Question

solve for x. −ax 3b > 5 x > the quantity 3 times b minus 5 all over a x > the quantity 5 minus 3 times b all over negative a x < the quantity 3 times b plus 5 all over a x < the quantity negative 3 times b plus 5 all over negative a

Answer 1

-ax + 3b > 5

Subtract 3b: -ax > 5 -3b

Diuvide by - a, which implies to change the sign > to <:

x < [5 -3b]/ (-a) or x < [3b - 5]/a ..... both are equivalent

That is second option.

Answer 2

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