Question

33 times a number xxx, subtracted from 181818, is less than -90.?90.Minus, 90, point Write an inequality for the statement above. Find the solution set of the inequality. Write the solution using a fraction or integer.

Answer 1

Answer:

Let the number be x.

Given: 33 times a number x, subtracted from 18, is less than -90.

we can write this statement in inequality form, i.e,

$18-33x<-90$

Now, to find the solution set for this inequality:-

$18-33x<-90$

Subtraction poperty of equality states that you subtract the same number from both sides of an equation.

Subtract  18 from both sides,

$18-33x-18<-90-18$

Simplify:

$-33x<-108$

Multiply both sides by -1 (reverse the inequality)

$(-1)(-33x)>(-108)(-1)$ or

$33x>108$

Divide both sides by 33, we get

$\frac{33x}{33}= \frac{108}{33}$

Simplify:

$x>\frac{36}{11}$

Therefore, the solution set for this inequality is,  $(\frac{36}{11} ,\infty )$ [

The solution using a fraction or integer is, tex]x>\frac{36}{11}[/tex] 0r $x>3\frac{3}{11}$