Question

Define a variable, write an inequality, and solve each problem. Check your solution.

Three times a number increased by 8 is no more than the number decreased by 4.

Short answer

The variable is x is and the inequality equation is $3x+8\le x-4$. The solution for the inequality is $x\le -6$.

Step-by-step solution

Step 1. State the concept for inequality.

To graph the endpoint with strict inequality < or >, use the open parenthesis or a hollow circle at that endpoint.

To graph the endpoint with inequality symbol $\le \text{or}\ge$, use the bracket or a solid circle at that endpoint.

Properties of inequality:

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Step 2. Write the inequality for the given condition.

Let the number be x.

Three times a number increased by 8 is no more than the number decreased by 4:

$3x+8\le x-4$

Step 3. Solve for x.

Solve for xin the inequation $3x+8\le x-4$

$\begin{array}{c}3x+8\le x-4\\ 3x+8-8\le x-4-8\\ 3x\le x-12\\ 3x-x\le x-12-x\\ 2x\le -12\\ \frac{2x}{2}\le \frac{-12}{2}\\ x\le -6\end{array}$

Step 4. Check the solution.

Put the value of xas $-6$.

$$\begin{gathered} \begin{array}{c}3x+8=x-4\\ 3\left(-6\right)+8=-6-4\end{array} \\ -18+8=-10 \\ -10=-10 \\ \text{LeftHandSide}=\text{RightHandSide} \end{gathered}$$

Put the value of xas $-10$.

$$\begin{gathered} 3x+8\le x-4 \\ 3\left(-10\right)+8\le -10-4 \\ -30+8\le -14 \\ -22\le -14\left[\text{TrueStatement}\right] \end{gathered}$$

Thus, the solution of the inequality equation $3x+8\le x-4$is $x\le -6$.