Question

1. représente la situation suivante à l’aide d’un système linéaire : le périmètre d’un rectangle mesure 234 pi. si la longueur double, le périmètre augmente de 58 pi.

Explanation:

Step1: Define variables

Let \( l \) be the length and \( w \) be the width of the rectangle. The perimeter formula for a rectangle is \( P = 2(l + w) \). Initially, \( 2(l + w)=234 \), so \( l + w=\frac{234}{2}=117 \).

Step2: Analyze the change

When the length is doubled, the new length is \( 2l \). The new perimeter \( P'=2(2l + w) \). We know that \( P' - P = 58 \). Substitute \( P = 234 \), so \( P'=234 + 58 = 292 \).

Step3: Set up the new equation

\( 2(2l + w)=292 \), divide both sides by 2: \( 2l + w = 146 \).

Step4: Solve the system of equations

We have the system:

1. \( l + w = 117 \)
2. \( 2l + w = 146 \)

Subtract the first equation from the second: \( (2l + w)-(l + w)=146 - 117 \), which simplifies to \( l = 29 \).

Then substitute \( l = 29 \) into \( l + w = 117 \), we get \( 29+w = 117 \), so \( w = 117 - 29 = 88 \). (If we need to find length and width, but the problem might be to set up the system or find dimensions. Assuming we need to find length and width.)

Answer:

Length \( l = 29 \) pi, Width \( w = 88 \) pi (or depending on the exact question, if it's to set up the system, the equations are \( 2(l + w)=234 \) and \( 2(2l + w)=234 + 58 \))