Question

How do you solve the system \[{{x}^{2}}+{{y}^{2}}=18\] and \[x-y=0\] ?

Answer 1

Hint: To solve this problem we have to consider the given equations as equation (1) and equation (2). Now by observing the equations we can see that the power of degree varies in equations. Now by squaring the equation (2) and substituting in equation (1), we can solve this equation.

Complete step by step solution:
For the given problem we are given to solve the system of equations.
Let us consider the given equations as equation (1) and equation (2).
\[{{x}^{2}}+{{y}^{2}}=18.............\left( 1 \right)\]
\[x-y=0.............\left( 2 \right)\]
Now we have to find the values from the equation (1) and equation (2). But in the given two equations it is difficult to find the values of x and y because in the equation (1) x and y are squared but in the equation (2), x and y are in like normal.
So, let us square the terms of x and y in equation (2).
Add the term ‘y’ on both sides of equation (2), we get
\[\Rightarrow x=y\]
By squaring the above equation, we get
\[\Rightarrow {{x}^{2}}={{y}^{2}}\]
Let us consider the equation as equation (3), we get
\[\Rightarrow {{x}^{2}}={{y}^{2}}..........\left( 3 \right)\]
Let us substitute the equation (3) in equation (1), we get
\[\Rightarrow {{y}^{2}}+{{y}^{2}}=18\]
By simplifying a bit, we get
\[\Rightarrow {{y}^{2}}=9\]
For finding the value of ‘y’ let us root the above equation, we get
\[\Rightarrow y=\pm 3\]
The feasible solutions when the value of y substituted in equation (1), we get
\[x=3,y=3\text{ and }x=-3,y=-3\]. Their feasibility was checked against the equation \[x-y=0\].

Note: We can have the solutions \[x=-3,y=3\] and \[x=3,y=-3\]. But they will satisfy only the first equation. But we are given two equations and our solutions should satisfy both the solutions. Students should solve this type of problem more to get a grip on this topic.