$\left(\frac{x}{-3}\right)-10\ge \:-1\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-27\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-27]\end{bmatrix}$

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 = $\frac{x}{-3}$

Therefore,

10 subtracted from $\frac{x}{-3}$ is at least -1

at least means "less than or equal to"

Which is translated into algebraic expression as:

$(\frac{x}{-3}) - 10\geq -1$

Solve the above inequality

$\left(\frac{x}{-3}\right)-10\ge \:-1\\\\\mathrm{Add\:}10\mathrm{\:to\:both\:sides}\\\\\frac{x}{-3}-10+10\ge \:-1+10\\\\\mathrm{Simplify}\\\\\frac{x}{-3}\ge \:9\\\\Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)$

When we multiply or divide by negative number. we have to flip the inequality sign

$\frac{x\left(-1\right)}{-3}\le \:9\left(-1\right)\\\\\mathrm{Simplify}\\\\\frac{x}{3}\le \:-9\\\\\mathrm{Multiply\:both\:sides\:by\:}3\\\\\frac{3x}{3}\le \:3\left(-9\right)\\\\\mathrm{Simplify}\\\\x\le \:-27$

Thus the solution of inequality is found

4 0

4 years ago