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

The points $\left( 2,1 \right),\left( 2,-1 \right),\left( -2,1 \right)$ and $\left( -2,-1 \right)$ are the solutions of both equations.
The given set of equations is\begin{align} & 2{{x}^{2}}-3{{y}^{2}}=5 \\ & 3{{x}^{2}}+4{{y}^{2}}=16 \end{align}. Calculation: The provided equations are, $3{{x}^{2}}+4{{y}^{2}}=16$ (I) $2{{x}^{2}}-3{{y}^{2}}=5$ (II) Simplify as follows: Step 1: Let’s eleminate the variable ${{y}^{2}}$, multiply 3 in equation (I) and multiply 4 in equation (2), by addition method in equation (I) and (II) simplify as follows: \begin{align} & 9{{x}^{2}}+12{{y}^{2}}+\left( 8{{x}^{2}}-12{{y}^{2}} \right)=48+20 \\ & 17{{x}^{2}}=68 \\ & {{x}^{2}}=4 \end{align} So, $x=\pm 2$ are solutions of both equations. Step 2: Put the value of $x$ in equation (I) to find out the value of $y$ simplify as follows: When$x=\pm 2$, \begin{align} & 3{{x}^{2}}+4{{y}^{2}}=16 \\ & 3{{\left( \pm 2 \right)}^{2}}+4{{y}^{2}}=16 \\ & 12+4{{y}^{2}}=16 \\ & 4{{y}^{2}}=4 \end{align} Simplify further, ${{y}^{2}}=1$ So, $y=\pm 1$ are solutions of this equations. Step 3: Check the proposed solutions in both of the equations, now begin by check$\left( \pm 2,\pm 1 \right)$, and replace $x$ with $\pm 2$ and $y$ with$\pm 1$. \begin{align} & 3{{x}^{2}}+4{{y}^{2}}=16 \\ & 3{{\left( \pm 2 \right)}^{2}}+4{{\left( \pm 1 \right)}^{2}}=16 \\ & 12+4=16 \\ & 16=16 \end{align} $\text{True}$ And, \begin{align} & 2{{x}^{2}}-3{{y}^{2}}=5 \\ & 2\times {{\left( \pm 2 \right)}^{2}}-3\times {{\left( \pm 1 \right)}^{2}}=5 \\ & 8-3=5 \\ & 5=5 \end{align} $\text{True}$ Hence, the points $\left( 2,1 \right),\left( 2,-1 \right),\left( -2,1 \right)$ and $\left( -2,-1 \right)$ are the solution of the system.