\[9\left| 9-8x \right|=2x+3\]
well, you will get two results: \(9\left| 9-8x \right|=2x+3\) \(9(9-8x )=2x+3\) OR \(9(9-8x )=-2x-3\)
solve each case individually.
i dont know where to start. will you walk me through it?
well, expand the right side in case 1
I mean the left side (not the right side)
would you move the x to the left side first?
\(9(9-8x )=2x+3\) \(9\cdot9-9\cdot8x =2x+3\) \(81-72x =2x+3\)
So in case two you would have the same, but the right side would be -2x-3
Solve for each: \(81-72x =2x+3\) \(81-72x =-2x-3\)
x would go on the left side?
you can add 72x to both sides in case 2 and case 1
leaving us with 81=74x+3
two cases: \(81-72x =2x+3\) \(81-72x+72x =2x+72x+3\) \(81=74x+3\) right!
then subtract 3 from both sides.
yes, so x=?
what about in the second case?
Now in case 2: \(81-72x =-2x-3\)
you also add 72x to both sides, THEN, add 3 to both sides
oh the first one is incorrect, and this one too
when in case one you have: 78=74x you divide by 74 on both sides, and thus x=78/74 (not 74/78)
And in case two you made the same mistake
so it would be 84/70?
Yes, for case 2.
And they are not extraneous.
let me say correctly what would have been extraneous:
Any x-solution, in this case, that is equal to \(C\) that satisfies: \(C<-2/3\)
because if x is less than -2/3 then 2x+3 is negative, and absolute value can NOT e equal to negative (since absolute value is really a distance)