Answer:
If you cancel out all of the x terms via addition or subtraction, and you get something along the lines of 1 = 2, then you have no solution.
11x + 4 = 11x + 7 Subtract 11x from both sides
4 = 7 No value for x will satisfy this equation.
If you cancel out all the x terms via addition or subtraction and you get something along the lines of 1 = 1, then you have infinite solutions.
2(x + 1) = 2x + 2 Expand the left side using the distributive property
2x + 2 = 2x + 2 Subtract 2x from both sides
2 = 2 Every value for x will satisfy this equation
If you can’t cancel out all the x terms with addition or subtraction, you probably have 1 solution.
5x + 2 = 3x + 100 Subtract 3x from both sides
2x + 2 = 100 Subtract 2 from both sides
2x = 98 Divide by 2 on both sides
x = 49 The only x value that satisfies this equation is 49
There are other cases where functions of x aren’t injective, meaning there’s more than one x value that satisfies the equation. Here’s what I mean.
x^2 = 4
x = 2 or -2
x^3 = 1
x = 1, -1/2 + isqrt(3)/2, -1/2 - isqrt(3)/2
sin(x) = 0
x = 2n*pi, where n is some integer