4.7 Systems of Linear and Nonlinear Equations

4.7 Systems of Linear and Nonlinear Equations

Quick Summary: A nonlinear system pairs a linear equation with a nonlinear equation (such as a quadratic parabola or circle). Graphically, the solutions to the system are the points of intersection $(x, y)$ between the line and the curve. A linear-quadratic system can have $0$ solutions (no intersection), $1$ solution (tangent line), or $2$ solutions (secant line).

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Geometric Configurations of Linear-Nonlinear Systems

   CASE 1: 2 Solutions (Secant)     CASE 2: 1 Solution (Tangent)      CASE 3: 0 Solutions (No Touch)
           y                                y                                y
           |  /                             |                                |        /
           | /  * (x_2, y_2)                |      * Point of                |       / Line
       ----+*-------/---x               ----+-----*-- Tangency -x        ----+------/--------x
           | \     /                        |    / \                         |     /   /\
           |  \   /                         |   /   \                        |    /   /  \
           |   \ /                          |  / Line\                       |       /    \
         Parabola & Line                  Parabola & Line                  Parabola & Line

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Solving Linear-Quadratic Systems via Substitution

Step-by-Step Algebraic Protocol

1. Isolate a variable in the linear equation (usually $y = mx + d$ or $x = \dots$).
2. Substitute this linear expression into the quadratic equation in place of that variable.
3. Rearrange all terms to form a single standard quadratic equation: $Ax^2 + Bx + C = 0$.
4. Solve for $x$ using factoring or the quadratic formula.
5. Back-Substitute each $x$-value into the linear equation to determine its matching $y$-value.
6. Write solutions as coordinate pairs: $(x_1, y_1)$ and $(x_2, y_2)$.

Worked Example 1: Solving a Linear-Quadratic System

$\begin{cases} y = x^2 - 3x - 4 \\ y = x + 1 \end{cases}$

• Step 1: Set the two expressions for $y$ equal to each other: $x^2 - 3x - 4 = x + 1$
• Step 2: Collect all terms on one side: $x^2 - 4x - 5 = 0$
• Step 3: Factor the quadratic: $(x - 5)(x + 1) = 0 \implies x = 5 \quad \text{or} \quad x = -1$
• Step 4: Back-substitute into the linear equation $y = x + 1$:
  • For $x = 5$: $y = 5 + 1 = 6 \implies (5, 6)$
  • For $x = -1$: $y = -1 + 1 = 0 \implies (-1, 0)$

The solutions to the system are $(5, 6)$ and $(-1, 0)$.

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Determining the Number of Solutions via the Discriminant

When a linear-quadratic system is combined into $Ax^2 + Bx + C = 0$, the discriminant $\Delta = B^2 - 4AC$ dictates the number of intersection points:

Worked Example 2: Finding a Parameter for Tangency (1 Solution)

Problem: For what value of $k$ does the system below have exactly one real solution? $\begin{cases} y = 2x + k \\ y = -x^2 + 6x - 1 \end{cases}$

• Step 1: Substitute $y = 2x + k$ into the quadratic: $2x + k = -x^2 + 6x - 1$
• Step 2: Set the equation to zero in standard form: $x^2 - 4x + (k + 1) = 0$ Here $A = 1, B = -4, C = k + 1$.
• Step 3: Set the discriminant equal to zero for exactly one solution: $\Delta = B^2 - 4AC = 0$ $(-4)^2 - 4(1)(k + 1) = 0$ $16 - 4k - 4 = 0$ $12 - 4k = 0 \implies 4k = 12 \implies k = 3$

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Systems Involving Circles and Lines

A line can also intersect a circle $(x - h)^2 + (y - k)^2 = r^2$ at $0, 1,$ or $2$ points.

Worked Example 3: Circle-Line Intersection

$\begin{cases} (x - 2)^2 + (y + 1)^2 = 25 \\ y = 3 \end{cases}$

• Substitute $y = 3$ directly into the circle equation: $(x - 2)^2 + (3 + 1)^2 = 25$ $(x - 2)^2 + 4^2 = 25$ $(x - 2)^2 + 16 = 25 \implies (x - 2)^2 = 9$
• Apply the square root property: $x - 2 = \pm 3 \implies x = 2 \pm 3$ $x = 5 \quad \text{or} \quad x = -1$
• The points of intersection are $(5, 3)$ and $(-1, 3)$ ($2$ solutions).

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Digital SAT Desmos Graphical Verification Strategy

Desmos handles non-linear systems effortlessly:

1. Enter both equations exactly as written:
   • Line 1: y = x^2 - 3x - 4
   • Line 2: y = x + 1
2. Look for Intersections: Desmos automatically computes and highlights intersections with gray dots.
3. Click the Dots: Click each gray dot to display the exact coordinates $(5, 6)$ and $(-1, 0)$.
4. Check the Question Prompt: Ensure you provide what the question specifically requests (e.g., the $x$-value, the $y$-value, $x_1 + x_2$, or $xy$).

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Common Pitfalls & Exam Traps

• Pairing the wrong $x$ and $y$ values: When you obtain $x = 5$ and $x = -1$, calculate $y(5)$ and $y(-1)$ individually. Never mix the coordinates into $(5, 0)$ or $(-1, 6)$.
• Back-substituting into the quadratic instead of the linear: Back-substituting into the linear equation $y = mx + b$ is faster and eliminates potential extraneous arithmetic.
• Misinterpreting "single solution" in context: A single solution to a linear-quadratic system means the line is tangent to the curve ($B^2 - 4AC = 0$).
• Missing second intersection on circles: For circle equations $(x - h)^2 + (y - k)^2 = r^2$, remember that solving $(x - h)^2 = c$ yields two distinct $x$-values ($\pm \sqrt{c}$).