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:
90 clockwise (or counterclockwise) rotation and then a reflection over the axis between the two shape (those two steps go in any order)
Step-by-step explanation:
for this lets mark the innermost point of each shape a (blue or A) and a' (red or B)* and the second point b and b'
here we see that the two shapes are in a position to where they seem reflected over a non-existent third diagonal axis, though this is not the case, we need to bring the shape into a position where it can be transformed to the quadrant of shape B and overlap the shape
so when you have a reflection over a diagonal axis, we can rotate or reflect the shape to a new quadrant, and perform the step thats not the first, so say we made a reflection over the X-axis, the shape is now in the lower half of the graph with shape B, from here we perform our last step wich is to rotate the shape into the quadrant of shape B in a clockwise motion, now a and a' overlap and b and b' overlap, same for c, c',d and d'
(*the ' in this case is called a prime symbol, when used, distinguishes two points or lines on a graph, A' = A prime)
Answer:
About 8.6 units.
If you count up and sideways, it would be 12, but then there is some stuff we need to calculate at the end because it's not fully.
Answer:
(3, - 1)
Step-by-step explanation:
5x - 4y = -11
2x + 3y = -9
Multiply the top equation by 2 and the bottom equation by - 5 to cancel out the x's.
10x - 8y = - 22
- 10x - 15y = 45
Cancel out x's.
-8y = - 22
- 15y = 45
Add like terms.
- 23y = 23
Can't have a negative variable to flip them.
-23 = 23y
Divide.
y = - 1
Input y into one of the equations and solve for x.
2x + 3(-1) = -9
Simplify.
2x -3 = -9
Cancel out -3 by adding 3.
2x = -6
Divide.
x - -3
Answer:
65
Step-by-step explanation:
