SOLUTION:
Step 1:
In this question, we are given the following:
Write an equation of a line that is parallel to the line whose equation is
[tex]\text{3 y = x + 6}[/tex]
and that passes through the point (-3,4)
Step 2:
From the question, we can see that the given equation is given as:
[tex]\begin{gathered} 3\text{ y = x + 6} \\ \text{Divide both sides by 3, we have that:} \\ y\text{ = }\frac{1}{3}x\text{ + 2} \end{gathered}[/tex]
Comparing this, with the equation of a line, we have that:
[tex]\begin{gathered} y\text{ = mx + c} \\ \text{Then, the gradient of line, m = }\frac{1}{3} \end{gathered}[/tex]
Step 3:
Now, using the equation of a line:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{where (x }_1,y_1)\text{ = ( -3 , 4 )} \\ m\text{ = }\frac{1}{3} \end{gathered}[/tex][tex]\begin{gathered} y\text{ - }4\text{ = }\frac{1}{3}(\text{ x -- 3)} \\ y\text{ - 4 =}\frac{1}{3}(\text{ x+ 3)} \end{gathered}[/tex]
Multiply through by 3, we have that:
[tex]\begin{gathered} \text{3 ( y - 4 ) = ( x + 3)} \\ 3y\text{ - 12 = x + 3} \\ \text{Hence, we have that:} \\ 3y\text{ = x + 3 +1 2} \\ 3\text{ y = x + 15} \end{gathered}[/tex]
CONCLUSION:
The equation of the line that is parallel to the given line is:
[tex]3y\text{ = x + 15}[/tex]
