let's bear in mind that an absolute value expression is in effect a piece-wise expression, namely it has a ± versions of the same expression.
![\bf 5|3x-4| = x+1\implies |3x-4|=\cfrac{x+1}{5}\implies \begin{cases} +(3x-4)=\cfrac{x+1}{5}\\[1em] -(3x-4)=\cfrac{x+1}{5} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(3x-4)=\cfrac{x+1}{5}\implies 3x-4=\cfrac{x+1}{5}\implies 15x-20=x+1 \\\\\\ 14x-20=1\implies 14x=21\implies x = \cfrac{21}{14}\implies \boxed{x=\cfrac{3}{2}} \\\\[-0.35em] ~\dotfill\\\\ -(3x-4)=\cfrac{x+1}{5}\implies -3x+4=\cfrac{x+1}{5}\implies -15x+20=x+1 \\\\\\ 20=16x+1\implies 19=16x\implies \boxed{\cfrac{19}{16}=x}](https://tex.z-dn.net/?f=%5Cbf%205%7C3x-4%7C%20%3D%20x%2B1%5Cimplies%20%7C3x-4%7C%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%2B%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5C%5C%5B1em%5D%20-%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%2B%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%203x-4%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%2015x-20%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%2014x-20%3D1%5Cimplies%2014x%3D21%5Cimplies%20x%20%3D%20%5Ccfrac%7B21%7D%7B14%7D%5Cimplies%20%5Cboxed%7Bx%3D%5Ccfrac%7B3%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%20-3x%2B4%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%20-15x%2B20%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%2020%3D16x%2B1%5Cimplies%2019%3D16x%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B19%7D%7B16%7D%3Dx%7D)
The first part to question 8 is 2.25 and the answer to part b is 12
<span>You have mentioned that two of the angles of ΔKSD are
∠K=16° and ∠D=34°. So that would mean that ∠S=130°,
because∠K+∠D+∠S=180°. That's because all the angles
of any triangle must total 180°.
But you also say that ∠T=51°. But since no angle of
ΔKSD can equal 51°, neither can any angle of ΔWAT
equal to 51°. Therefore ∠T=51° is impossible!</span>
Answer:
B. a decimal that does not terminate and does
not repeat.
the answer would be x=−0.870452,0.712370