Chapter 11, Problem 22CT

To determine

The solution of the system of equations, $\{\begin{array}{l}2y^2-3x^2=5\\ y-x=1\end{array}$.

Answer to Problem 22CT

Solution:

The solution of system of equation $\{\begin{array}{l}2y^2-3x^2=5\\ y-x=1\end{array}$ are $x=1,y=2$ and $x=3,y=4$ or $\left(1,2\right),\left(3,4\right)$.

Explanation of Solution

Given information:

The system of equations, $\{\begin{array}{l}2y^2-3x^2=5\\ y-x=1\end{array}$.

Explanation:

Consider the system of equation,

$2y^2-3x^2=5$ … (1)

$y-x=1$… (2)

From equation (2), $y=1+x$

Substitute the value of $y=1+x$ in the equation (1),

$\Rightarrow 2{\left(1+x\right)}^2-3x^2=5$

$\Rightarrow 2\left(1+2x+x^2\right)-3x^2=5$

$\Rightarrow 2+4x+2x^2-3x^2=5$

$\Rightarrow 2+4x-x^2=5$

$\Rightarrow x^2-4x+5-2=0$

$\Rightarrow x^2-4x+3=0$

$\Rightarrow \left(x-1\right)\left(x-3\right)=0$

By using zero product property,

$\Rightarrow x-1=0$ or $x-3=0$

$\Rightarrow x=1$ or $x=3$

Now substitute $x=1$ in equation $y=1+x$,

$\Rightarrow y=1+1=2$

Now plug $x=3$ in equation $y=1+x$,

$\Rightarrow y=1+3=4$

The solution of the system is $x=1,y=2$ and $x=3,y=4$.

Hence, the solution of system of equation $\{\begin{array}{l}2y^2-3x^2=5\\ y-x=1\end{array}$ are $x=1,y=2$ and $x=3,y=4$ or $\left(1,2\right),\left(3,4\right)$.