[tex]\boxed{y=3x+12}[/tex]
In this problem we know the equation of a line, which is:
[tex]3x-y=7[/tex]
We also can write this equation as:
[tex]y=3x-7[/tex]
This line has a slope [tex]m=3[/tex] which is also the slope of the line we are looking for because they're parallel. We also have a point [tex](-5,-3)[/tex]. Therefore, we can write this equation as follows:
[tex]y-y_{0}=m(x-x_{0}) \\ \\ y-(-3)=3(x-(-5)) \\ \\ y+3=3(x+5) \\ \\ y+3=3x+15 \\ \\ \boxed{y=3x+12}[/tex]
From the figures below, the line in red is [tex]3x-y=7[/tex] while the line in blue is [tex]y=3x+12[/tex] and this line passes through the point (-5, -3)!
Answer:
The equation of the line is y =3x+12
Step-by-step explanation:
The first step is to rewrite the equation of the line in the slope-intercept form in order to identify the slope of the line. The re-written equation is; y = 3x-7. This implies that the slope of the line is 3. Since the two lines will be parallel they will be having equal slopes of 3. The slope-intercept form of the equation of this line will be; y =3x+c. Since the line passes through (-5, -3), substitute x with -5 and y with -3 to solve for c; -3 =3(-5)+c. Then c =12 and the equation of the line becomes; y =3x+12
