Nonlinear System of Equations by Substitution

Problem 200: Solve the system by using substitution $\begin{cases} x^2 + 9y^2 = 9 \\ y = \frac{1}{3}x - 3\end{cases}$ .

Solution: Using equation two and plugging it into equation one we have

(1) $\begin{align*} x^2 + 9y^2 & = 9 \\ x^2 + 9(1/3x - 3)^2 & = 9 \\ x^2 + 9(1/9x^2 - 2x + 9) & = 9 \\ x^2 + x^2 - 18x + 81 & = 9 \\ 2x^2 - 18x + 72 & = 0 \\ x^2 - 9x + 36 & = 0 \\(x-12)(x-3) & = 0 \\ & \Rightarrow x = 12, x = 3. \end{align*}$

Using this $x$ ‘s values we can find the $y$ ‘s values. That is,

$x= 12$ :

(2) $\begin{align*} y = 1/3(12) - 3 = 4 - 3 = 1.\end{align*}$

$x= 3$ :

(3) $\begin{align*} y = 1/3(3) - 3 = 1 - 3 = -2.\end{align*}$

Thus the potential solutions are $(12, 1)$ and $(3, -2)$ . However, we need to check that this potential solutions satisfy the two equations. Hence,

$(12, 1)$ :

$\begin{cases} x^2 + 9y^2 = 9 \quad \Rightarrow \quad (12)^2 + 9(1)^2 \ne 9 \\ y = \frac{1}{3}x - 3 \quad \Rightarrow \quad 1 = \frac{1}{3}(12) - 3 \end{cases}$

Since the point $(12,1)$ does not satisfies both equations, this point is not a solution of the system.

$(3, -2)$ :

$\begin{cases} x^2 + 9y^2 = 9 \quad \Rightarrow \quad (3)^2 + 9(-2)^2 \ne 9 \\ y = \frac{1}{3}x - 3 \quad \Rightarrow \quad -2 = \frac{1}{3}(3) - 3 \end{cases}$

Since the point $(3,-2)$ does not satisfies both equations, this point is not a solution of the system.

Therefore, the system has no solutions.

One Problem at a Time

Nonlinear System of Equations by Substitution

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