## Intermediate Algebra for College Students (7th Edition)

Step 1: Make the either equation in the form of $y^2$ in order to this will multiply either equation by $2$. Step 2: After this, add the original equation and new equation. Step 3: We will get the solution in only one variable. Step 4: we will solve for the value of the either $x$ or $y$ variable that satisfies the equation. Step 5: Finally, we will get the solution set in the form of $(x,y)$ Let us take an example: After adding the given two equations, we get $\dfrac{11}{x^2}=11$ when we multiply the first equation by $2$. or, $x^2=1$; or $x=\pm 1$ From first equation $\dfrac{6}{x^2}+\dfrac{2}{y^2}=14$ when $x=1$, we have $y=\pm \dfrac{1}{2}$ From first equation $\dfrac{6}{x^2}+\dfrac{2}{y^2}=14$ when $x=-1$, we have $y=\pm \dfrac{1}{2}$ Thus, solution set is {$(1,\dfrac{1}{2}),(1,-\dfrac{1}{2}),(-1,\dfrac{1}{2}),(-1,-\dfrac{1}{2})$}