How do you solve #10u + 6= 4( 2u - 1) + 2u + 10#?

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Answer 1 · 2016-09-20T10:10:36

#x# has any value. #x in RR#

Multiply out the brackets and then re-arrange the terms of the equation.

#10u +6 = 4(2u-1) +2u +10#

#10u+6 = 8u-4 +2u +10#

#10u -8u -2u = -4+10-6#

Simplifying both sides gives

#0=0" "larr# you might ask "What now?#

The statement is TRUE, but there is no #x# left to solve for.

This is a special type of equation called an identity which is true for any value of x.

#x# has any value.