You need to determine an equation that passes through the point (-3,6) and has a slope of m=-5/2
To determine said equation you have to use the point slope form:
[tex]y-y_1=m(x-x_1)[/tex]Where
m is the slope
(x₁,y₁) are the coordinates of one point of the line.
Replace the formula with the information we have:
[tex]\begin{gathered} y-6=-\frac{5}{2}(x-(-3)) \\ y-6=-\frac{5}{2}(x+3) \end{gathered}[/tex]Apply the distributive probability of multiplications to the rigth side of the equation:
[tex]\begin{gathered} y-6=-\frac{5}{2}\cdot x+(-\frac{5}{2})\cdot3 \\ y-6=-\frac{5}{2}x-\frac{15}{2} \end{gathered}[/tex]Pass the x-related term to the left side and the constant to the rigth side of the equation
[tex]\begin{gathered} y-6+\frac{5}{2}x=-\frac{15}{2} \\ y+\frac{5}{2}x=-\frac{15}{2}+6 \\ y+\frac{5}{2}x=-\frac{3}{2} \end{gathered}[/tex]Multiply both sides of the equation by 2 to cancel the denominator of the fractions:
[tex]\begin{gathered} 2(y+\frac{5}{2}x)=2(-\frac{3}{2}) \\ 2\cdot y+2\cdot\frac{5}{2}x=2\cdot-\frac{3}{2} \\ 2y+5x=-3 \end{gathered}[/tex]Now reorder the terms to determine the standard form:
[tex]5x+2y=-3[/tex]The correct option is B.