How do you write an inequality and solve given "three fourths of a number decreased...

Step 1
First, let's call "a number" ${\displaystyle n}$
"three fourths of a number" then can be written as:
${\displaystyle \frac{3}{4}n}$
This, "decreased by nine" then can be written as:
${\displaystyle \frac{3}{4}n-9}$
"is at least" is the same as ${\displaystyle \ge }$ so we can now write:
${\displaystyle \frac{3}{4}n-9\ge }$
and what it is "at least" is "forty two" so:
${\displaystyle \frac{3}{4}n-9\ge 42}$
To solve this we first add ${\displaystyle 9}$ to each side of the inequality to isolate the n term while keeping the inequality balanced:
${\displaystyle \frac{3}{4}n-9+9\ge 42+9}$
${\displaystyle \frac{3}{4}n-0\ge 51}$
${\displaystyle \frac{3}{4}n\ge 51}$
Now, we multiply each side of the inequality by ${\displaystyle \frac{4}{3}}$ tosolve for n while keeping the inequality balanced:
${\displaystyle \frac{4}{3}\times \frac{3}{4}n\ge \frac{4}{3}\times 51}$
${\displaystyle \frac{\cancel{4}}{\cancel{3}}\times \frac{\cancel{3}}{\cancel{4}}n\ge 4\times 17}$
${\displaystyle n\ge 68}$

Step 1
Let's use some punctuation so show exactly what is meant: note how the placement of a comma gives a different meaning.
${\displaystyle \text{three fourths of a number},\text{decreased by 9}\text{is at least 42}}$
Let the number be x
${\displaystyle \frac{3}{4}x-9\ge 42}$
Solving gives:
${\displaystyle \frac{3}{4}x\ge 51}$
${\displaystyle \frac{4}{3}\times \frac{3}{4}x\ge 51\times \frac{4}{3}}$
${\displaystyle x\ge 56}$
However, if we interpret it differently we could have:
${\displaystyle \text{three fourths of,}\text{a number decreased by 9,}\text{is at least 42}}$
${\displaystyle \frac{3}{4}(x-9)\ge 42}$
Solving gives:
${\displaystyle \frac{4}{3}\times \frac{3}{4}(x-9)\ge 42\times \frac{4}{3}}$
${\displaystyle x-9\ge 56}$
${\displaystyle x\ge 65}$