Q8. - Algebra 1

Question

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

Twice a number increased by 3 is less than the number decreased by 4.

Step-by-Step Solution

Answer

The variable is x is and the inequality equation is $2x+3<x-4$. The solution for the inequality is $x<-7$.

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:

$$\begin{gathered} 1.b\le c\Rightarrow b\pm a\le c\pm a \\ 2.b\le c\Rightarrow ab\le ac,\text{ if }a>0 \\ 3.b\le c\Rightarrow ab\ge ac,\text{ if }a<0 \end{gathered}$$ 

Step 2. Write the inequality for the given condition.

Let the number be x.

Twice a number increased by 3 is less than the number decreased by 4:

$2x+3<x-4$  

Step 3. Solve for x.

Solve for x  in the inequation $2x+3<x-4$.

$$\begin{gathered} 2x+3<x-4 \\ 2x+3-3<x-4-3 \\ 2x<x-7 \\ 2x-x<x-7-x \\ x<-7 \end{gathered}$$ 

Step 4. Check the solution when x = - 10 .

Put the value of x as $-10$.

$$\begin{gathered} 2x+3<x-4 \\ 2\left(-10\right)+3<-10-4 \\ -20+3<-14 \\ -17<-14 \left[\text{True Statement}\right] \end{gathered}$$ 

Thus, the solution of the inequality equation $2x+3<x-4$ is $x<-7$.