by applying substitution method ....√2x+√3y=0 √3x-√8y=0
Answers
Answer:
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Given:
√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}
2
x+
3
y=0.......(a)
3
x−
8
y=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}
→
2
x+
3
y=0
→
2
x=−
3
y
→x=−
2
3
y
→x=−
2
3
y
×
2
2
→x=−
2
3
2
y
→x=−
2
6
y
equation (b):
\begin{gathered}\to \sqrt{3}x-\sqrt{8}y=0\\\\\end{gathered}
→
3
x−
8
y=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
(−
2
6
y
)−
8
y=0
→−
3
(
2
3
×
2
y
)−
8
y=0
→−
2
3
2
y
−2
2
y=0
→
2
−3
2
y−4
2
y
=0
→
2
−3
2
y−4
2
y
=0
→−
2
y(
2
3+4
)=0
→−
2
y(
2
7
)=0
→y=−
7
2
2
→y=−
7
2
2
×
2
2
→y=−
7×2
2
2
→y=−
7
2
put the value of y into the equation (a):
Equation (a):
\to \sqrt{2}x+\sqrt{3}y=0→
2
x+
3
y=0
\begin{gathered}\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}\\\\\end{gathered}
→
2
x+
3
×(−
7
2
)=0
→
2
x−
7
2
3
=0
→
2
x−
7
3
2
=0
→
2
x=
7
3
2
→x=
7
2
3
2
→x=
7
3
The final value of x and y is: \begin{gathered}\bold{-\frac{\sqrt{2}}{7} \ \ _{and} \ \ \frac{\sqrt{3}}{7}}\\\end{gathered}
−
7
2
and
7
3