One Problem at a Time

NonLinear System of Equations by Substitution II

Problem 201: Solve the system by using substitution: \begin{cases} y^2 - x = 0 \\ y = 3x - 2 \end{cases}

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

(1)   \begin{align*} y^2 - x & = 0 \\ (3x-2)^2 - x & = 0 \\ 9x^2 -12x + 4 - x & = 0 \\ 9x^2 - 13x + 4 & = 0 \\ & \Rightarrow x = 1, x = \frac{4}{9}. \end{align*}

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

x= 1:

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

x= \frac{4}{9}:

(3)   \begin{align*} y = 3(4/9) - 2 = \frac{4}{3} - 2 = -\frac{2}{3}.\end{align*}

Thus the potential solutions are (1, 1) and \left(\frac{4}{9}, -\frac{2}{3} \right). However, we need to check that this potential solutions satisfy the two equations. Hence,

(1, 1):

    \[\begin{cases} y^2 -x = 0 \quad \Rightarrow \quad (1)^2 - 1 = 0 \\ y = 3x -2 \quad \Rightarrow \quad 1 = 3(1) - 2 \end{cases}\]

Since the point (1,1) does satisfies both equations, this point is a solution of the system.

\left(\frac{4}{9}, -\frac{2}{3}\right):

    \[\begin{cases} y^2 - x = 0 \quad \Rightarrow \quad (-2/3)^2 - 4/9 = 0 \\ y = 3x - 2 \quad \Rightarrow \quad -2/3 = 3(4/9) - 2 \end{cases}\]

Since the point \left(\frac{4}{9}, -\frac{2}{3} \right) does satisfies both equations, this point is a solution of the system.

Therefore, the system’s solutions are (1, 1) and \left(\frac{4}{9}, -\frac{2}{3} \right).

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