Question 986002
{{{10-abs(2x-1)=4}}}....here, you are asked to find a value of {{{x}}} that makes {{{10-abs(2x-1)}}} equal to {{{4}}}, or the intersection points of the {{{10-abs(2x-1)}}} and horizontal line {{{y=4}}}

{{{10-4=abs(2x-1)}}}

{{{6=abs(2x-1)}}}..........since {{{abs(2x-1)=sqrt((2x-1)^2)}}} => we will have {{{(2x-1)}}} and {{{-(2x-1)}}}}

solutions:

{{{6=2x-1}}}=>{{{6+1=2x}}}=>{{{7=2x}}}=>{{{highlight(x=7/2)}}}

{{{6=-(2x-1)}}}=>{{{6=-2x+1}}}=>{{{2x=1-6}}}=>{{{2x=-5}}}=>{{{highlight(x=-5/2)}}}


{{{drawing( 600, 600, -10, 10, -10, 15,
circle(7/2,4,.12),locate(7/2,4,p(7/2,4)),
circle(-5/2,4,.12),locate(-5/2,4,p(-5/2,4)), graph( 600, 600, -10,10, -10, 15, 10-abs(2x-1),4)) }}}