Question

How do you find the exact solutions to the system \[{{x}^{2}}+{{y}^{2}}=25\ and\ {{y}^{2}}+9{{x}^{2}}=25\]?

Answer 1

Hint: In the given question, we have been asked to solve a system of equation i.e. \[{{x}^{2}}+{{y}^{2}}=25\ and \ {{y}^{2}}+9{{x}^{2}}=25\]. In order to solve the equations we will use elimination methods. We need to either add both the equations or subtract both the equations to get the equation in one variable. Then we solve the equation in one variable in a way we solve the general linear equation. Then substituting it for the value of another variable.

Complete step by step solution:
We have given that
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=25\ \]------ (1)
\[\Rightarrow \ {{y}^{2}}+9{{x}^{2}}=25\]
Rewrite as,
\[\Rightarrow 9{{x}^{2}}+\ {{y}^{2}}=25\]----- (2)
Multiply the equation (1) by 9, we get
\[\Rightarrow 9{{x}^{2}}+9{{y}^{2}}=225\ \]----- (3)
Subtracting equation (2) from equation (3), we get
\[\Rightarrow 9{{x}^{2}}+9{{y}^{2}}-\left( 9{{x}^{2}}+\ {{y}^{2}} \right)=225\ -25\]
Simplifying the above equation, we get
\[\Rightarrow 9{{x}^{2}}+9{{y}^{2}}-9{{x}^{2}}-{{y}^{2}}=225\ -25\]
Combining the like terms, we get
\[\Rightarrow 8{{y}^{2}}=200\]
Dividing both the sides of the equation by 8, we get
\[\Rightarrow {{y}^{2}}=25\]
Transposing power 2 to the RHS, we get
\[\Rightarrow y=\sqrt{25}=\pm 5\]
Therefore,
\[\Rightarrow y=\pm 5\]
Now,
Taking y = 5
Substitute the value of \[y=5\] in equation (1), we get
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=25\ \]
\[\Rightarrow {{x}^{2}}+{{5}^{2}}=25\ \]
Simplifying the numbers in the above equation, we get
\[\Rightarrow {{x}^{2}}+25=25\ \]
Subtracting 25 from both the side of the equation, we get
\[\Rightarrow x=0\]
Therefore, we get
\[\Rightarrow \left( x,y \right)=\left( 0,5 \right)\]
Now,
Taking y = -5
Substitute the value of \[y=-5\] in equation (1), we get
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=25\ \]
\[\Rightarrow {{x}^{2}}+{{\left( -5 \right)}^{2}}=25\ \]
Simplifying the numbers in the above equation, we get
\[\Rightarrow {{x}^{2}}+25=25\ \]
Subtracting 25 from both the side of the equation, we get
\[\Rightarrow x=0\]
Therefore, we get
\[\Rightarrow \left( x,y \right)=\left( 0,-5 \right)\]

Therefore, the system of the equations has 2 solutions (0, 5) and (0, -5).
Hence, it is the required solution.

Note: In an elimination method for solving a system of equations we will either add both the equations or subtract both the equations to get the equation in one variable. To eliminate the variable, you will add the two given equations if the coefficients of one variable are opposites and you will subtract the two equations if the coefficient of one variable is exactly the same. If both the situations do not satisfy then we will need to multiply one of the equations by a number that will lead to the same leading coefficient.