Solve a simple system consisting of a linear equation and a quadratic equation in two variables, algebraically and graphically (TN A1.A.REI.C.7).

What this topic is asking

Standard A1.A.REI.C.7 asks you to solve a simple system of one linear and one quadratic equation in two variables, both algebraically (substitution) and graphically (intersection of a line and a parabola). The solutions are the points the two graphs share, and there can be two, one, or none.

The substitution method

Because both equations give $y$ (or can be solved for $y$), set the right-hand sides equal and solve the resulting quadratic.

Zero, one, or two intersections

A line meets a parabola in at most two points. After substituting, the discriminant of the resulting quadratic tells you how many:

• Two real solutions (discriminant positive): the line is a secant, crossing twice.
• One real solution (discriminant zero): the line is tangent, touching once.
• No real solution (discriminant negative): the line misses the parabola.

This mirrors the discriminant's role for a single quadratic, because counting intersections is the same as counting real roots of the combined equation.

How TNReady examines this topic

• Numeric response. Solve the system and enter the $x$-values or a point.
• Multiple choice. Choose the solution points, or how many times the graphs intersect.
• Graphing. Identify or mark the intersection points of a given line and parabola.

A clarifying idea is that this standard is where the solving quadratics skills pay off inside a systems context: the system reduces to a quadratic, and your factoring or formula work finishes it.

Why a line and a parabola behave differently from two lines

Two lines meet at most once (or coincide), but a line and a parabola can meet twice because the parabola curves back. That extra possibility is exactly why substituting produces a quadratic, whose degree two allows up to two roots, instead of the linear equation two lines produce. The geometry and the algebra agree: the highest degree in the combined equation caps the number of intersection points. Recognizing this stops a common error of expecting a single answer; when an item asks for "the solutions" of a line-and-parabola system, plan for the possibility of two ordered pairs, and report both unless the discriminant says otherwise.

When the line is not solved for y

Sometimes one equation is in standard form, so you solve it for $y$ first, then substitute. For $\begin{cases} y = x^2 - 1 \\ x + y = 1 \end{cases}$, rearrange the linear equation to $y = 1 - x$, then set equal to the parabola: $x^2 - 1 = 1 - x$, giving $x^2 + x - 2 = 0$, which factors as $(x + 2)(x - 1) = 0$, so $x = -2$ or $x = 1$, with points $(-2, 3)$ and $(1, 0)$. The extra first step, isolating $y$ in the linear equation, is the only difference; once both equations read $y = \dots$, the substitution proceeds as usual.

Try this

Q1. Solve $\begin{cases} y = x^2 \\ y = 4 \end{cases}$. [2 points]

• Cue. $x^2 = 4 \Rightarrow x = \pm 2$; points $(2, 4)$ and $(-2, 4)$.

Q2. How many times does $y = x + 5$ meet $y = x^2 + 5$? [1 point]

• Cue. $x^2 + 5 = x + 5 \Rightarrow x^2 - x = 0 \Rightarrow x(x - 1) = 0$; two points, at $x = 0$ and $x = 1$.