Question

√2x+√3y=0
√3x-√8y=0
Solve the following pair of linear equations by the substitution method

Answer 1

Given:

√2x+√3y=0

√3x-√8y=0

To find:

solve by substitution method.

Solution:

Let,

$\sqrt{2}x+\sqrt{3}y=0.......(a)\\\\\sqrt{3}x-\sqrt{8}y=0......(b)$

Solve equation (a) and put the value in the equation (b):

$\to \sqrt{2}x+\sqrt{3}y=0\\\\\to \sqrt{2}x=-\sqrt{3}y\\\\\to x=-\frac{\sqrt{3} y}{\sqrt{2}}\\\\\to x=-\frac{\sqrt{3} y}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\\\\to x=-\frac{\sqrt{3}\sqrt{2} y}{2} \\\\\to x=-\frac{\sqrt{6} y}{2}$

equation (b):

$\to \sqrt{3}x-\sqrt{8}y=0$

$\to \sqrt{3}(-\frac{\sqrt{6}y}{{2}})-\sqrt{8}y=0\\\\\to - \sqrt{3}(\frac{\sqrt{3} \times \sqrt{2} y}{{2}})-\sqrt{8}y=0\\\\\to -\frac{3 \sqrt{2} y}{{2}}-2\sqrt{2}y=0\\\\\to \frac{-3 \sqrt{2}y-4\sqrt{2}y}{2}=0\\\\\to \frac{-3 \sqrt{2}y-4\sqrt{2}y}{2}=0\\\\\to -\sqrt{2}y(\frac{3+4}{2})=0\\\\\to -\sqrt{2}y(\frac{7}{2})=0\\\\\to y= - \frac{2}{7\sqrt{2}}\\\\\to y= - \frac{2}{7\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\\\\to y= - \frac{2\sqrt{2}}{7\times 2} \\\\\to y= - \frac{\sqrt{2}}{7}$

put the value of y into the equation (a):

Equation (a):

$\to \sqrt{2}x+\sqrt{3}y=0$

$\to \sqrt{2}x+\sqrt{3} \times (-\frac{\sqrt{2}}{7})=0\\\\\to \sqrt{2}x- \frac{\sqrt{2}\sqrt{3}}{7}=0\\\\\to \sqrt{2}x- \frac{\sqrt{3}\sqrt{2}}{7}=0\\\\\to \sqrt{2}x = \frac{\sqrt{3}\sqrt{2}}{7}\\\\\to x = \frac{\sqrt{3}\sqrt{2}}{7 \sqrt{2}}\\\\\to x = \frac{\sqrt{3}}{7}$

The final value of x and y is: $\bold{-\frac{\sqrt{2}}{7} \ \ _{and} \ \ \frac{\sqrt{3}}{7}}$

Answer 2

plz mark as BRAINLIEST

See attachment for answer