Ejercicio 180 Algebra De Baldor Official

[ \frac2x-1 + \frac3x+2 = 1 ] Denominators: (x-1), (x+2) Step 2 – LCM [ \textLCM = (x-1)(x+2) ] Step 3 – Multiply both sides by LCM [ (x-1)(x+2) \cdot \frac2x-1 + (x-1)(x+2) \cdot \frac3x+2 = 1 \cdot (x-1)(x+2) ] Step 4 – Simplify [ 2(x+2) + 3(x-1) = (x-1)(x+2) ] [ 2x + 4 + 3x - 3 = x^2 + 2x - x - 2 ] [ 5x + 1 = x^2 + x - 2 ] Step 5 – Rearrange [ 0 = x^2 + x - 2 - 5x - 1 ] [ x^2 - 4x - 3 = 0 ]

Ejercicio 180 indeed has denominators with (x). For example: ejercicio 180 algebra de baldor

Wait – that has no (x) in denominator – that’s from another exercise. Let me recall: [ \frac2x-1 + \frac3x+2 = 1 ] Denominators: