### Systems of linear equations: Two equations with two unknowns

### Systems of linear equations

Systems of linear equations

Assume that for two given unknown numbers #x# and #y#, it is known that they are solutions to the following linear equations:

\[\lineqs{3 x +2 y +1&=&0\cr -2 x -3 y -6&=&0\cr}\]

We call this a** system of two equations with two** **unknowns**. The **solution** to the system is the point #\rv{x,y}# which is a solution to both linear equations. Hence, the solution to the system is the intersection point of the two lines which represent the linear equations.

We can also write the system with the sign for "and", which is #\land#. It then looks like this:

\[\begin{array}{rcl} 3 x +2 y +1=0 &\land&-2 x -3 y -6=0\end{array}\]

geogebra plaatje

No

If #\rv{-2, 6}# is a solution, then both equations have to be true if we substitute #x=-2# and #y=6# in the equations.

In this case, if we substitute #x=-2# and #y=6# in the system, we get:

\[\lineqs{-6\cdot -2-5\cdot 6+2=-16 \ne 0 \cr 9\cdot -2+2\cdot 6+8=2 \ne 0 \cr}\]

Both equations are incorrect, hence #\rv{-2, 6}# is not a solution of the system.

If #\rv{-2, 6}# is a solution, then both equations have to be true if we substitute #x=-2# and #y=6# in the equations.

In this case, if we substitute #x=-2# and #y=6# in the system, we get:

\[\lineqs{-6\cdot -2-5\cdot 6+2=-16 \ne 0 \cr 9\cdot -2+2\cdot 6+8=2 \ne 0 \cr}\]

Both equations are incorrect, hence #\rv{-2, 6}# is not a solution of the system.

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