Question

5 root x + 6 root Y equal to 16, 7 x minus 3 root Y equal to 11​

Answer 1

Answer:

$x=\cfrac {-5 \pm \sqrt{2153}}{196}$

Step-by-step explanation:

Given equations are

$5\sqrt{x} + 6\sqrt{y} = 16$                         ...(1)

$7x-3\sqrt{y}= 11$                            ...(2)

From Eqn. (1),

$\sqrt{y} = \cfrac{16-5\sqrt{x}}{6}$

Submitting value of $\sqrt{y}$ in Eqn. (2), we get

$7x-\left( \cfrac{16-5\sqrt{x}}{6} \right) = 11$

⇒ $42x - (16 - 5\sqrt{x}) = 66$

⇒ $42x - 16 + 5\sqrt(x) = 66$

⇒ $42x + 5\sqrt{x} - 16 - 66 = 0$

⇒ $42x + 5\sqrt{x} - 82 = 0$                    ...(3)

Let $\sqrt{x} = m$

Squaring both sides,

$x = m^2$

Substituting value of x in Eqn. (3), we get

$42m^2 + 5m - 82 = 0$

Using quadratic formula,

$m = \cfrac{-b \pm \sqrt{b^2 - 4ac}}{a^2}$

$= \cfrac{-5 \pm \sqrt{5^2 - 4(14)(-38)}}{(14)^2}$

$= \cfrac{-5 \pm \sqrt{25+2128}}{196}$

$=\cfrac {-5 \pm \sqrt{2153}}{196}$