Solve the given equations and select the corr | vedclass

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

Question

(D) From equation $I$: $(x+y)^{3} = 1331 \implies x+y = \sqrt[3]{1331} = 11$. So,$y = 11-x$. Substitute this into equation $III$: $x(11-x) = 28 \impli

Vedclass · Accepted Answer

if $x = y$ or relationship between $x$ and $y$ cannot be established.

Vedclass · Answer

(D) From equation $I$: $(x+y)^{3} = 1331 \implies x+y = \sqrt[3]{1331} = 11$. So,$y = 11-x$. Substitute this into equation $III$: $x(11-x) = 28 \implies 11x - x^{2} = 28$. Rearranging gives the quadratic equation: $x^{2} - 11x + 28 = 0$. Factoring the quadratic: $x^{2} - 7x - 4x + 28 = 0 \implies x(x-7) - 4(x-7) = 0 \implies (x-7)(x-4) = 0$. Thus,$x = 7$ or $x = 4$. If $x = 7$,then $y = 11-7 = 4$. If $x = 4$,then $y = 11-4 = 7$. Now,using equation $II$: $x-y+z = 0 \implies z = y-x$. Case $1$: If $x = 7, y = 4$,then $z = 4-7 = -3$. Case $2$: If $x = 4, y = 7$,then $z = 7-4 = 3$. Since $x$ can be greater than $y$ or less than $y$,no fixed relationship between $x$ and $y$ can be established.

Solve the given equations and select the correct answer from the given options.
$I. (x+y)^{3} = 1331$
$II. x-y+z = 0$
$III. xy = 28$

Similar Questions

The sum of the squares of two natural consecutive odd numbers is $394$. The sum of the numbers is

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

If the equations $x^2 + bx - 1 = 0$ and $x^2 + x + b = 0$ have a common root different from $-1$,then $|b|$ is equal to

Solve the given two equations and select the correct answer from the given options.
$I.$ $(441)^{\frac{1}{2}} x^{2} - 111 = (15)^{2}$
$II.$ $\sqrt{121} y^{2} + (6)^{3} = 260$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-7x+12=0$
$II.$ $y^{2}+y-12=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given equations and select the correct answer from the given options.$I. (x+y)^{3} = 1331$$II. x-y+z = 0$$III. xy = 28$