# Solving systems of equations by addition

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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.

1. 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.
2. 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.
3. 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.
4. 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 $$x= 27\land y = 11$$.

$$x= 27\land y = 11$$