Answer:
Option (d) is correct.
x < the quantity negative 3 times b plus 5 all over negative a
Step-by-step explanation:
Given : -ax + 3b > 5
We have to solve for x and choose for the correct option from the given options.
Consider -ax + 3b > 5
Subtract 3b both side, we have,
-ax + 3b - 3b > 5 - 3b
Simplify , we have,
-ax > 5 - 3b
Divide both side by -1,
\(\left(-ax\right)\left(-1\right)<5\left(-1\right)-3b\left(-1\right)\)
Divide both side by a , we get,
\(\frac{ax}{a}<\frac{3b-5}{a}\)
\(x<\frac{3b-5}{a}\)
or same as \(x<\frac{-3b+5}{-a}\)
Thus, x < the quantity negative 3 times b plus 5 all over negative a
