### Systems of linear equations: An equation of a line

### The equation of a line

We have *seen* that solutions of the form #\blue p \cdot x + \green q\cdot y+\purple r=0# have a line as solutions. We have also *seen* that the linear formula #y = a\cdot x+b# has a line as a graph. Hence, there are two ways of describing the equation of a line.

#y=-{{5\cdot x}\over{2}}+2#

Because the coefficient of #y# is not equal to zero in the given equation, it is possible to reduce the equation to the form #y=a\cdot x+b#. We achieve this form through reduction:

\[\begin{array}{rcl}

-5\cdot x-2\cdot y&=&-4 \\&&\phantom{xxx}\blue{\text{the given equation}}\\

-2\cdot y&=&5\cdot x-4 \\&&\phantom{xxx}\blue{\text{added }5\cdot x\text{ left and right}}\\

y&=&\displaystyle -{{5\cdot x}\over{2}}+2\\&&\phantom{xxx}\blue{\text{left and right divided by the coefficient of }y}

\end{array}\]

Because the coefficient of #y# is not equal to zero in the given equation, it is possible to reduce the equation to the form #y=a\cdot x+b#. We achieve this form through reduction:

\[\begin{array}{rcl}

-5\cdot x-2\cdot y&=&-4 \\&&\phantom{xxx}\blue{\text{the given equation}}\\

-2\cdot y&=&5\cdot x-4 \\&&\phantom{xxx}\blue{\text{added }5\cdot x\text{ left and right}}\\

y&=&\displaystyle -{{5\cdot x}\over{2}}+2\\&&\phantom{xxx}\blue{\text{left and right divided by the coefficient of }y}

\end{array}\]

Unlock full access

Teacher access

Request a demo account. We will help you get started with our digital learning environment.

Student access

Is your university not a partner?
Get access to our courses via

Or visit omptest.org if jou are taking an OMPT exam.

**Pass Your Math**independent of your university. See pricing and more.Or visit omptest.org if jou are taking an OMPT exam.