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{ b x + c y = a + b a x ( 1 a − b − 1 a + b ) + c y ( 1 b − a − 1 b + a ) = 2 a a + b \begin{cases} bx+cy = a+b \\ ax \left(\dfrac{1}{a-b}-\dfrac{1}{a+b}\right)+cy\left(\dfrac{1}{b-a}-\dfrac{1}{b+a}\right) =\dfrac{2a}{a+b} \end{cases}

In the system of equations above, a a , b b , and c c are constants, while x x and y y are variables. What is the value of c y x cyx ?

a a a c \frac{a}{c} b c \frac{b}{c} b b a b \frac{a}{b} c − a c-a It is a numeric value c c

Zakir Husain by Zakir Husain , 41, India

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2 solutions

Given b x + c y = a + b . . . ( 1 ) bx + cy = a + b \quad ...(1) and

a x ( 1 a − b − 1 a + b ) + c y ( 1 b − a − 1 a + b ) = 2 a a + b 2 a b x − 2 c a y ( a − b ) ( a + b ) = 2 a a + b b x − c y = a − b . . . ( 2 ) \begin{aligned} ax \left(\frac 1{a-b} - \frac 1{a+b} \right) + cy \left(\frac 1{b-a} - \frac 1{a+b} \right) & = \frac {2a}{a+b} \\ \frac {2abx - 2cay}{(a-b)(a+b)} & = \frac {2a}{a+b} \\ bx - cy & = a -b & ...(2) \end{aligned}

{ ( 1 ) + ( 2 ) : 2 b x = 2 a ⟹ x = a b ( 1 ) − ( 2 ) : 2 c y = 2 b ⟹ c y = b ⟹ c y x = b ⋅ a b = a \begin{cases} (1)+(2): & 2bx = 2a & \implies x = \dfrac ab \\ (1)-(2): & 2cy = 2b & \implies cy = b \end{cases} \implies cyx = b \cdot \dfrac ab = \boxed a

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The second equation is the same as

b x − c y = a − b bx-cy=a-b .

Solving this with the first equation we get x = a b , c y = b x=\frac{a}{b}, cy=b

⟹ c y x = b × a b = a \implies cyx =b\times \frac{a}{b}=\boxed a .

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