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Question: How do you write \(4x-2y=-1\) in standard form?...

How do you write 4x−2y=−14x-2y=-1 in standard form?

Explanation

Solution

We recall the three forms of writing a linear equation: the general form ax+by+c=0ax+by+c=0, the slope intercept form y=mx+cy=mx+c and the standard form ax+by=cax+by=c. We compare the given equation 4x−2y=−14x-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,zx,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,yx,y has the general form
ax+by+c=0ax+by+c=0
Here a,b,ca,b,c have to be real numbers and a,ba,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−x-axis at an angle θ\theta then its slope is given by m=tan⁡θm=\tan \theta and if it cuts y−y-axis at a point (0,c)\left( 0,c \right) from the origin the y−y-intercept is given by cc. The slope-intercept form of equation is given by
y=mx+cy=mx+c
We know that the standard form of linear equation otherwise also known as intercept form is written with constant cc on the right of equality sign as
ax+by=cax+by=c
It is called intercept form too because the xx and y−y- are obtained as ca,cb\dfrac{c}{a},\dfrac{c}{b} at the points(ca,0),(0,cb)\left( \dfrac{c}{a},0 \right),\left( 0,\dfrac{c}{b} \right). We are given in the question the following equation
4x−2y=−14x-2y=-1
We see that the above equation is in the form ax+by=cax+by=c where a=4,b=−2,c=−1a=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′=0ax+by+{{c}^{'}}=0 to slope point y=mx+cy=mx+c from using m=−ab,c=−c′bm=\dfrac{-a}{b},c=\dfrac{-{{c}^{'}}}{b}.