Q45. - Algebra 2

Question

Define a variable and write an inequality for each problem and then solve.

45. Twice the sum of a number and 5 is no more than 3 times that same number increased by 11.

Step-by-Step Solution

Answer

The inequality is: $2\left(\text{x}+5\right)\le 3\text{x}+11$

Value of the number is $\text{x}\ge -\mathbf{1}$

Step 1 - Define the variable.

A variable is a term that is used to represent an unknown value.

Step 2 - Write the inequality.

From given, let the number added to 5 be $x$. 

As twice the sum is no more than 3 times $x$ increased by 11.

So, the inequality is:

$2\left(x+5\right)\le 3x+11$

Step 3 - Solve the inequality.

To simplify this inequality, use subtraction property 

$\begin{array}{c}2\left(x+5\right)\le 3x+11\\ 2x+10\le 3x+11\\ 10-11\le 3x-2x\\ x\ge -1\end{array}$

Therefore, the inequality is $2\left(x+5\right)\le 3x+11$ and the value of $x$ is $x\ge -1$.