### Functions: Lines and linear functions

### Systems of equations

A number of notions introduced in *Linear equations with a single unknown* will be extended to systems of equations.

With a **system of equations** we mean one or more equations with one or more unknowns.

A **solution** to the system of equations is a list of values of the unknowns that, when entered in each equation of the system, makes all equalities true.

To **solve** a system of equations is to determine all solutions. The set of all solutions is called **the solution set**.

Generally speaking, we order the unknowns and write solutions as lists with values of the unknowns in the predefined order.

Two systems of equations are called **equivalent** if they have the same solution set.

A typical example is the system \[\lineqs{6\cdot x^2- y^2 &=& 14 \cr 2\cdot x^2 + 3\cdot y^2 &=& 18 \cr}\] with unknowns #x# and #y#, that can also be written as \[{6\cdot x^2- y^2 = 14 \quad \land\quad 2\cdot x^2 + 3\cdot y^2 = 18 }\tiny.\]

In the input field of exercises, this system is entered as \[\left[6x^2-y^2 = 14, 2x^2+3y^2 = 18\right] \tiny. \]

If the order of the unknowns is defined as #x,y#, then the list #\left[\sqrt{3},2\right]# is a solution. This can be verified by substituting #x=\sqrt{3}# and #y=2# in the equations and observing that the result is a true statement: \[\lineqs{6\cdot 3- 4 &=& 14 \cr 2\cdot 3 + 3\cdot 4 &=& 18 \cr}\tiny.\]

These equalities are true, so #\rv{x,y}=\rv{\sqrt{3},2}# is a solution. To solve this system is to find all solutions. In this case, these are, besides the solution already found, #\left[\sqrt{3},-2\right]#, #\left[-\sqrt{3},2\right]# en #\left[-\sqrt{3},-2\right]#.

In case the system of equations consists of a single equation with a single unknown, the definition of equivalence coincides with the definition of *Linear equations with a single unknown*.

After all, substitution of the right hand side of the second equation by #p# in the first equation gives

\[ y=4\cdot \left({{4}\over{x}}+2\right)-2 \tiny.\]

Expanding of the right hand side then gives #y={{6\cdot x+16}\over{x}}#.

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