Villaretq0

Answered

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"?

Answer & Explanation

tennispopj8

Expert

2022-06-29Added 20 answers

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:

$\frac{{4}}{{3}}\times \frac{3}{4}n\ge \frac{{4}}{{3}}\times 51$

$\frac{\overline{){4}}}{\overline{){3}}}\times \frac{{\overline{){3}}}}{{\overline{){4}}}}n\ge {4}\times 17$

$n\ge 68$

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:

$\frac{{4}}{{3}}\times \frac{3}{4}n\ge \frac{{4}}{{3}}\times 51$

$\frac{\overline{){4}}}{\overline{){3}}}\times \frac{{\overline{){3}}}}{{\overline{){4}}}}n\ge {4}\times 17$

$n\ge 68$

Taniyah Estrada

Expert

2022-06-30Added 5 answers

Step 1

Let's use some punctuation so show exactly what is meant: note how the placement of a comma gives a different meaning.

$\text{three fourths of a number}},{\phantom{\rule{1ex}{0ex}}\text{decreased by 9}}\textcolor[rgb]{}{\phantom{\rule{1ex}{0ex}}\text{is at least 42}$

Let the number be x

$\frac{3}{4}x}{-9}\textcolor[rgb]{}{\ge 42$

Solving gives:

$\frac{3}{4}x\ge 51$

$\frac{4}{3}\times \frac{3}{4}x\ge 51\times \frac{4}{3}$

$x\ge 56$

However, if we interpret it differently we could have:

$\text{three fourths of,}}{\phantom{\rule{1ex}{0ex}}\text{a number decreased by 9,}}\textcolor[rgb]{}{\phantom{\rule{1ex}{0ex}}\text{is at least 42}$

$\frac{3}{4}}{(x-9)}\textcolor[rgb]{}{\ge 42$

Solving gives:

$\frac{4}{3}\times \frac{3}{4}(x-9)\ge 42\times \frac{4}{3}$

$x-9\ge 56$

$x\ge 65$

Let's use some punctuation so show exactly what is meant: note how the placement of a comma gives a different meaning.

$\text{three fourths of a number}},{\phantom{\rule{1ex}{0ex}}\text{decreased by 9}}\textcolor[rgb]{}{\phantom{\rule{1ex}{0ex}}\text{is at least 42}$

Let the number be x

$\frac{3}{4}x}{-9}\textcolor[rgb]{}{\ge 42$

Solving gives:

$\frac{3}{4}x\ge 51$

$\frac{4}{3}\times \frac{3}{4}x\ge 51\times \frac{4}{3}$

$x\ge 56$

However, if we interpret it differently we could have:

$\text{three fourths of,}}{\phantom{\rule{1ex}{0ex}}\text{a number decreased by 9,}}\textcolor[rgb]{}{\phantom{\rule{1ex}{0ex}}\text{is at least 42}$

$\frac{3}{4}}{(x-9)}\textcolor[rgb]{}{\ge 42$

Solving gives:

$\frac{4}{3}\times \frac{3}{4}(x-9)\ge 42\times \frac{4}{3}$

$x-9\ge 56$

$x\ge 65$

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