To answer this question, we need to know that the x- and y-intercepts are:
• The x-intercept: (5, 0). The point where the line passes through the x-axis.
,
• The y-intercept: (0, 3). The point where the line passes through the y-axis.
To find the equation of the line, we can use the two-point form equation of the line, and then we will find the standard form of the line, which is of the form:
![Ax+By=C](https://img.qammunity.org/2023/formulas/mathematics/high-school/75j0pzqy8f6gtpgzjampxw030qc85p0hp7.png)
We still need to label both points:
• (5, 0) ---> x1 = 5, y1 = 0.
,
• (0, 3) ---> x2 = 0, y2 = 3.
The two-point form of the line is given by:
![y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/college/hkbzvop4iz62zgm93u190774353c4ig6id.png)
Then, substituting these values into this equation, we have:
![y-0=(3-0)/(0-5)(x-5)\Rightarrow y=(3)/(-5)(x-5)\Rightarrow y=-(3)/(5)(x-5)](https://img.qammunity.org/2023/formulas/mathematics/college/8rmd4dnybqxtyzvy9mt8uq72rh3c23dhaf.png)
Then, we have:
![y=-(3)/(5)x+(3)/(5)\cdot5\Rightarrow y=-(3)/(5)x+3](https://img.qammunity.org/2023/formulas/mathematics/college/8re76d8j3bgqax05csn4rm81dznky17kzs.png)
This is the slope-intercept form of the line. To find the standard form of the line, we can multiply the equation by 5 as follows:
![5(y=-(3)/(5)x+3)\Rightarrow5y=5(-(3)/(5))x+5\cdot3\Rightarrow5y=-3x+15](https://img.qammunity.org/2023/formulas/mathematics/college/j63jgn28im9256pxzt03rm2vf9v5agbgr3.png)
Adding 3x to both sides of the equation, we have:
![5y+3x=-3x+3x+15\Rightarrow5y+3x=15](https://img.qammunity.org/2023/formulas/mathematics/college/dkw6st5i5fv26oi7ezn0i8m7z9x4rdo922.png)
Finally, the standard form of the line is given by 3x + 5y = 15.