ca ) /[ ( b - a ) ( b + a )]