let's multiply both sides in each equation by the LCD of all fractions in it, thus doing away with the denominator.
![\begin{cases} \cfrac{1}{2}x+\cfrac{1}{3}y&=7\\\\ \cfrac{1}{4}x+\cfrac{2}{3}y&=6 \end{cases}\implies \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6\left( \cfrac{1}{2}x+\cfrac{1}{3}y \right)=6(7)}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{1}{4}x+\cfrac{2}{3}y\right)=12(6)} \end{cases}\implies \begin{cases} 3x+2y=42\\ 3x+8y=72 \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20%5Ccfrac%7B1%7D%7B2%7Dx%2B%5Ccfrac%7B1%7D%7B3%7Dy%26%3D7%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B4%7Dx%2B%5Ccfrac%7B2%7D%7B3%7Dy%26%3D6%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B6%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7Dx%2B%5Ccfrac%7B1%7D%7B3%7Dy%20%5Cright%29%3D6%287%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B12%7D%7D%7B12%5Cleft%28%20%5Ccfrac%7B1%7D%7B4%7Dx%2B%5Ccfrac%7B2%7D%7B3%7Dy%5Cright%29%3D12%286%29%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%203x%2B2y%3D42%5C%5C%203x%2B8y%3D72%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{using elimination}}{ \begin{array}{llll} 3x+2y=42&\times -1\implies &\begin{matrix} -3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2y=&-42\\ 3x+8y-72 &&~~\begin{matrix} 3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8y=&72\\ \cline{3-4}\\ &&~\hfill 6y=&30 \end{array}} \\\\\\ y=\cfrac{30}{6}\implies \blacktriangleright y=5 \blacktriangleleft \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20elimination%7D%7D%7B%20%5Cbegin%7Barray%7D%7Bllll%7D%203x%2B2y%3D42%26%5Ctimes%20-1%5Cimplies%20%26%5Cbegin%7Bmatrix%7D%20-3x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~-2y%3D%26-42%5C%5C%203x%2B8y-72%20%26%26~~%5Cbegin%7Bmatrix%7D%203x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%2B8y%3D%2672%5C%5C%20%5Ccline%7B3-4%7D%5C%5C%20%26%26~%5Chfill%206y%3D%2630%20%5Cend%7Barray%7D%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B30%7D%7B6%7D%5Cimplies%20%5Cblacktriangleright%20y%3D5%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \stackrel{\textit{substituting \underline{y} on the 1st equation}~\hfill }{3x+2(5)=42\implies 3x+10=42}\implies 3x=32 \\\\\\ x=\cfrac{32}{3}\implies \blacktriangleright x=10\frac{2}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left(10\frac{2}{3}~~,~~5 \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20%5Cunderline%7By%7D%20on%20the%201st%20equation%7D~%5Chfill%20%7D%7B3x%2B2%285%29%3D42%5Cimplies%203x%2B10%3D42%7D%5Cimplies%203x%3D32%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B32%7D%7B3%7D%5Cimplies%20%5Cblacktriangleright%20x%3D10%5Cfrac%7B2%7D%7B3%7D%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cleft%2810%5Cfrac%7B2%7D%7B3%7D~~%2C~~5%20%5Cright%29~%5Chfill)
Answer:
4 hours and 30 minutes
Step-by-step explanation:
I'll assume you have a basic knowledge of time and know the sequence. So all you do is count to 1:00 and see how many steps you took. In this case 4, but then you have 30 minutes left. Add that to the 4 hours, and you have a total of 4 hours and 30 minutes of elapsed time.
Hope this helps :)
The answer would be 32. take your 26 and add 18 then subtract 12. then rework it forward if you don't understand.
ANSWER
See attachment
EXPLANATION
The given inequality is

This implies that,

Multiply both sides of the second inequality by -1 and reverse the inequality sign.

The graphical solution to this inequality is shown in the attachment.