Mega Linear Equations Solver

Take 3(x − 2) + 5 = 2x + 1. Expand the left: 3x − 6 + 5, which collects to 3x − 1. The right is already 2x + 1. So the tidied equation is 3x − 1 = 2x + 1. Now move everything to one side by subtracting the right from the left: (3x − 2x) + (−1 − 1) = x − 2, so x − 2 = 0. The coefficient of x is 1, which is not zero, so there is exactly one solution: x = 2. Check it in the original rather than the tidied version, which is the point of checking: 3(2 − 2) + 5 = 3(0) + 5 = 5, and 2(2) + 1 = 5. Both sides are 5. Now a case with fractions: (x + 1)/3 − (2x − 1)/4 = 5. Everything is over 12, so the left becomes (4(x + 1) − 3(2x − 1))/12 = (4x + 4 − 6x + 3)/12 = (−2x + 7)/12. Setting that equal to 5 and multiplying up gives −2x + 7 = 60, so −2x = 53 and x = −53/2. Note the −3 multiplying the whole of (2x − 1), including the −1 that becomes +3; missing that second sign is the commonest slip in the whole topic. Now the two cases that look like failures. 2x + 1 = 2x + 3. Subtracting the right from the left gives 0x − 2, so the x term is gone and −2 = 0 remains. That is false for every x, so there is no solution. On a graph the two sides are parallel lines a distance apart, and parallel lines do not meet. 2(x + 1) = 2x + 2. Expanding the left gives 2x + 2, which is the right exactly. Subtracting leaves 0 = 0, true for every x. Every number is a solution, and on a graph the two sides are the same line drawn twice. Finally, what this will not do. Type x(x + 1) = 0 and it stops and says the expression is quadratic. That is deliberate: a solver that expanded and dropped the x² would report x = 0, which is a real root but not the only one — x = −1 is lost without a word. Refusing is the honest answer where a number would not be.