Linear Simultaneous Equations Revision Notes for Leaving Cert Mathematics

Linear Simultaneous Equations

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Simultaneous linear equations are a set of two or more linear equations with multiple variables, solved together to find values of the variables that satisfy all equations simultaneously (at the same time)

Let's consider two linear equations :

$x+2y=10$
$3x-y=5$ On a graph, the two linear equations look like this :

You'll notice that both lines share a single point, the point of intersection, this is the point that satisfies both line simultaneously.

There are two ways to solve simultaneous equations

Elimination
Substitution

Method 1 : Elimination

Example

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Solve the pair of simultaneous equations :

2x + 3y = 10 \quad \text{(Equation A)} \\ 5x - 4y = 7 \quad \text{(Equation B)}

Eliminate one variable, we can multiply Equation A by $5$

10x + 15y = 50 \quad \text{(Equation A)}

Multiply Equation B by $2$

10x - 8y = 14 \quad \text{(Equation B)}

Subtract the equations :

10x + 15y = 50\quad \\ \underline{-(10x - 8y = 14)} \quad \\ 23y=36

Solve for $y$ :

\therefore \boxed {y = \frac{36}{23}}

Substitute y back into one of the original equations to find $x$ :

2x + 3\left(\frac{36}{23}\right) = 10

2x = 10 - \frac{108}{23}

2x = \frac{122}{23}

\therefore \boxed {x = \frac{61}{23}}

Method 2 : Substitution

Example

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Solve the pair of simultaneous equations :

2x + 3y = 10 \quad \text{(Equation A)} \\ 5x - 4y = 7 \quad \text{(Equation B)}

Solve for one variable :

2x + 3y = 10

2x = 10 - 3y

x = \frac{10 - 3y}{2}.

Substitute into the other equation

5x - 4y = 7

5\left(\frac{10 - 3y}{2}\right) - 4y = 7

Simplify:

\frac{5(10 - 3y)}{2} - 4y = 7

\frac{50 - 15y}{2} - 4y = 7

50 - 15y - 8y = 14

50 - 23y = 14

-23y = 14 - 50

-23y = -36

y = \frac{-36}{-23} = \frac{36}{23}.

Substitute Back to Solve for $x$ :

x = \frac{10 - 3y}{2}

x = \frac{10 - 3\left(\frac{36}{23}\right)}{2}

x = \frac{122}{46}.

x = \frac{61}{23}.

Linear Simultaneous Equations (Leaving Cert Mathematics): Revision Notes