Question

How do you write $y = \dfrac{7}{2}x + \dfrac{1}{4}$ in standard form?

Answer 1

Hint: The equation of a straight line in slope-intercept form is: $y = mx + b$. Where m is the value of slope and b is the y-intercept. Here, m and b are constants, and x and y are variables. Since x and y are variables that describe the position of specific points on the graph, m and b describe features of the function. In this question, a linear equation is given. We will convert this equation into the standard form of a straight-line equation. For that, we have to take LCM. The LCM is the smallest positive number that all of the numbers divide into evenly.
List the prime factors of each number.
Multiply each factor the greatest number of times it occurs in either number.

Complete step by step solution:
In this question, the straight line is
$ \Rightarrow y = \dfrac{7}{2}x + \dfrac{1}{4}$
To convert the above equation in the standard form, we will first take LCM of the right-hand side.
The LCM is the smallest positive number that all of the numbers divide into evenly.
The first step to find LCM is, list the prime factors of each number.
Factors of 2 are 1, and 2. Factors of 4 are 1, 2, and 2.
The second step to find LCM is to multiply each factor by the greatest number of times it occurs in either number.
Here, the greater number is 4.
Hence, the LCM will be 4.
Therefore,
$ \Rightarrow y = \dfrac{{14x + 1}}{4}$
Let us multiply 4 on both sides to remove the denominator from the right-hand side.
$ \Rightarrow y \times 4 = \dfrac{{14x + 1}}{4} \times 4$
That is equal to,
$ \Rightarrow 4y = 14x + 1$
Let us subtract 14x on both sides.
 $ \Rightarrow - 14x + 4y = 14x + 1 - 14x$
That is equal to,
$ \Rightarrow - 14x + 4y = 1$

Hence, the standard form of the given equation is $ - 14x + 4y = 1$.

Note:
A straight line is a linear equation of the first order. If we put the value of x is equal to 0 to find the y-intercept and put the value of y is equal to 0 to find the x-intercept. So, the point on the x-axis is in the form of $\left( {x,0} \right)$, and the point on the y-axis is in the form of $\left( {0,y} \right)$.