Alliic select the equations that are equivalent to the equation y+5 = -3(x-1) select all that apply y = -3x - 2 y = -4x -5 3x + y = -2 -4x + y = - 5

First, we need to solve the first equation in order to see what the options need to be equivalent to. our first step to solving this is to distribute -3 through the parenthesis 
y + 5 = -3x + 3
next, move the constant to the right side and change its sign
y = -3x + 3 - 5
calculate the difference
y = -3x - 2
since we cant find x, your final solution for y is y = -3x - 2, x ∈ R
now that we know the solution to that equation, we can now solve the options listed to see which ones are equivalent.
first is y = -3x - 2, which is very simple to solve because y has already been found. since we know y and there is no solution for x, our answer to this equation is y = -3x -2, x ∈ R. since this is the exact same answer as our first equation, we know that the A is equivalent to y + 5 = -3(x - 1)
the next equation is also already solved for y and has no x, so the answer to this equation is y = -4x - 5, x ∈ R. since this is does not have the same answer that y + 5 = -3(x - 1) had, option B is not equivalent.
our next equation is 3x + y = -2 does not provide y for us, so we must solve it.
first, you need to move the variable to the right side and change its sign
3x = -2 - y
now divide both sides of the equation by 3
x = - 2/3 = y/3
write all the numerators above their common denominators
x = - 2 + y/3 
since we dont know what y is, our final solution is
x = - , y ∈ R
since this answer does not match the answer we are looking for, option C is not equivalent.  
finally, we must solve -4x + y = -5
the first step to solving this is to move the variable to the right side and change its sign
-4x = -5 - y
divide both sides of the equation by -4
x = 5/4 + y/4
now, write all the numerators above their common denominators 
x = 5 + y/4
since we dont know what y is, our answer is x =  , y ∈ R
since this is also not equivalent to y + 5 = -3(x - 1), our only answer is option A.
let me know if you have any further questions 
:)