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 \(x= 27\land y = 11\).

Answer:

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

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