Chapter 8.7, Problem 17E

Solution

To solve: the system of equations by substitution.

No solution.

Given:

  $\begin{array}{c}9x^2+4y^2=36\\ -x+y=-4\end{array}$

Calculation:

Consider,

  $-x+y=-4$

By solving it for $y$ , it follows that

  $y=x-4$

Now put $y=x-4$ in $9x^2+4y^2=36$ , it follows that

  $\begin{array}{c}9x^2+4{\left(x-4\right)}^2=36\\ 9x^2+4\left(x^2-8x+16\right)=36\\ 9x^2+4x^2-32x+64=36\\ 13x^2-32x+28=0\end{array}$

It is not possible for any real values of $x$ .

Hence there is no solution.