Simultaneous Equations Revision Notes for Grade 12 NSC Matric Mathematics

Simultaneous Equations

What are simultaneous equations?

Simultaneous equations are sets of two or more equations that must be solved together to find the values of the unknown variables (typically x and y). These equations share the same variables and must all be true at the same time.

There are two main types you'll encounter:

Linear-linear: Both equations are straight line equations
Linear-quadratic: One equation is linear (straight line) and one is quadratic (parabola or hyperbola)

The solutions represent the points where the graphs of these equations intersect.

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Understanding the type of simultaneous equations you're dealing with is crucial because it determines which solution method will be most effective. Linear-linear systems can often be solved using elimination, while linear-quadratic systems typically require substitution.

Solving linear-quadratic simultaneous equations

When you have one linear equation and one quadratic equation, use the substitution method. This involves four clear steps:

Step 1: Express one variable in terms of the other using the linear equation
Step 2: Substitute this expression into the quadratic equation
Step 3: Solve the resulting quadratic equation for one variable
Step 4: Substitute the found values back into the linear equation to find the second variable

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Always use the linear equation for substitution in Step 1. It's much easier to express a variable from a linear equation than from a quadratic equation, and this approach will save you time and reduce errors.

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Worked Example: Solving Linear-Quadratic System

Solve the system: $y + 2x - 2 = 0$ (Equation 1)
$2x^2 + y^2 = 3yx$ (Equation 2)

Step 1: Express y in terms of x from Equation 1 $y = 2 - 2x$

Step 2: Substitute into Equation 2 $2x^2 + (2 - 2x)^2 = 3x(2 - 2x)$

Step 3: Expand and solve for x

$2x^2 + (4 - 8x + 4x^2) = 6x - 6x^2$
$2x^2 + 4 - 8x + 4x^2 = 6x - 6x^2$
$6x^2 - 8x + 4 = 6x - 6x^2$
$12x^2 - 14x + 4 = 0$
$6x^2 - 7x + 2 = 0$

Factorise: $(3x - 2)(2x - 1) = 0$

Therefore: $x = \frac{2}{3}$ or $x = \frac{1}{2}$

Step 4: Find corresponding y values

For $x = \frac{2}{3}$ : $y = 2 - 2(\frac{2}{3}) = \frac{2}{3}$

For $x = \frac{1}{2}$ : $y = 2 - 2(\frac{1}{2}) = 1$

Final solutions: $(\frac{2}{3}, \frac{2}{3})$ and $(\frac{1}{2}, 1)$

Finding intersection points of graphs algebraically

To find where two graphs intersect, you need to solve their equations simultaneously. This gives you the exact coordinates of intersection points.

The method involves three steps:

Set the equations equal to each other to find x-values
Solve for x (this may involve factorisation)
Substitute the x-values into either original equation to find corresponding y-values

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Finding intersections algebraically is more accurate than reading coordinates from a graph. While graphs give you a visual understanding, algebraic solutions provide exact values that are essential for precise mathematical work.

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Worked Example: Finding Graph Intersections

Find the intersection points of $y = \frac{6}{x}$ and $y = x - 1$

Step 1: Set the equations equal $x - 1 = \frac{6}{x}$

Step 2: Multiply both sides by x and solve $x(x - 1) = 6$
$x^2 - x = 6$
$x^2 - x - 6 = 0$

Factorise: $(x - 3)(x + 2) = 0$
Therefore: $x = 3$ or $x = -2$

Step 3: Find corresponding y-values

For $x = 3$ : $y = 3 - 1 = 2$

For $x = -2$ : $y = -2 - 1 = -3$

Final intersection points: $(3, 2)$ and $(-2, -3)$

Solving linear-linear simultaneous equations

When both equations are linear, you can use either the substitution method or the elimination method.

Substitution method steps:

Eliminate one variable by expressing it in terms of the other
Solve for the remaining variable
Substitute back to find the second variable

Elimination method steps:

Add or subtract the equations to eliminate one variable
Solve for the remaining variable
Substitute back to find the second variable

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Choose your method based on the structure of the equations. If one equation already has a variable with coefficient 1 (like $y = 3x + 2$ ), substitution is usually quicker. If the coefficients are similar in both equations, elimination might be more efficient.

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Worked Example: Linear-Linear System

Solve the system:

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

First, rearrange Equation 2 to standard form:

$x^2 + y + x - y^2 = 0$
$x^2 + x + y - y^2 = 0$

Step 1: Express y from Equation 1

$y = 3 - 2x$

Step 2: Substitute into the rearranged Equation 2

$x^2 + x + (3 - 2x) - (3 - 2x)^2 = 0$
$x^2 + x + 3 - 2x - (9 - 12x + 4x^2) = 0$
$x^2 + x + 3 - 2x - 9 + 12x - 4x^2 = 0$
$-3x^2 + 11x - 6 = 0$
$3x^2 - 11x + 6 = 0$

Factorise: $(3x - 2)(x - 3) = 0$
Therefore: $x = \frac{2}{3}$ or $x = 3$

Step 3: Find corresponding y values

For $x = \frac{2}{3}$ : $y = 3 - 2(\frac{2}{3}) = \frac{5}{3}$
For $x = 3$ : $y = 3 - 2(3) = -3$

Final solutions: $(\frac{2}{3}, \frac{5}{3})$ and $(3, -3)$

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Common Mistakes to Avoid:

Forgetting to substitute back to find the second variable
Making algebraic errors during expansion - always double-check your working
Not factorising correctly - practice your factorisation skills as this is often the key step
Mixing up which equation to use for substitution

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Key Exam Tips and Methods:

Always label your equations (Equation 1, Equation 2) before you start solving
For linear-quadratic systems: Use substitution - express the linear equation in terms of one variable
For linear-linear systems: Choose either substitution or elimination based on which looks simpler
Check your solutions by substituting back into both original equations
When finding intersections: Set the equations equal and solve systematically
Don't forget factorisation - it's often the key step in solving quadratic equations
Write coordinates clearly as (x, y) pairs for your final answer

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Key Points to Remember:

Simultaneous equations must be solved together to find where graphs intersect
Linear-quadratic systems typically give two solution points
Always substitute your x-values back to find the corresponding y-values
Factorisation is usually required when solving the resulting quadratic equation
Check your answers make sense by substituting back into the original equations

Simultaneous Equations (Grade 12 NSC Matric Mathematics): Revision Notes