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A general method of solving two linear equations with two unknowns is by eliminating one unknown. Here we discuss a method that is also suitable for larger systems of linear equations, with more unknowns.

The goal is to give the first equation the form \(x = a\) and the second equation the form \(y = b\).

- The strategy is to edit the equations in such a way that the new system will be equivalent to the old;
- the new system looks more like a solution than the old one.

Steps that will occur are multiplying all terms in the same equation by the same non-zero number and subtracting one equation over the other.

**The addition method for linear equations**

A system of two linear equations with unknowns \(x\) and \(y\) can be solved as follows.

- Make sure \(x\) occurs in the first equation: if this is not the case, then switch the two equations; this way, \(x\) will occur in the first equation.
- Replace the second equation by the difference of this equation with a suitable multiple of the first equation, in such a way that \(x\) no longer occurs in the second equation.
- Replace the first equation by the difference of this equation with a suitable multiple of the second equation in such a way that \(y\) no longer occurs in the first equation.
- The first equation is now a linear equation with \(x\) as the only unknown, and the second is a linear equation with \(y\) as the only unknown. These equations can be solved with the theory of linear equations with one unknown.

**Let’s practice!**

Solve the following system of equations with unknowns \(x\) and \(y\).

\(\left\{\begin{array}{l}-5·\left\{x\right\}+12·\left\{y\right\}+3=0\\ 2·\left\{x\right\}-5·y+1=0\end{array}\right.\)

Give the answer in the form \(x=a\land y=b\) for suitable values of \(a\) and \(b\).

There are many ways to get to this solution. We will describe one.

- To make sure that the unknown \(x\) is present in the first equation, we switch the two equations if this was not the case in the original system: \(-5\cdot x+12\cdot y+3=0\land 2\cdot x-5\cdot y+1=0\).
- Next we get rid of the term with \(x\) from the second equation by multiplying the first equation with \(\frac{2}{-5}\) and subtracting from the second: \(-5\cdot x+12\cdot y+3=0\land -{{y}\over{5}}=0\).
- By dividing (left and right hand side) in the second equation by \(-\frac{1}{5}\) we find \(y = 11\). We now have the system \(-5\cdot{x}+12\cdot y+3=0\land y=11\).
- If we enter the solution of \(y\) (the second equation) in the first equation (or stated different: we subtract \(12\) times the second equation from the first), then we find the system: \(135-5\cdot x=0\land y=11\).
- The first equation can be solved with the theory of linear equations with one unknown. The result is \(x= 27\land y = 11\).

**Answer:**

\(x= 27\land y = 11\)