Question

How do you write $y - \dfrac{1}{4} = - 3\left( {x + \dfrac{1}{4}} \right)$ in standard form?

Answer 1

Hint: In this question, we want to write the given expression in standard form. The standard form of the linear equation of two variables is $ax + by = c$. Where ‘a’ is the coefficient of x, ‘b’ is the coefficient of y, and ‘c’ is the constant term. Here, a, b, and c are integers, and ‘a’ is non-negative. And x and y are variables. The linear equation has only the first power.

Complete step-by-step solution:
In this question, we want to write the given expression in standard form, and the given expression is:
$ \Rightarrow y - \dfrac{1}{4} = - 3\left( {x + \dfrac{1}{4}} \right)$
Let us remove the bracket on the right-hand side.
$ \Rightarrow y - \dfrac{1}{4} = - 3x - \dfrac{3}{4}$
Let us add 3x on both sides.
$ \Rightarrow 3x + y - \dfrac{1}{4} = - 3x - \dfrac{3}{4} + 3x$
That is equal to,
$ \Rightarrow 3x + y - \dfrac{1}{4} = - \dfrac{3}{4}$
Now, let us add $\dfrac{1}{4}$ on both sides.
$ \Rightarrow 3x + y - \dfrac{1}{4} + \dfrac{1}{4} = - \dfrac{3}{4} + \dfrac{1}{4}$
Simplify the above expression. For that, let us take LCM on the right-hand side.
$ \Rightarrow 3x + y = \dfrac{{ - 3 + 1}}{4}$
That is equal to,
$ \Rightarrow 3x + y = \dfrac{{ - 2}}{4}$
Therefore,
$ \Rightarrow 3x + y = - \dfrac{1}{2}$
Now, multiply both sides by 2.
$ \Rightarrow 2\left( {3x + y} \right) = - \dfrac{1}{2}\left( 2 \right)$
Now, remove the bracket on the left-hand side.
$ \Rightarrow 6x + 2y = - 1$
As we know the standard form of the linear equation is $ax + by = c$.

In this linear equation, the value of ‘a’ is 6, the value of ‘b’ is 2 and the value of ‘c’ is -1.

Note: In the linear equation a, b, and c have no common factors other than 1. The standard form of an equation is useful for finding the x and y-intercepts of a graph. That is the point where the graph crosses the x-axis and the point where it crosses the y-axis. The standard form of an equation is useful for finding the points where two or more functions intersect.