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)
If N = 8
if M = 7
then the equation is 8(2) - 7
do the multiplication first 8(2) = 16
now do the subtraction 16 - 7 = 9
Answer:
Two solutions
Step-by-step explanation:
The number of points of intersections represents the number of solutions to the system of equations. Since the parabola intersects the circle at two points, there are two solutions to the circle.
Furthermore, these two points of intersection are exactly the solutions to the system of equations. Finding the coordinates of the points of intersection will give you the solutions to the system of equations.
Answer:
1.)B
2.)C
I apologize if this answer is wrong.
