If 5x + 9y = 5 and 125x^3 + 729y^3 = 120,then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equations are:
$5x + 9y = 5$ ---$(1)$
$125x^3 + 729y^3 = 120$ ---$(2)$
We know the algebraic identity: $(a + b)^3 = a^3 + b^3 + 3ab(a + b

Vedclass · Accepted Answer

$\frac{1}{135}$

Vedclass · Answer

(B) Given equations are:
$5x + 9y = 5$ ---$(1)$
$125x^3 + 729y^3 = 120$ ---$(2)$
We know the algebraic identity: $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.
Let $a = 5x$ and $b = 9y$. Then $a^3 = 125x^3$ and $b^3 = 729y^3$.
Cubing equation $(1)$ on both sides:
$(5x + 9y)^3 = 5^3$
$(5x)^3 + (9y)^3 + 3(5x)(9y)(5x + 9y) = 125$
$125x^3 + 729y^3 + 135xy(5x + 9y) = 125$
Substitute the values from equation $(1)$ and $(2)$ into the expanded form:
$120 + 135xy(5) = 125$
$120 + 675xy = 125$
$675xy = 125 - 120$
$675xy = 5$
$xy = \frac{5}{675}$
$xy = \frac{1}{135}$

If $5x + 9y = 5$ and $125x^3 + 729y^3 = 120$,then the value of the product of $x$ and $y$ is:

Similar Questions

Let $S$ be the set of all real roots of the equation,$3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|$. Then $S$

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4\sqrt{ab}x + 1 = 0$ will be

Let $P(x) = x^3 - ax^2 + bx + c$ where $a, b, c \in R$ has integral roots such that $P(6) = 3$,then '$a$' cannot be equal to

The value of $k$ for which the quadratic equation $(k + 11)x^2 - (k + 3)x + 1 = 0$ has real and equal roots is:

Solve the given two equations and select the correct option.
$I.$ $12x^2 - 47x + 40 = 0$
$II.$ $4y^2 + 3y - 10 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module