Question

How do you graph $5x+y=-5$ using intercepts?

Answer 1

We are asked to draw the graph of the equation $5x+y=-5$. The degree of an equation is the highest power of the variable present in it. So, for this equation, the highest power present $x$ is 1, the degree is also 1. From this, it can be said that this is a linear equation. The graph of a linear equation represents a straight line.

Complete step by step answer:
The general equation of a straight line is $ax+by+c=0$, where $a,b,c$ are any real numbers. The given equation is $5x+y=-5$, the equation can also be written as $5x+y+5=0$, comparing with the general equation of straight line, we get $a=5,b=1\And c=5$.
To plot the graph of an equation of the straight line, we should know at least two points, through which the line passes.
To make things simple, let’s take the X-intercept and Y-intercept as the two points. X-intercept is the point where the line crosses X-axis, this means that the Y-coordinate will be $0$, similarly Y-intercept is the point where the line crosses Y-axis, so X-coordinate will be $0$. We will use this property now.
We substitute $y=0$ in the equation $5x+y+5=0$, we get

& \Rightarrow 5x+0+5=0 \\\ & \Rightarrow 5x+5=0 \\\ \end{aligned}$$ Subtracting 5 from both sides of the equation we get, $$\begin{aligned} & \Rightarrow 5x+5-5=0-5 \\\ & \Rightarrow 5x=-5 \\\ \end{aligned}$$ We divide 5 to both sides we get, $$\begin{aligned} & \Rightarrow \dfrac{5x}{5}=\dfrac{-5}{5} \\\ & \therefore x=-1 \\\ \end{aligned}$$ So, the coordinates of the X-intercept are $$\left( -1,0 \right)$$. Similarly, now we substitute $$x=0$$ in the equation, we get $$\begin{aligned} & \Rightarrow 5(0)+y+5=0 \\\ & \Rightarrow y+5=0 \\\ \end{aligned}$$ Subtracting 5 from both sides of the equation, we get $$\begin{aligned} & \Rightarrow y+5-5=0-5 \\\ & \therefore y=-5 \\\ \end{aligned}$$ So, the coordinates of the Y-intercept are $$\left( 0,-5 \right)$$. Using these two points we can plot the graph of the equation as follows:  **Note:** Here, we found the two points which are X-intercept and Y-intercept by substituting either x or $$y$$ to be zero, one at a time. We can also find these values by converting the straight-line equation to the equation in intercept form which is, $$\dfrac{x}{a}+\dfrac{y}{b}=1$$. Here, $$a\And b$$are X-intercept and Y-intercept respectively.