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Sagot :
Answer:
5
Step-by-step explanation:
2x²y³ + 4y³ = 149 + 3x²
[tex] 2x^2y^3 - 3x^2 = 149 - 4y^3 [/tex]
[tex] x^2(2y^3 - 3) = 149 - 4y^3 [/tex]
[tex] x^2 = \dfrac{149 - 4y^3}{2y^3 - 3} [/tex]
[tex] x = \pm \sqrt{\dfrac{149 - 4y^3}{2y^3 - 3}} [/tex]
Try y = 1
[tex]x = \pm \sqrt{\dfrac{149 - 4(1)}{2(1)^3 - 3}} = \pm \sqrt{-145} = i\sqrt{145}[/tex]
For y = 1, x is imaginary.
Try y = 2
[tex] x = \pm \sqrt{\dfrac{149 - 4(2)^3}{2(2)^3 - 3}} = \pm \sqrt{9} = \pm 3[/tex]
Since x and y are positive integers, ignore x = -3.
When x = 3, y = 2.
x + y = 3 + 2 = 5
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