Chapter 8, Problem 37REAE

Solution

To Find: Solution of the system of equation.

Hence, the solution of system is $\left(2,0\right)\text{and }(-1,12i)$

Given:

  $\begin{array}{c}16x^2-4y^2=64\\ 4x^2+9y^2-40x=-64\end{array}$

Concept Used: Substitution method of solving system of equations

Calculation:

  $\begin{array}{c}16x^2-4y^2=64\mathrm{......}(i)\\ 4x^2+9y^2-40x=-64\mathrm{......}(ii)\end{array}$

In equation (ii):

  $\begin{array}{c}4y^2=16x^2-64\\ y^2=\frac{16x^2-64}{4}\\ y^2=4x^2-16\end{array}$

Substituting the value of $y^2$ from equation (i) to equation (ii):

  $\begin{array}{c}4x^2+9\left(4x^2-16\right)-40x=-64\\ 4x^2+36x^2-144-40x=-64\\ 40x^2-40x-144+64=0\\ 40x^2-40x-80=0\\ 40\left(x^2-x-2\right)=0\\ x^2-x-2=0\\ x^2-2x+x-2=0\\ x\left(x-2\right)+1\left(x-2\right)=0\\ \left(x-2\right)\left(x+1\right)=0\end{array}$

Using zero product rule:

  $\begin{array}{c}x-2=0\text{or }x+1=0\\ x=2\text{or}x=-1\end{array}$

Now

  $\begin{array}{c}x=2\text{or}x=-1\\ y^2=4\times 2^2-16\text{or}{y}^2=4\times {\left(-1\right)}^2-16\\ y=\sqrt{16-16}\text{or}y=\sqrt{4-16}\\ y=0\text{or}y=12i\left\{\because i=\sqrt{-1}\right\}\end{array}$

Conclusion:

Hence, the solution of system is $\left(2,0\right)\text{and }(-1,12i)$