# How do I solve equations with modulus functions on both sides?

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Whilst it is possible to do it algebraically, it's usually easier to solve it graphically. For example: for which values of x is |x+2| = |3x-6|. By sketching the graphs of y=|x+2| and y=|3x-6|, it is easy to see that there are 2 intersections: one for positive values of x+2 and negative values of 3x-6, the other for positive values of x+2 and positive values of 3x-6. To find the first x value solve x+2 = -3x+6, to find the second x value solve x+2 = 3x-6.

If you prefer to work algebraically, simply ask "for which values of x is x+2 negative, and for which is it positive". Do the same for 3x-6. Now solve for x in each section (with the 3 sections being: both negative, one positive one negative, both positive). Check each value you've found to see if it makes sense. Demonstrated below.

For both negative:
-x-2 = -3x+6, x<-2
x = 4, x<-2. This is a clear contradiction, so we ignore this value.

For x+2 positive and 3x-6 negative:
x+2 = -3x+6, -2 x = 1, -2

For both positive:
x+2 = 3x-6, 2 x = 4, 2

Therefore the answer to |x+2| = |3x-6| is x = 1, x = 4

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