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Exercises Systems of equations

Example 1 from Algebra and Trigonometry, 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables

Solving a System of Nonlinear Equations Representing a Parabola and a Line

Solve the system of equations.

\begin{array}{l} x - y = −1 \\ y = x^{2} + 1 \end{array}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Solve the first equation for $x$ and then substitute the resulting expression into the second equation.

$\begin{array}{l} y = {(y −1)}^{2} + 1 \\ = (y^{2} −2 y + 1) + 1 \\ = y^{2} −2 y + 2 \\ 0 = y^{2} −3 y + 2 \\ = (y −2) (y −1) \end{array}$
Expand the equation and set it equal to zero.

$\begin{array}{l} y = {(y −1)}^{2} + 1 \\ = (y^{2} −2 y + 1) + 1 \\ = y^{2} −2 y + 2 \\ 0 = y^{2} −3 y + 2 \\ = (y −2) (y −1) \end{array}$
Solving for $y$ gives $y = 2$ and $y = 1.$ Next, substitute each value for $y$ into the first equation to solve for $x .$ Always substitute the value into the linear equation to check for extraneous solutions.
$\begin{array}{l} x - y = −1 \\ x - (2) = −1 \\ x = 1 \\ x - (1) = −1 \\ x = 0 \end{array}$
The solutions are $(1, 2)$ and $(0, 1),$ which can be verified by substituting these $(x, y)$ values into both of the original equations. See Figure 3.

Figure 3

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the given system of equations by substitution.

\begin{matrix} 3 x - y & = & -2 \\ 2 x^{2} - y & = & 0 \end{matrix}

How did it go?

PreviousFinding the Solution to a Dependent System of Equations Next Finding the Intersection of a Circle and a Line by Substitution

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.3, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.