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Mathematics

Solve:

ax + by = c
bx + ay = 1 + c

Linear Equations

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Answer

Given:

ax + by = c ……………….(1)
bx + ay = 1 + c ……………….(2)

Multiplying equation (1) by b:

(ax + by = c) x b

⇒ abx + b2y = bc ……………….(3)

Multiplying equation (2) by a:

(bx + ay = 1 + c) x a

⇒ abx + a2y = a + ac ……………….(4)

Subtract Equation (4) from Equation (3),

abx+b2y=bcabx+a2y=a+ac−−−(b2−a2)y=bc−(a+ac)⇒(b2−a2)y=bc−a−ac\begin{matrix} & abx & + & b^2y & = & bc \ & abx & + & a^2y & = & a + ac \ & - & - & & & - \ \hline & & & (b^2 - a^2)y & = & bc - (a + ac) \ \Rightarrow & & & (b^2 - a^2)y & = & bc - a - ac \ \end{matrix}

⇒ y = bc−a−acb2−a2\dfrac{bc - a - ac}{b^2 - a^2}

Substituting the value of y in equation (3), we get:

⇒ abx + b2 ×bc−a−acb2−a2\times \dfrac{bc - a - ac}{b^2 - a^2} = bc

⇒ abx = bc - b2(bc−a−ac)b2−a2\dfrac{b^2(bc - a - ac)}{b^2 - a^2}

⇒ abx = bc(b2−a2)b2−a2−b3c−b2a−ab2cb2−a2\dfrac{bc(b^2 - a^2)}{b^2 - a^2} - \dfrac{b^3c - b^2a - ab^2c}{b^2 - a^2}

⇒ abx = b3c−a2bc−(b3c−b2a−ab2c)b2−a2\dfrac{b^3c - a^2bc - (b^3c - b^2a - ab^2c)}{b^2 - a^2}

⇒ abx = b3c−a2bc−b3c+b2a+ab2cb2−a2\dfrac{b^3c - a^2bc - b^3c + b^2a + ab^2c}{b^2 - a^2}

⇒ abx = −a2bc+b2a+ab2cb2−a2\dfrac{- a^2bc + b^2a + ab^2c}{b^2 - a^2}

⇒ x = ab(b+bc−ac)ab(b2−a2)\dfrac{ab(b + bc - ac)}{ab(b^2 - a^2)}

⇒ x = b+bc−acb2−a2\dfrac{b + bc - ac}{b^2 - a^2}

⇒ x = ac−b−bca2−b2\dfrac{ac - b - bc}{a^2 - b^2}

Hence, x = ac−b−bca2−b2\dfrac{ac - b - bc}{a^2 - b^2} and y = bc−a−acb2−a2\dfrac{bc - a - ac}{b^2 - a^2}.

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