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For an expression like:
2x+4+6x+6=26 we figure out the bounds for 2x+4 2x+4= 2x+4 where 2x+4 >=0, (2x+4) where 2x+4 < 0 2x+ 4>=0 2x > 4 x >2 so the expression 2x+4 is 2x+4= 2x+4 where x>=2, (2x+4) where x < 2
similarly for 6x+6 we get 6x+6= 6x+6 where x>=1, (6x+6) where x < 1
so we look at our values under our smallest < bound and the work our way up starting at x <2, we have 2x+4=2x4 6x+6=6x6 , because if x<2 then x <1 must also be true so we have 2x4 + 6x6 where x<2 or simplified 8x10 where x <2 so we now we can solve the equation 8x10=26 8x=36 x=36/8=9/2 this is consistent with x<2, so we accept this as an answer
next smallest bound is x<1 so we look at numbers between the last < bound and this < bound i.e. where x>=2 and x<1 2x+4=2x+4 6x+6=6x6 so we have 2x+4 + 6x6 where x>=2 and x<1 or simplified 4x2 where x>=2 and x<1 now we solve 4x2=26 4x=28 x=28/4=7 but this is inconsistent with our bounds x>=2 and x<1, so we reject this as an answer As an illustration I will show what happens if we substitute x=7 into the expression 2x+4+6x+6=2(7)+4+6(7)+6=14+4+42+6=10+36=10+36=46 Notice that it would have worked if it was 10+36 instead of 10+36
now we try numbers greater than the largest < bound x >=1 2x+4=2x+4 6x+6=6x+6 2x+4+6x+6=2x+4+6x+6 2x+4+6x+6=8x+10 where x >1 we solve this equation now 8x+10=26 8x=16 x=2 x=2 is consistent with x>1 so we accept this solution
so we have two solutions x=9/2 and x=2
substitute the values in to check 2x+4+6x+6=26 for x=9/2 2(9/2)+4+6(9/2)+6=9+4+27+6=5+21=5+21=26 for x=2 2x+4+6x+6=2(2)+4+6(2)+6=4+4+12+6=8+18=26
Hope this makes sense 