Question

Solve the following system of equations. \[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\], \[{{x}^{2}}+{{y}^{2}}=160\]

Answer 1

Hint: We can solve a linear equation in two variables using the elimination method and substitution method. In this question we have taken the help of the first equation to find some value and using that value in the second equation, we have solved our question. We have also added and subtracted a variable with opposite terms to form a perfect square in the second equation.

Complete step by step answer:
A linear equation is the equation of two variables that represents the equation of a straight line on a coordinate plane. The linear equations that have the highest power one are known as the linear equation of first order and the equation with highest power two is known as the linear equation of second-order and so on. The linear equations with one variable can be solved easily but for solving the linear equation in two variables, we need two equations to find the value of a variable.
We have to solve the following equations.
\[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\]……eq(1)
\[{{x}^{2}}+{{y}^{2}}=160\]…..eq(2)
First, we will take eq(1).
\[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\]
Now we will take the LCM of the above equation and solve it.
\[\begin{align}
  & \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3} \\
 & \Rightarrow \dfrac{y+x}{xy}=\dfrac{1}{3} \\
 & \Rightarrow x+y=\dfrac{xy}{3} \\
\end{align}\]
……..eq(4)
So from eq(4), we have obtained the value of \[x+y\].
Now we will take eq(2).
\[{{x}^{2}}+{{y}^{2}}=160\]
We will add and subtract \[2xy\] on the left-hand side of the equation which is as shown below.
\[{{x}^{2}}+{{y}^{2}}-2xy+2xy=160\]……..eq(5)
As we know that, \[{{(x+y)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\].
So on putting this value in eq(5). We get the following results.
\[{{(x+y)}^{2}}-2xy=160\]…….eq(6)
Now we will put the value of \[x+y\], which we have obtained in eq(4) in eq(6). \[{{\left( \dfrac{xy}{3} \right)}^{2}}-2xy=160\]
On solving this equation, we get.
\[\begin{align}
  & \dfrac{{{x}^{2}}{{y}^{2}}}{9}-2xy=160 \\
 & \Rightarrow \dfrac{{{x}^{2}}{{y}^{2}}-18xy}{9}=160 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{x}^{2}}{{y}^{2}}-18xy=1440 \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}-18xy-1440=0 \\
\end{align}\]
Let us consider \[xy=t\] in the above equation.
\[{{t}^{2}}-18t-1440=0\]….eq(7)
So this is in the form of a quadratic equation and now we will solve it.
On solving the above quadratic, we get
\[{{t}^{2}}-(48-30)t-1440=0\]
\[\Rightarrow {{t}^{2}}-48t+30t-1440=0\]
\[\Rightarrow t(t-48)+30(t-48)=0\]
\[\Rightarrow (t-48)(t+30)=0\]
So from above,
\[t-48=0\]
\[\Rightarrow t=48\]
\[\begin{align}
  & t+30=0 \\
 & \Rightarrow t=-30 \\
\end{align}\]
Now we will put the original value of t in the above results.
\[\begin{align}
  & xy=48 \\
 & xy=-30 \\
\end{align}\]
So from the above values of \[xy\], we come to know that the above equations can have an infinite number of solutions.

Note:
Equations can be of two types, linear equations, and nonlinear equations. Linear equations represent the straight line but non-linear equations do not form a straight line. They just represent the curve. Linear equations can be represented in three forms- standard form, slope-intercept form, and point-slope form.