4.3 Linear-Quadratic Systems: Parabolas and Circles

4.3 Systems of Equations: Linear-Quadratic and 3-Variable

A system of equations consists of two or more equations sharing common variables. In Regents Algebra II, students advance beyond basic two-variable linear systems to solve non-linear systems (lines intersecting parabolas or circles) and multi-variable linear systems in three dimensions.

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Linear-Quadratic Systems in Two Variables

A linear-quadratic system pairs a linear equation of degree 1 with a quadratic equation of degree 2:

$$\begin{cases} \text{Linear Equation:} & Ax + By = C \quad \text{or} \quad y = mx + b \\[6pt] \text{Quadratic Equation:} & y = ax^2 + bx + c \quad \text{(Parabola)} \\[3pt] & \text{or} \quad (x - h)^2 + (y - k)^2 = r^2 \quad \text{(Circle)} \end{cases}$$

Geometric Classifications of Solutions

The number of real solutions corresponds to the number of physical intersection points between the line and the conic curve:

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The Algebraic Method: Substitution

Because quadratic equations involve squared terms, substitution is the primary algebraic strategy:

1. Isolate a variable: Solve for $y$ (or $x$) in the linear equation.
2. Substitute: Replace that variable in the quadratic equation with the linear expression, creating a single-variable quadratic equation.
3. Solve the quadratic: Write the equation in standard form $Ax^2 + Bx + C = 0$ and solve using factoring or the quadratic formula.
4. Back-substitute: Substitute each resulting $x$-value back into the linear equation to compute the corresponding $y$-value.
5. Write coordinate pairs: Pair each $x$ with its associated $y$ as an ordered pair $(x, y)$, and verify in the quadratic equation.

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Step-by-Step Worked Example: Line-Parabola System

Task: Solve the system algebraically for all values of $x$ and $y$:

$$\begin{cases} y - 2x = 1 \\[4pt] y = x^2 - x - 9 \end{cases}$$

• Step 1: Isolate $y$ in the linear equation

  $$y = 2x + 1$$

• Step 2: Substitute into the quadratic equation

  $$2x + 1 = x^2 - x - 9$$

• Step 3: Collect all terms to form a standard quadratic equation Subtract $2x$ and subtract $1$ from both sides:

  $$x^2 - 3x - 10 = 0$$

• Step 4: Solve for $x$ by factoring

  $$(x - 5)(x + 2) = 0$$ $$x = 5 \quad \text{or} \quad x = -2$$

• Step 5: Back-substitute into the linear equation to determine $y$

  • For $x = 5$: $$y = 2(5) + 1 = 10 + 1 = 11 \implies (5, 11)$$
  • For $x = -2$: $$y = 2(-2) + 1 = -4 + 1 = -3 \implies (-2, -3)$$

• Step 6: Verify in the quadratic equation

  • Point $(5, 11)$: $11 = 5^2 - 5 - 9 = 25 - 14 = 11 \quad \checkmark$
  • Point $(-2, -3)$: $-3 = (-2)^2 - (-2) - 9 = 4 + 2 - 9 = -3 \quad \checkmark$

• Final Answer: The solution set is $\{(5, 11), (-2, -3)\}$.

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Graphical Solutions and Standard AII-A.REI.11

Under New York standard AII-A.REI.11, the $x$-coordinates of the intersection points of two functions $y = f(x)$ and $y = g(x)$ represent the solutions to the equation $f(x) = g(x)$.

When a Regents prompt directs you to solve a system graphically, you must:

1. State a clear coordinate scale and label both axes ($x$ and $y$).
2. Plot the parabola with at least 5 key points: the vertex $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$, axis of symmetry, $y$-intercept, and symmetric partner points.
3. Plot the straight line using its $y$-intercept and slope ($m = \frac{\Delta y}{\Delta x}$).
4. Circle and label the intersection coordinates clearly on the grid.

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Three-Variable Linear Systems (Enrichment - Not on the Algebra II Blueprint)

A three-variable linear system consists of three equations of the form:

$$\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\[4pt] a_2 x + b_2 y + c_2 z = d_2 \\[4pt] a_3 x + b_3 y + c_3 z = d_3 \end{cases}$$

The Three-Dimensional Geometry of Planes

In three-dimensional coordinate space $\mathbb{R}^3$, the graph of a linear equation $ax + by + cz = d$ is a flat two-dimensional plane.

No common intersection occurs when planes are parallel, or when the three planes intersect pairwise along three parallel lines, forming a triangular prism.

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Systematic Elimination Protocol for $3 \times 3$ Systems

Attempting to solve three equations haphazardly often leads to circular algebra. Follow this disciplined five-step elimination algorithm:

Step 1: Choose a Target Variable  ───> Select the variable with the simplest coefficients (±1)
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Step 2: Eliminate from Pair (1,2) ───> Multiply and add to eliminate target; label Equation (4)
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Step 3: Eliminate from Pair (1,3) ───> Eliminate the SAME target variable; label Equation (5)
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Step 4: Solve 2x2 System (4,5)    ───> Solve Equations (4) and (5) for the two remaining variables
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Step 5: Back-Substitute           ───> Substitute known values into an original equation for target

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Step-by-Step Worked Example: Solving a $3 \times 3$ System

Task: Solve the system algebraically for $x$, $y$, and $z$:

$$\begin{aligned} (1) \quad x + y + z &= 6 \\[3pt] (2) \quad 2x - y + 3z &= 12 \\[3pt] (3) \quad x + 2y - z &= 1 \end{aligned}$$

• Step 1: Choose a target variable to eliminate Notice that $x$ has coefficient $1$ in Equation (1) and Equation (3). Let us eliminate $x$.

• Step 2: Eliminate $x$ using Equations (1) and (3) Subtract Equation (3) from Equation (1):

  $$(x + y + z) - (x + 2y - z) = 6 - 1$$ $$x - x + y - 2y + z - (-z) = 5$$ $$-y + 2z = 5 \quad \text{--- [Equation 4]}$$

• Step 3: Eliminate $x$ using Equations (1) and (2) Multiply Equation (1) by $2$:

  $$2(x + y + z) = 2(6) \implies 2x + 2y + 2z = 12$$

  Subtract Equation (2) from this result:

  $$(2x + 2y + 2z) - (2x - y + 3z) = 12 - 12$$ $$2x - 2x + 2y - (-y) + 2z - 3z = 0$$ $$3y - z = 0 \implies z = 3y \quad \text{--- [Equation 5]}$$

• Step 4: Solve the $2 \times 2$ system of Equations (4) and (5) Substitute $z = 3y$ into Equation (4):

  $$-y + 2(3y) = 5$$ $$-y + 6y = 5$$ $$5y = 5 \implies y = 1$$

  Now compute $z$ using Equation (5):

  $$z = 3(1) = 3$$

• Step 5: Back-substitute into an original equation to find $x$ Substitute $y = 1$ and $z = 3$ into Equation (1):

  $$x + 1 + 3 = 6$$ $$x + 4 = 6 \implies x = 2$$

• Step 6: Verify the ordered triple $(2, 1, 3)$ in all three equations

  • Equation (1): $2 + 1 + 3 = 6 \quad \checkmark$
  • Equation (2): $2(2) - 1 + 3(3) = 4 - 1 + 9 = 12 \quad \checkmark$
  • Equation (3): $2 + 2(1) - 3 = 2 + 2 - 3 = 1 \quad \checkmark$

• Final Answer: The unique solution is the ordered triple $(2, 1, 3)$.

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Common Regents Traps and Strategic Advice

• Eliminating Different Variables: In $3 \times 3$ systems, you must eliminate the same variable from both pairs. Eliminating $x$ from the first pair and $y$ from the second pair leaves three variables, making progress impossible.
• Sign Errors During Subtraction: When subtracting an entire equation, distribute the negative sign to every single term on both sides.