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प्रश्न
Solve each of the following systems of equations by the method of cross-multiplication
`x/a + y/b = 2`
`ax - by = a^2 - b^2`
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उत्तर
The system of the given equations may be written as
`1/a x xx + 1/b xx y - 2 = 0`
`ax - by + b^2 - a^2 = 0`
here
`a_1 = 1/a, b_1 = 1/b, c_1 = -2`
`a_2 = a, b_2= -b, c_2 = b^2 - a^2`
By cross multiplication, we get
`=> x/(1/b xx (b^2 - a^2) - (-2) xx (-b)) = (-y)/(1/a xx (b^2 - a^2) - (-2) xx a) = 1/((-bxx1)/a - (a xx1)/b)`
`=> x/((b^2 - a^2)/b - 2b) = (-y)/((b^2 - a^2)/b + 2b) = 1/((-b)/a - a/b)`
`=> x/((b^2 -a^2 - 2b^2)/b) = (-y)/((b^2 - a^2 + 2b^2)/a) = 1/((-b^2 - a^2)/(ab)`
`=> x/((a^2 - b^2)/b) = (-y)/((b^2 + a^2)/a) = 1/((-b^2 -a^2)/(ab)`
Now
`x/((-a^2 -b^2)/b) = 1/((-b^2 - a^2)/(ab)`
`=> x = (-a^2 - b^2)/b xx (ab)/(-b^2 - a^2)`
And
`(-y)/((b^2 + a^2)/a)= 1/((-b^2 -a^2)/(ab))`
`=> -y = (b^2 + a^2)/a xx (ab)/(-b^2 - a^2)`
`=> -y = ((b^2 + a^2)xxb)/(-(b^2 + a^2)`
=> y = b
Hence, , x = a, y = b is the solution of the given system of the equations.
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