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2.2 Solve simultaneously for \( a \) and \( b \) : \( 2 a-b=7 \) and \( a^{2}+a b+b^{2}=7 \) 2.3 The area of a room is \( 20 \mathrm{~m}^{2} \). If the length is increased by 3 m and the width is increased by 1 m , the room will double in area. Determine the original dimensions of the room.

Ask by Barker Vaughan. in South Africa
Mar 14,2025

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**Problem 2.2 Solutions:** - \( a = 2 \), \( b = -3 \) - \( a = 3 \), \( b = -1 \) **Problem 2.3 Solutions:** - Original dimensions: \( 12 \, \text{m} \times \frac{5}{3} \, \text{m} \) or \( 5 \, \text{m} \times 4 \, \text{m} \)

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To solve for \( a \) and \( b \) in the equations \( 2a - b = 7 \) and \( a^2 + ab + b^2 = 7 \), we can first express \( b \) in terms of \( a \) using the first equation: \( b = 2a - 7 \). Substituting this expression into the second equation results in a quadratic in terms of \( a \). Upon solving the quadratic equation, we find suitable values for \( a \) and subsequently determine \( b \). For the area problem, let’s denote the original dimensions as length \( l \) and width \( w \). We know that \( lw = 20 \). When we increase the length by 3 m and the width by 1 m, the new area becomes \( (l + 3)(w + 1) = 40 \). By plugging in the original area equation into this new equation, we can derive a system of equations to find the values of \( l \) and \( w \). Solving these equations gives us the room's original dimensions!

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