Question

Solve the following systems of equations.
${10}^{3log(x-y)}=250,\sqrt{x-y}+\frac{1}{2}\sqrt{x+y}=\frac{26-y}{\sqrt{x-y}}.$

Answer 1

Solution: Given,

=> 10³ˡᵒᵍˣ⁻ʸ = 250

Taking log both side having base 10 ,we have

=> log₁₀10³ˡᵒᵍ₁₀ˣ⁻ʸ = log₁₀250

=> 3log(x-y)log₁₀10 = log₁₀5²10 [• logaⁿ = nloga ]

=> 3log₁₀(x-y) = 2log₁₀10/2 + log₁₀10 [• logₐᵃ = 1 ]

=> 3log₁₀(x-y) = 2(log₁₀10 - log₁₀2)+1

=> 3log₁₀(x-y) = 2 - 2log₁₀2 + 1

=> 3log₁₀x-y = 3 - 2log₁₀2

=> 3log₁₀x-y = log₁₀10³ - log₁₀2² [• nloga = logaⁿ ]

=> log₁₀(x-y)³ = log₁₀(10³/2²)

=> (x - y)³ = (10³/2²) [• loga = logb, a = b ]

=> x - y = 10/₃√4 [• cube root of 4 is 1.587401052]

=> x - y = 10/1.587401052

=> x = 6.299605249 + y equation (1)

Now, √(x-y) + 1/2•√(x+y) = (26 - y)/√(x-y)

=> √(x-y)•√(x-y) + 1/2•√(x+y)•√(x-y)=26 - y

=> x - y + 1/2•√(x²-y²) = 26 - y

=> x + 1/2•√(x-y) = 26

=> 1/2•√(x-y) = 26 - x

• squaring both side ,we have

=> 1/4•(x²-y²) = 676 + x² - 52x

=> 1/4•((6.29960524 + y) - y²)=676 + (6.299605249 + y)²- 52(6.299605249 + y)

=> 1/4•(39.68502629 + y² -y² + 12.5992105y) = 676 + 39.68502629 + 12.5992105y + y² - 327.5794729 - 52y [• value of x from equation (1)]

=> 9.9212565 + 3.14980262y = 705.7637697 - 39.4007895y + y² - 327.5794729

=> y² - 42.55059212y + 378.1842968 = 0

This is quadratic equation, roots are given by

=> X = [ - b ± √(b²-4ac)]/2

=> Y = [42.55059212 ± √(42.55059212)² - 4•1•378.1842968)]/2

=> Y = [42.55059212 ± 17.25733766]/2

=> Y₁ = 29.903 , => Y₂ = 12.646

But,Argument cannot be negative

So , Y = 12.646 be consider

=> X = 6.299 + y

=> X₁ = 6.299 + 29.903 , => X₂ = 6.299 + 12.646

=> X₁ = 36.202, => X₂ = 18.945

Therefore solution of systems of equations are

Y = 12.646 , X₁ = 36.202 , X₂ = 18.945