Villaretq0

2022-06-28

How do you write an inequality and solve given "three fourths of a number decreased by nine is at least forty two"?

tennispopj8

Expert

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

Expert

Step 1
Let's use some punctuation so show exactly what is meant: note how the placement of a comma gives a different meaning.
$\text{three fourths of a number},\phantom{\rule{1ex}{0ex}}\text{decreased by 9}\phantom{\rule{1ex}{0ex}}\text{is at least 42}$
Let the number be x
$\frac{3}{4}x-9\ge 42$
Solving gives:
$\frac{3}{4}x\ge 51$
$\frac{4}{3}×\frac{3}{4}x\ge 51×\frac{4}{3}$
$x\ge 56$
However, if we interpret it differently we could have:
$\text{three fourths of,}\phantom{\rule{1ex}{0ex}}\text{a number decreased by 9,}\phantom{\rule{1ex}{0ex}}\text{is at least 42}$
$\frac{3}{4}\left(x-9\right)\ge 42$
Solving gives:
$\frac{4}{3}×\frac{3}{4}\left(x-9\right)\ge 42×\frac{4}{3}$
$x-9\ge 56$
$x\ge 65$

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