Question

How do you write $4x-2y=-1$ in standard form?

Answer 1

Hint: We recall the three forms of writing a linear equation: the general form $ax+by+c=0$, the slope intercept form $y=mx+c$ and the standard form $ax+by=c$. We compare the given equation $4x-2y=-1$ and try to match the different forms equation with it.

Complete step by step answer:
We know that equation is a mathematical statement which involves equality between two algebraic expressions. The algebraic expressions contain unknowns called variables like $x,y,z$ and known constants. The highest power on any variable is called its degree.
We know that a linear equation is an equation with degree 1 with two variables $x,y$ has the general form
\[ax+by+c=0\]
Here $a,b,c$ have to be real numbers and $a,b$ cannot be zero. We know from the Cartesian coordinate system that every linear equation can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and if it cuts $y-$axis at a point $\left( 0,c \right)$ from the origin the $y-$intercept is given by $c$. The slope-intercept form of equation is given by
\[y=mx+c\]
We know that the standard form of linear equation otherwise also known as intercept form is written with constant $c$ on the right of equality sign as
\[ax+by=c\]
It is called intercept form too because the $x$ and $y-$ are obtained as $\dfrac{c}{a},\dfrac{c}{b}$ at the points$\left( \dfrac{c}{a},0 \right),\left( 0,\dfrac{c}{b} \right)$. We are given in the question the following equation
\[4x-2y=-1\]
We see that the above equation is in the form $ax+by=c$ where $a=4,b=-2,c=-1$. So the given equation is already in standard from and we do not need to convert it. \[\]

Note:
We note that we need at least 2 linear equations in two variables to find a unique solution. The standard form of the equation is useful while using elimination methods to solve the equations. We can convert an equation in general from $ax+by+{{c}^{'}}=0$ to slope point $y=mx+c$ from using $m=\dfrac{-a}{b},c=\dfrac{-{{c}^{'}}}{b}$.