Q. 22 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $21$ and $22$, solve each system of equations.

$\left\{\begin{array}{r}2y^2-3x^2=5\\ y-x=1\end{array}\right.$

Step-by-Step Solution

Answer

Solution of system equations is $(1,2) \& (3,4)$

Step 1. Given information

A system of equations is given as

$\left\{\begin{array}{r}2y^2-3x^2=5\\ y-x=1\end{array}\right.$

Step 2. Solving for x

First we will find the value of y in terms of x as,

$$\begin{gathered} y-x=1 \\ y=1+x \end{gathered}$$

$$\begin{gathered} 2y^2-3x^2=5 \\ 2(1+x{)}^2-3x^2=5 \\ 2\left(x^2+2x+1\right)-3x^2=5 \\ -x^2+4x+2-5=0 \\ {x}^2-4x+3=0 \\ (x-3\left)\right(x-1)=0 \\ x=1,3 \end{gathered}$$

Now substitute the values of x as,

$$\begin{gathered} y=1+1 \\ y=2 \\ \& \\ y=1+3 \\ y=4 \end{gathered}$$