Simultaneous Equations – One Linear, One Quadratic Revision Notes for Leaving Cert Mathematics

Simultaneous Equations – One Linear, One Quadratic

What are simultaneous equations with one linear and one quadratic?

When we have a system of two equations where one equation is linear (degree 1) and one equation is quadratic (degree 2), we call these simultaneous equations with one linear and one quadratic. The goal is to find the values of x and y that satisfy both equations at the same time.

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These equations represent different shapes when graphed:

Linear equations form straight lines (e.g., $x + y = 3$ )
Quadratic equations form curves like circles, parabolas, or hyperbolas (e.g., $x^2 + y^2 = 17$ or $y = x^2$ )

The solutions represent the intersection points where the line meets the curve.

The substitution method

To solve these types of simultaneous equations, we use the method of substitution. This is the most effective approach when dealing with one linear and one quadratic equation.

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The substitution method works particularly well here because linear equations are easy to rearrange, and substituting into the quadratic equation creates a single-variable quadratic that we can solve using familiar techniques.

Step-by-step process

Step 1: From the linear equation, express one variable in terms of the other

Choose the simpler variable to isolate (usually the one without a coefficient)
For example: if $x + y = 3$ , then $x = 3 - y$

Step 2: Substitute this expression into the quadratic equation

Replace the isolated variable in the quadratic equation
Solve the resulting quadratic equation (usually by factorising)
This gives you the values for one variable

Step 3: Find the corresponding values for the other variable

Substitute each solution back into the linear equation
This gives you the complete coordinate pairs

Worked examples

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Worked Example: Line and circle intersection

Problem: Solve the simultaneous equations:

$x + y = 3$ ... (1)
$x^2 + y^2 = 17$ ... (2)

Solution:

From equation (1): $x = 3 - y$

Substituting $(3 - y)$ for x in equation (2): $(3 - y)^2 + y^2 = 17$

Expanding: $9 - 6y + y^2 + y^2 = 17$ Simplifying: $2y^2 - 6y - 8 = 0$ Dividing by 2: $y^2 - 3y - 4 = 0$

Factorising: $(y - 4)(y + 1) = 0$ Therefore: $y = 4$ or $y = -1$

Finding the x-values:

When $y = 4$ : $x = 3 - 4 = -1$
When $y = -1$ : $x = 3 - (-1) = 4$

The solutions are: $(-1, 4)$ and $(4, -1)$

The diagram shows that $x + y = 3$ represents a straight line and $x^2 + y^2 = 17$ represents a circle. The intersection points (-1, 4) and (4, -1) are where the line crosses the circle.

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Worked Example: Line and parabola intersection

Problem: Find the coordinates of the intersection points A and B of the line $y = 2x + 3$ and the curve $y = x^2$ .

Solution:

We have:

$y = x^2$ ... (1)
$y = 2x + 3$ ... (2)

Substituting $(2x + 3)$ for y in equation (1): $2x + 3 = x^2$

Rearranging: $x^2 - 2x - 3 = 0$

Factorising: $(x + 1)(x - 3) = 0$ Therefore: $x = -1$ or $x = 3$

Finding the y-values using equation (2):

When $x = -1$ : $y = 2(-1) + 3 = 1$
When $x = 3$ : $y = 2(3) + 3 = 9$

The line intersects the curve at points $A(-1, 1)$ and $B(3, 9)$

This diagram shows how a straight line can intersect a parabola at two distinct points.

Graphical interpretation

The number of solutions depends on how the line and curve interact:

Two solutions: The line intersects the curve at two points (most common case)
One solution: The line is tangent to the curve (touches at exactly one point)
No real solutions: The line and curve don't intersect

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Always remember that the solutions you find algebraically correspond to actual points where the graphs intersect. If you get complex or no real solutions, this means the line and curve don't meet on the real coordinate plane.

Key formulas and concepts

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Essential equations to remember:

Circle equation: $x^2 + y^2 = r^2$ (where r is the radius) Parabola equation: $y = ax^2 + bx + c$ or $y = x^2$ Linear equation: $ax + by = c$ or $y = mx + c$

Important algebraic skills needed:

Expanding brackets: $(a - b)^2 = a^2 - 2ab + b^2$
Factorising quadratic expressions
Solving quadratic equations
Substitution techniques

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Key Points to Remember:

Always start with the linear equation when using substitution - it's much easier to rearrange
Check your solutions by substituting back into both original equations
Two intersection points are typical when a line crosses a curve
The solutions represent coordinates where the line and curve meet on a graph
Factorisation is usually the key to solving the resulting quadratic equation after substitution

Simultaneous Equations – One Linear, One Quadratic (Leaving Cert Mathematics): Revision Notes