Question

How do you write $5x=10y+15$ in standard form?

Answer 1

Hint: In this question we have a linear equation which we have to convert into the standard form. To convert the given equation in the standard form we will use convert the equation in the form of the equation of a straight line which is $ax+by+c=0$. We will then plot the line on the graph to get the required solution.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow 5x=10y+15$
Since this is a linear equation, we will convert this into the form of the equation of a straight line which is $ax+by+c=0$.
On transferring the terms $10y$ and $15$ from the right-hand side to the left-hand side, we get:
$\Rightarrow 5x-10y-15=0$
Now we can see that the term $5$ is common in all the terms, we will take it out as common as:
$\Rightarrow 5\left( x-2y-3 \right)=0$
On dividing both the sides of the equation with $5$, we get:
$\Rightarrow x-2y-3=0$
The above equation is in the form of an equation of a straight line. On drawing the line on the graph, we get:

Which is the required solution.

Note: We have used the equation of a straight line to solve this question but we can also use the slope-intercept form of a line which is $y=mx+c$ where $m$ is the slope of the line and $c$ is the $y$-intercept.
We have the expression as $5x=10y+15$.
On transferring the term $15$ from the right-hand side to the left-hand side, we get:
$\Rightarrow 5x-15=10y$
On taking $5$ common from both the sides of the equation, we get:
$\Rightarrow 5\left( x-3 \right)=5\left( 2y \right)$
On dividing equation with $5$, we get:
$\Rightarrow x-3=2y$
On transferring the term $2$ from the right-hand side to the left-hand side and rearranging the equation, we get:
$\Rightarrow y=\dfrac{x-3}{2}$
On splitting the denominator, we get:
$\Rightarrow y=\dfrac{x}{2}-\dfrac{3}{2}$
The above equation is in the form of the slope-intercept form of a line. On drawing it on the graph we get:

Which is the same line as the previous line therefore the solution is correct.