Question

In the following exercises, solve the system of equations by using elimination.

$$\begin{gathered} x^2+y^2=14 \\ {x}^2-y^2=16 \end{gathered}$$

Short answer

Solution :

$$\begin{gathered} \left(\begin{array}{cc}x=\sqrt{15},\hspace{0.17em}& y=i\\ x=\sqrt{15},\hspace{0.17em}& y=-i\\ x=-\sqrt{15},\hspace{0.17em}& y=i\\ x=-\sqrt{15},\hspace{0.17em}& y=-i\end{array}\right) \\ (\sqrt{15},i) \\ (\sqrt{15},-i) \\ (-\sqrt{15},i) \\ (-\sqrt{15},-i) \end{gathered}$$

Step-by-step solution

Given information 

Two equations are given

$$\begin{gathered} x^2+y^2=14 \\ {x}^2-y^2=16 \end{gathered}$$

Eliminate the variable y 

$$\begin{gathered} x^2+y^2=14 \\ {x}^2-y^2=16 \\ --------- \\ 2x^2=30 \\ {x}^2=15 \\ x=\pm \sqrt{15} \end{gathered}$$

solve for y

Substitute the x values and find out y

$$\begin{gathered} x=-\sqrt{15} \\ {\left(\sqrt{15}\right)}^2+y^2=14 \\ 15+y^2=14 \\ {y}^2=-1 \\ y=\pm i \\ x=+\sqrt{15} \\ {(-\sqrt{15})}^2+y^2=14 \\ 15+y^2=14 \\ {y}^2=-1 \\ y=\pm i \end{gathered}$$

verify the solutions

$$\begin{gathered} x^2+y^2=14 \\ {x}^2-y^2=16 \\ x=\sqrt{15},y=i \\ {\left(\sqrt{15}\right)}^2+{\left(i\right)}^2=14,true \\ {\left(\sqrt{15}\right)}^2-{\left(i\right)}^2=16,true \\ x=-\sqrt{15},y=i \\ {(-\sqrt{15})}^2+{\left(i\right)}^2=14,true \\ {(-\sqrt{15})}^2-{\left(i\right)}^2=16,true \\ x=\sqrt{15},y=-i \\ {\left(\sqrt{15}\right)}^2+{(-i)}^2=14,true \\ {\left(\sqrt{15}\right)}^2-{(-i)}^2=16,true \\ x=-\sqrt{15},y=-i \\ {(-\sqrt{15})}^2+{(-i)}^2=14,true \\ {(-\sqrt{15})}^2-{(-i)}^2=16,true \end{gathered}$$