Question

Solve the following simultaneous equation to calculate the value of x and y.
\[\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6};\dfrac{3}{x}+\dfrac{2}{y}=0\]

Answer 1

Hint: To solve this question, we will first make the two equations in terms of x and y and then by using the method of substitution we will calculate the value of x and y separately by using the obtained two linear equations in two variables.

Complete step-by-step answer:
We are given
\[\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}......\left( i \right)\]
\[\dfrac{3}{x}+\dfrac{2}{y}=0.....\left( ii \right)\]
Let us consider equation (i) first. We have (i) as \[\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}.\]
\[\Rightarrow \dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}\]
Taking LCM of the terms on the left, we have
\[\Rightarrow \dfrac{2\times 3y+2x}{3xy}=\dfrac{1}{6}\]
\[\Rightarrow \dfrac{6y+2x}{3xy}=\dfrac{1}{6}\]
Cross multiplying the above equation, we have,
\[\Rightarrow 6\left( 6y+2x \right)=3xy\]
\[\Rightarrow 2\left( 6y+2x \right)=xy\]
\[\Rightarrow 12y+4x-xy=0.....\left( iii \right)\]
Now consider equation (ii), then we have,
\[\dfrac{3}{x}+\dfrac{2}{y}=0\]
Taking tan of the terms on the left we have,
\[\Rightarrow \dfrac{3y+2x}{xy}=0\]
On cross multiplying, we get,
\[\Rightarrow 3y+2x=0.....\left( iv \right)\]
Now, finally, we will solve the equations (iii) and (iv) to get the value of x and y.
\[\Rightarrow 3y+2x=0\]
\[\Rightarrow 3y=-2x\]
\[\Rightarrow y=\dfrac{-2x}{3}\]
From equation (iii), we have,
\[\Rightarrow 12y+4x-xy=0\]
Substituting this value of \[y=\dfrac{-2x}{3}\] obtained above in this equation, we have,
\[\Rightarrow 12\left( \dfrac{-2x}{3} \right)+4x-x\left( \dfrac{-2x}{3} \right)=0\]
\[\Rightarrow \dfrac{-24x}{3}+4x+\dfrac{2{{x}^{2}}}{3}=0\]
\[\Rightarrow \dfrac{-24x+12x+2{{x}^{2}}}{3}=0\]
On cross multiplying, we get,
\[\Rightarrow -24x+12x+2{{x}^{2}}=0\]
\[\Rightarrow 2\left( {{x}^{2}}+6x-12x \right)=0\]
As \[2\ne 0\]
\[\Rightarrow {{x}^{2}}+6x-12x=0\]
Finally, we will use the splitting by middle terms to get the value of x. Applying the splitting by the middle term, we have,
\[\Rightarrow {{x}^{2}}+6x-12=0\]
\[\Rightarrow {{x}^{2}}-6x=0\]
\[\Rightarrow x\left( x-6 \right)=0\]
\[\Rightarrow x=0;x=6\]
When, x = 0, y = 0
When x= 6, \[y=\dfrac{-2\times 6}{3}=-4\]
(x, y) = (0, 0) is not possible as x and y are in the denominator of the equation (i) and (ii). Hence, (x, y) = (6, – 4) is the answer to the given question.

Note: We can also solve by using \[x=\dfrac{-3y}{2}\] and substituting this value of x in the equation (iii) would give
\[\Rightarrow 12y+4\left( \dfrac{-3y}{2} \right)-\left( \dfrac{-3y}{2} \right)y=0\]
\[\Rightarrow 12y+\left( -6 \right)y+\dfrac{3}{2}{{y}^{2}}=0\]
\[\Rightarrow 24y-12y+3{{y}^{2}}=0\]
\[\Rightarrow 3{{y}^{2}}+12y=0\]
\[\Rightarrow {{y}^{2}}+4y=0\]
\[\Rightarrow y\left( y+4 \right)=0\]
\[\Rightarrow y=0;x=0\]
\[\Rightarrow y=-4;x=-3\times \dfrac{-4}{2}=+6\]
x, y = 0 is not possible so (x, y) = (6, – 4) is the answer.