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gizmo_the_mogwai [7]
3 years ago
9

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 negat

ive 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
Mathematics
2 answers:
kodGreya [7K]3 years ago
8 0

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)

Divide both side by a , we get,

\frac{ax}{a}

x

or same as x

Thus,  x < the quantity negative 3 times b plus 5 all over negative a

arsen [322]3 years ago
3 0
-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.
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