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NEXUS

**Linear simultaneous equations** are a set of equations with multiple variables that are solved together. Each equation represents a straight line in a coordinate system, and the solution is the point(s) where the lines intersect.

We will cover two methods, elimination and substitution.

The elimination method requires the terms of one of the variable to have the same cooeficient.

We will start by multiplying equations to align coefficients. Then we will add or subtract equations to eliminate one variable.

\[\begin{align} 2x + 3y &= 12 \quad \text{(Equation 1)} \\ 4x - 2y &= 10 \quad \text{(Equation 2)} \end{align}\]

Multiply Equation 1 by 2:

\[4x + 6y = 24\]

Now subtract Equation 2:

\[(4x + 6y) - (4x - 2y) = 24 - 10\]

This simplifies to:

\[8y = 14 \quad \Rightarrow \quad y = \frac{14}{8} = \frac{7}{4}\]

Substitute \(y\) back into either original equation to find \(x\).

\[2x + 3\left(\frac{7}{4}\right) = 12\]

\[2x = 12 - \frac{21}{4}\]

\[x = \frac{27}{8}\]

So, the solution is:

\[\left( \frac{27}{8}, \frac{7}{4} \right)\]

If we make a variable the subject, we can substitute it into the other equation to solve.

\[\begin{align} y &= 2x + 1 \quad \text{(Equation 1)} \\ 3x + y &= 9 \quad \text{(Equation 2)} \end{align}\]

Substitute \(y\) in Equation 2:

\[3x + (2x + 1) = 9\]

This simplifies to:

\[5x + 1 = 9 \quad \Rightarrow \quad 5x = 8 \quad \Rightarrow \quad x = \frac{8}{5}\]

Substitute \(x\) back into Equation 1 to find \(y\):

\[y = 2\left(\frac{8}{5}\right) + 1 = \frac{16}{5} + 1 = \frac{21}{5}\]

So, the solution is:

\[\left( \frac{8}{5}, \frac{21}{5} \right)\]

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