## anonymous one year ago 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

1. misty1212

hi!!

2. misty1212

i will provide the answer for you so long as you tell me where this question comes from (my guess FLVS)

3. gibbs

x<−5−3b/a

4. misty1212

because in fact you cannot solve an absolute value inequality with variable coefficients

5. misty1212

@gibbs is straight up wrong

6. anonymous

Its from FLVS @misty1212

7. misty1212

ok lets solve it the way they want you to, even though it is wrong

8. anonymous

ok

9. misty1212

$−ax + 3b > 5$ subtract $$3b$$ get $-ax>5-3b$ then divide by $$-a$$

10. misty1212

now here is the mistake they want you to change the inequality, because they want you to think that $$-a$$ is negative that is sheer nonsense you do not know if $$-a$$ is negative or positive

11. misty1212

in any case they want you to write $x<\frac{5-3b}{-a}$ or $x<\frac{3b-5}{a}$

12. misty1212

it is totally incorrect, you should write them and tell them that anyone who thinks $$-a$$ is negative (because it has a minus sign) does not understand what a variable is and has no business trying to teach math or write math questions

13. anonymous

My answers ar this though... 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 @misty1212