Q25. - Algebra 1

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.

Step-by-Step Solution

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 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:

width="189" style="max-width: none; vertical-align: -38px;" $$\begin{gathered} 1.b\le c\Rightarrow b\pm a\le c\pm a \\ \begin{array}{l}2.b\le c\Rightarrow ab\le ac,\text{ if }a>0\\ 3.b\le c\Rightarrow ab\ge ac,\text{ if }a<0\end{array} \end{gathered}$$

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 x in 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 x as $-6$.

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

Put the value of x as $-10$.

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

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