Check the picture below.
so the picture has a rectangle that is 8 units high and 12 units wide, and it has a couple of "empty" trapezoids, with a height of 5 and "bases" of 9 and 3.
now, if we just take the whole area of the rectangle and then subtract the area of those two trapezoids, what's leftover is the blue area.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=9\\ b=3 \end{cases}\implies \begin{array}{llll} A=\cfrac{5(9+3)}{2}\implies A=30 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Areas}}{\stackrel{rectangle}{(12\cdot 8)}~~ -~~\stackrel{\textit{two trapezoids}}{2(30)}}\implies 96-60\implies 36](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D5%5C%5C%20a%3D9%5C%5C%20b%3D3%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B5%289%2B3%29%7D%7B2%7D%5Cimplies%20A%3D30%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20Areas%7D%7D%7B%5Cstackrel%7Brectangle%7D%7B%2812%5Ccdot%208%29%7D~~%20-~~%5Cstackrel%7B%5Ctextit%7Btwo%20trapezoids%7D%7D%7B2%2830%29%7D%7D%5Cimplies%2096-60%5Cimplies%2036)
Answer:
x = 3
Step-by-step explanation:
To find the x-coordinate, substitute y = 8 into the equation abd solve for x
5x - 7 = 8 ( add 7 to both sides )
5x = 15 ( divide both sides by 5 )
x = 3
Thus (3, 7) is a solution to the equation
2 is the answer!!!!!!!!!!
You multiply the whole number and multiply the denominator (the bottom one of the fraction) then you take the product and add the numerator (the top one of the fraction). That's how you get an improper fraction. Then once you get and improper fraction, you divide the numerator and denominator by their common factors. That's how you simply a mixer number.
