Question

Solve the following pair of linear equations by the substitution method: (v)
$\sqrt{2}x + \sqrt{3} y= 0$
$\sqrt{3} x - \sqrt{8} y = 0$
​

Answer 1

Answer:

2x+√3y=0

√3x-√8y=0

To find:

solve by substitution method.

Solution:

Let,

\begin{gathered}\sqrt{2}x+\sqrt{3}y=0.......(a)\\\\\sqrt{3}x-\sqrt{8}y=0......(b)\\\end{gathered}2x+3y=0.......(a)3x−8y=0......(b)

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

\begin{gathered}\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} \\\\\end{gathered}→2x+3y=0→2x=−3y→x=−23y→x=−23y×22→x=−232y→x=−26y

equation (b):

\begin{gathered}\to \sqrt{3}x-\sqrt{8}y=0\\\\\end{gathered}→3x−8y=0

\begin{gathered}\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} \\\\\end{gathered}→3(−26y)−8y=0→−3(23×2y)−8y=0→−232y−22y=0→2−32y−42y=0→2−32y−42y=0→−2y(23+4)=0→−2y(27)=0→y=−722→y=−722

Answer 2

Solution: (i) x + y = 14 ; x – y = 4

solve first equation

x + y = 14

x = 14 - y …………..(1)

plug this value in equation second we get

x – y = 4

14 – y - y = 4

Add 14 both side we get

- 2 y = 4 - 14

- 2 y = - 10

Y = -10/-2

Y = 5

Plug y = 5 in equation first we get

X = 14 – y

X = 14 – 5

X = 9