Respuesta :
10 subtracted from the quotient of a number and -3 is at least -1 is:
[tex]\left(\frac{x}{-3}\right)-10\ge \:-1\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-27\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-27]\end{bmatrix}[/tex]
Solution:
Given statement is:
10 subtracted from the quotient of a number and -3 is at least -1
Quotient means reuslt of dividing two numbers
Let the number be "x"
quotient of a number and -3 = [tex]\frac{x}{-3}[/tex]
Therefore,
10 subtracted from [tex]\frac{x}{-3}[/tex] is at least -1
at least means "less than or equal to"
Which is translated into algebraic expression as:
[tex](\frac{x}{-3}) - 10\geq -1[/tex]
Solve the above inequality
[tex]\left(\frac{x}{-3}\right)-10\ge \:-1\\\\\mathrm{Add\:}10\mathrm{\:to\:both\:sides}\\\\\frac{x}{-3}-10+10\ge \:-1+10\\\\\mathrm{Simplify}\\\\\frac{x}{-3}\ge \:9\\\\Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)[/tex]
When we multiply or divide by negative number. we have to flip the inequality sign
[tex]\frac{x\left(-1\right)}{-3}\le \:9\left(-1\right)\\\\\mathrm{Simplify}\\\\\frac{x}{3}\le \:-9\\\\\mathrm{Multiply\:both\:sides\:by\:}3\\\\\frac{3x}{3}\le \:3\left(-9\right)\\\\\mathrm{Simplify}\\\\x\le \:-27[/tex]
Thus the solution of inequality is found
