Digital SAT Maths -Unit 2 - 2.4 Non-linear Equations- Study Notes- New Syllabus
Digital SAT Maths -Unit 2 – 2.4 Non-linear Equations- Study Notes- New syllabus
Digital SAT Maths -Unit 2 – 2.4 Non-linear Equations- Study Notes – per latest Syllabus.
Key Concepts:
Non-Linear Equations (Two Variables)
Non-Linear Equations (Two Variables)
A non-linear equation is any equation in \( x \) and \( y \) that does not form a straight line when graphed.
Common examples on the DIGITAL SAT include:
- Quadratics: \( y = x^2 + 3x + 1 \)
- Circles: \( x^2 + y^2 = 25 \)
- Absolute value: \( y = |x – 2| \)
When solving a system involving a non-linear equation and a linear equation, the solution represents the intersection points of the graphs.
Possible Number of Solutions
- 0 solutions → graphs do not touch
- 1 solution → graphs touch once (tangent)
- 2 solutions → graphs intersect twice
Main Method: Substitution
You almost always solve these systems using substitution.
Steps
- Solve the linear equation for one variable
- Substitute into the non-linear equation
- Solve the resulting quadratic equation
- Back-substitute to find the second variable
DIGITAL SAT Insight
The SAT often asks for only one value, such as the sum of the \( x \)-coordinates or the value of \( y \) at an intersection.
Example 1 (Line and Parabola):
Solve the system:
\( y = x + 1 \)
\( y = x^2 – 3 \)
▶️ Answer/Explanation
Set equations equal:
\( x + 1 = x^2 – 3 \)
Rearrange:
\( x^2 – x – 4 = 0 \)
Factor:
\( (x – 2)(x + 2) = 0 \)
\( x = 2 \) or \( x = -2 \)
Find \( y \):
If \( x=2 \), \( y=3 \)
If \( x=-2 \), \( y=-1 \)
Conclusion: The intersection points are \( (2,3) \) and \( (-2,-1) \).
Example 2 (Circle and Line):
Find the solutions:
\( x^2 + y^2 = 25 \)
\( y = x \)
▶️ Answer/Explanation
Substitute:
\( x^2 + x^2 = 25 \)
\( 2x^2 = 25 \)
\( x^2 = \dfrac{25}{2} \)
\( x = \pm \dfrac{5}{\sqrt{2}} \)
Since \( y=x \),
\( y = \pm \dfrac{5}{\sqrt{2}} \)
Conclusion: Two intersection points exist.
Example 3 (Number of Solutions):
How many solutions does the system have?
\( y = x^2 \)
\( y = -2 \)
▶️ Answer/Explanation
Substitute:
\( x^2 = -2 \)
No real solution exists.
Conclusion: The system has 0 real solutions.