21 Determine, without the use of a calculator, the value of \( x \) in each of the following: \( 2.1 .1 \quad 2 x^{2}-5 x+3=0 \) \( 212 \quad \frac{x+1}{3}-\frac{x-2}{5}-2=0 \) \( 2.2 x-3 y=3 \) and \( 2 x+y=7 \) Given that: \( \quad 2 x \) Determine the values of \( x \) and \( y \) simultaneously.

Let’s break it down into the three given equations step by step! 1. For the equation \( 2x^2 - 5x + 3 = 0 \), we can apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 2 \), \( b = -5 \), and \( c = 3 \). Plugging in those values gives: \[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1 \] Therefore, substituting back into the quadratic formula, we find: \[ x = \frac{5 \pm 1}{4} \] So, the solutions are \( x = \frac{6}{4} = \frac{3}{2} \) and \( x = \frac{4}{4} = 1 \). 2. For the next equation, \( \frac{x+1}{3} - \frac{x-2}{5} - 2 = 0 \), to eliminate the fractions, we can multiply through by \( 15 \) (the least common multiple of 3 and 5): \[ 15 \cdot \left(\frac{x+1}{3}\right) - 15 \cdot \left(\frac{x-2}{5}\right) - 30 = 0 \] This simplifies to \( 5(x + 1) - 3(x - 2) - 30 = 0 \), which expands to: \[ 5x + 5 - 3x + 6 - 30 = 0 \implies 2x - 19 = 0 \] Solving gives \( x = \frac{19}{2} \). 3. For the simultaneous equations \( 2x - 3y = 3 \) and \( 2x + y = 7 \), we can solve for \( y \) from the second equation: \[ y = 7 - 2x \] Now substitute this expression for \( y \) into the first equation: \[ 2x - 3(7 - 2x) = 3 \] This simplifies to: \[ 2x - 21 + 6x = 3 \implies 8x - 21 = 3 \implies 8x = 24 \implies x = 3 \] Now substitute \( x = 3 \) back into \( y = 7 - 2x \): \[ y = 7 - 2(3) = 7 - 6 = 1 \] Thus, \( x = 3 \) and \( y = 1 \) for the last pair of simultaneous equations! In summary: - From \( 2x^2 - 5x + 3 = 0 \), we get: \( x = \frac{3}{2}, 1 \). - From \( \frac{x+1}{3} - \frac{x-2}{5} - 2 = 0 \), we get: \( x = \frac{19}{2} \). - For \( 2x - 3y = 3 \) and \( 2x + y = 7 \), we find: \( x = 3, y = 1 \).