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Analyzing One-Variable Inequalities in Context

Analyzing One-Variable Inequalities in Context 1.5 - Solution

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The perimeter of any triangle is the sum of the lengths of its three sides. Since the given triangle is isosceles, two of its sides have the same length. Let's call the length of each congruent side xx and the third unknown side y.y. Now, we can write the expression representing the perimeter of the given triangle. 2x+y\begin{gathered} 2x+y \end{gathered} We are told that the perimeter is at most 1212 inches which means that the perimeter is less than or equal to 12.12. 2x+y12\begin{gathered} 2x + y \leq 12 \end{gathered} It is given that x=5.x={\color{#0000FF}{5}}. Let's substitute this value into the inequality and solve for y.y.
2x+y122x + y \leq 12
25+y122\cdot{\color{#0000FF}{5}} + y \leq 12
10+y1210 + y \leq 12
y2y \leq 2
This inequality tells us that the length of the remaining side must be less than or equal to 22 cm. Note that lengths are always positive, so the more precise way to express the possible lengths is by writing the following inequality. 0<y2\begin{gathered} 0< y\leq 2 \end{gathered} The length of the remaining side must be greater than 00 cm and less than or equal to 22 cm.