Answer:
x+y/4 = 1/2
x-3y/3 = 2
move variables to one side:
multiply the first equation by 4 to get: x+y =2
and the second equation by 3 to get: x-3y =6
then subtract the equations to cancel out x:
x+y = 2
- x-3y = 6
then u get
y--3y = 2-6
4y = -4
y=-1
substitute to solve for x:
x-1 / 4 =1/2
x-1 = 2
x=3
check:
3+-1
2/4= 1/2
correct!!!
Answer:
1: AAS, RQC 2: ASA, SRP
Step-by-step explanation:
(1) We are shown that two angles and one side are congruent in the order of AAS. Make sure you write the letters in terms of the corresponding angles. The angles and sides are congruent because the problem labels it for us. For example, B,A,C=R,Q,S. Answers: AAS, RQC.
(2) We are shown that two angles and one side are congruent in the order of ASA. Make sure you write the letters in terms of the corresponding angles again. For example, P,Q,R=P,S,R. The angles are congruent because the problem labels it for us. Side PR is congruent to side PR by reflexive property. Answers: ASA, SRP.
I hope this helped :) Good luck
![\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%204x%2B3y%3D-8%5C%5C%5C%5C%20-8x-6y%3D16%20%5Cend%7Bcases%7D~%5Chspace%7B10em%7D%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}](https://tex.z-dn.net/?f=4x%2B3y%3D-8%5Cimplies%203y%3D-4x-8%5Cimplies%20y%3D%5Ccfrac%7B-4x-8%7D%7B3%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-8x-6y%3D16%5Cimplies%20-6y%3D8x%2B16%5Cimplies%20y%3D%5Ccfrac%7B8x%2B16%7D%7B-6%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B8%7D%7B-6%7Dx%2B%5Ccfrac%7B16%7D%7B-6%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D)
one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.
since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.
