Precalculus Honors · Study guide

4.1 · Linear and nonlinear systems

Interpret solutions as intersections, use elimination efficiently, and distinguish inconsistent systems from dependent ones.

Precalculus Honors · Reading: Demana 10th ed. §§7.1, 7.3–7.4; Stewart 8th ed. §§9.1–9.2, 9.8–9.9

Learning objectives

Solve a system and verify the solution in every equation.
Classify one, no, or infinitely many linear-system solutions.
Find intersections of nonlinear graphs.

A solution must satisfy every relationship

For two equations in $\displaystyle x$ and $\displaystyle y$ , a solution is an intersection of their graphs. Elimination combines equations to remove a variable; substitution replaces one variable using the other equation.

Always check a resulting pair in the original equations. A nonlinear system may have multiple intersections or none, even when an algebraic method initially produces several candidates.

Parallel and coincident lines behave differently

If elimination produces a false statement such as $\displaystyle 0=5$ , the system is inconsistent. If it produces an identity and no independent restriction remains, the lines coincide and there are infinitely many solutions.

For inequalities, the solution is an intersection of shaded regions. Boundary inclusion depends on whether each inequality is strict or inclusive.

Substitution connects a line with a curve

For $\displaystyle y=x^2$ and $\displaystyle y=x+2$ , equate the outputs to get $\displaystyle x^2-x-2=0$ . Each real root supplies a candidate intersection, and substitution gives its matching y-coordinate.

Graphs support interpretation, but exact substitution verifies the coordinates. A graphing window may hide an intersection far from the origin.

Worked examples

Eliminate a variable

Solve $\displaystyle x+y=5$ and $\displaystyle x-y=1$ .

Add the equations to obtain $\displaystyle 2x=6$ .
Thus $\displaystyle x=3$ and $\displaystyle y=2$ .
Both original equations are satisfied.

A line and a parabola

Find intersections of $\displaystyle y=x^2$ and $\displaystyle y=x+2$ .

Solve $\displaystyle (x-2)(x+1)=0$ .
The x-values are $\displaystyle 2$ and $\displaystyle -1$ .
The intersections are $\displaystyle (2,4)$ and $\displaystyle (-1,1)$ .

Common mistakes

Checking a candidate in only one equation.
Treating a contradiction as an identity.
Assuming a nonlinear system has at most one solution.

Try it yourself

No calculator required.

1. Find all intersections of $\displaystyle y=x^2$ and $\displaystyle y=x+2$ .

The graphs y = x squared and y = x plus 2 meet twice. — Each solution must lie on both graphs.

Show worked solution

Equating gives $\displaystyle (x-2)(x+1)=0$ . The intersections are $\displaystyle (-1,1)$ and $\displaystyle (2,4)$ .

The graphs y = x squared and y = x plus 2 meet twice. The intersections are (minus 1, 1) and (2, 4). — The intersections are (minus 1, 1) and (2, 4).

2. Solve $\displaystyle 2x+y=7$ and $\displaystyle x-y=2$ .

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Add to obtain $\displaystyle 3x=9$ , so $\displaystyle x=3$ and $\displaystyle y=1$ .

3. Classify the system $\displaystyle x+y=3$ and $\displaystyle 2x+2y=8$ .

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Twice the first equation gives $\displaystyle 2x+2y=6$ , contradicting the second. The parallel lines have no common point, so the system has no solution.

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