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Zorluk: OrtaExponents, Radicals, and Algebraic Expressions

For all positive real numbers xx and yy, if x2y3=108x^2 y^3 = 108 and x3y2=72x^3 y^2 = 72, then x+y=5x + y = 5.

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Cevap

The statement is True.
Multiplying the two equations yields (xy)5=7776=65(xy)^5 = 7776 = 6^5, so xy=6xy = 6. Dividing the second equation by the first yields xy=23\frac{x}{y} = \frac{2}{3}, meaning x=23yx = \frac{2}{3}y. Substituting this into xy=6xy = 6 gives y=3y = 3 and x=2x = 2, making x+y=5x + y = 5.

Adım Adım Çözüm

1
Multiply the two system equations to find the product xyxy.
(x2y3)(x3y2)=x5y5=(xy)5=108×72=7776=65(x^2 y^3)(x^3 y^2) = x^5 y^5 = (xy)^5 = 108 \times 72 = 7776 = 6^5, so xy=6xy = 6.
Multiplying exponential expressions with common bases allows adding their exponents to form a unified power (xy)5(xy)^5.
2
Divide the second equation by the first equation to find the relationship between xx and yy.
\frac{x^3 y^2}{x^2 y^3} = \frac{x}{y} = \frac{72}{108} = \frac{2}{3}, which simplifies to x=23yx = \frac{2}{3}y.
Dividing exponential expressions subtracts their exponents, yielding the simple ratio of the variables.
3
Solve for individual values of xx and yy and evaluate x+yx + y.
Substituting x=23yx = \frac{2}{3}y into xy=6xy = 6 gives 23y2=6    y=3\frac{2}{3}y^2 = 6 \implies y = 3 and x=2x = 2. Therefore, x+y=5x + y = 5.
Solving the system confirms that x=2x = 2 and y=3y = 3 are the unique positive real solutions.

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