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densk [106]
3 years ago
14

Need help w 3 and 4 pls

Mathematics
1 answer:
zaharov [31]3 years ago
7 0
Since you are solving for x, your objective is to get x by itself

3a).
ax + b \geqslant cd
first subtract b on both sides
ax  \geqslant cd - b
then divide by a
x  \geqslant \frac{cd - b}{a}
3b).
\frac{a(x + 2)}{b}  > c
first multiply by b on both sides
a(x + 2) > bc
then divide by a
x + 2 >  \frac{bc}{a}
subtract by 2
x =  \frac{bc}{a}  - 2
4).
Since a is negative, when dividing the sign of the inequality would change such that > would be < and vice versa. The same applies to greater than or equal to and less than or equal to.

ax+b>d

first subtract b

ax>d-b

then divide by a to get x bu itself, but since a is negative the > sign would change to <

x  < \frac{d - b}{a}
So the answer choice 2 is the solution for question 4

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vampirchik [111]
Question 1:
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