![\bf \begin{cases} -\cfrac{2}{3}x+\cfrac{7}{3}y=-\cfrac{5}{3}\\\\ \cfrac{4}{5}x-\cfrac{14}{5}y=2 \end{cases}\qquad \qquad \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{-2x+7y=-5}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{4x-14y=10} \end{cases}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cbegin%7Bcases%7D%20-%5Ccfrac%7B2%7D%7B3%7Dx%2B%5Ccfrac%7B7%7D%7B3%7Dy%3D-%5Ccfrac%7B5%7D%7B3%7D%5C%5C%5C%5C%20%5Ccfrac%7B4%7D%7B5%7Dx-%5Ccfrac%7B14%7D%7B5%7Dy%3D2%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B3%7D%7D%7B-2x%2B7y%3D-5%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B5%7D%7D%7B4x-14y%3D10%7D%20%5Cend%7Bcases%7D%20)
if you ever wonder why we multiply both sides by the LCD, well is just to do away with the denominators.
now, notice something, if we multiply the first equation by say -2, we end up with
-2( -2x + 7y ) = -2( -5 )
4x - 14y = 10
well well well, low and behold the second equation is really the first equation in disguise, so both equations are really the same thing, namely is the same exact line, and so their graph is just one line pancaked on top of the other, which means every single point in the lines are solutions namely infinitely many solutions.
that said, let's proceed with the substitution
![\bf \stackrel{\textit{solving for \underline{y} in the first equation}}{-2x+7y=-5\implies 7y=2x-5}\implies y=\cfrac{2x-5}{7} \\\\\\ \stackrel{\textit{now plugging in that \underline{y} in the second equation}}{4x-14\left( \cfrac{2x-5}{7} \right)=10}\implies 4x-2(2x-5)=10 \\\\\\ 4x-4x+10=10\implies \stackrel{\stackrel{\textit{infinitely many solutions}}{\downarrow }}{0=0}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsolving%20for%20%5Cunderline%7By%7D%20in%20the%20first%20equation%7D%7D%7B-2x%2B7y%3D-5%5Cimplies%207y%3D2x-5%7D%5Cimplies%20y%3D%5Ccfrac%7B2x-5%7D%7B7%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bnow%20plugging%20in%20that%20%5Cunderline%7By%7D%20in%20the%20second%20equation%7D%7D%7B4x-14%5Cleft%28%20%5Ccfrac%7B2x-5%7D%7B7%7D%20%5Cright%29%3D10%7D%5Cimplies%204x-2%282x-5%29%3D10%20%5C%5C%5C%5C%5C%5C%204x-4x%2B10%3D10%5Cimplies%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Binfinitely%20many%20solutions%7D%7D%7B%5Cdownarrow%20%7D%7D%7B0%3D0%7D%20)
Step-by-step explanation:
Let one person earns $ x
Therefore other earns = $(x + 2000)
According to the given condition:
![x + (x + 2000) = 28000 \\ \\ \therefore \: x + x + 2000 = 28000 \\ \\ \therefore \: 2 x = 28000 - 2000\\ \\ \therefore \: 2 x = 26000\\\therefore \: x = \frac{26000}{2} \\ \\ \therefore \: x = 13000 \\ \implies \: x + 2000 \\ \: \: \: \: \: \: \: \: \: \: = 13000 + 2000 \\ \therefore \: x + 2000= 15000 \\](https://tex.z-dn.net/?f=x%20%2B%20%28x%20%2B%202000%29%20%3D%2028000%20%5C%5C%20%20%5C%5C%20%20%5Ctherefore%20%5C%3A%20x%20%2B%20x%20%2B%202000%20%3D%2028000%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%202%20x%20%3D%20%20%2028000%20%20-%20%202000%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%202%20x%20%3D%20%20%2026000%5C%5C%5Ctherefore%20%5C%3A%20%20x%20%3D%20%20%20%20%5Cfrac%7B26000%7D%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20%20x%20%3D%20%20%20%2013000%20%5C%5C%20%20%5Cimplies%20%5C%3A%20x%20%2B%202000%20%5C%5C%20%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%3D%2013000%20%2B%202000%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20x%20%2B%202000%3D%2015000%20%5C%5C%20)
Hence, smaller income is $13000.
The answer to this problem is x=210
Answer:
There is one "special" factoring type that can actually be done using the usual methods ... "Perfect square trinomials" are quadratics which are the results of squaring binomials. ... This means that I'll have a "plus" sign between the x and the 5. ... it had a middle term of –25x, and this does not match what the pattern requires.
Step-by-step explanation:
Answer: he invested $46062.5 at 6% and $23031.25 at 10%
Step-by-step explanation:
Let x represent the amount which he invested in the account paying 6% interest.
Let y represent the amount which he invested in the account paying 10% interest.
He puts twice as much in the lower-yielding account because it is less risky.. This means that
x = 2y
The formula for determining simple interest is expressed as
I = PRT/100
Considering the account paying 6% interest,
P = $x
T = 1 year
R = 6℅
I = (x × 6 × 1)/100 = 0.06x
Considering the account paying 10% interest,
P = $y
T = 1 year
R = 10℅
I = (y × 10 × 1)/100 = 0.1y
His annual interest is $7370dollars. it means that
0.06x + 0.2y = 7370 - - - - - - - - - -1
Substituting x = 2y into equation 1, it becomes
0.06 × 2y + 0.2y = 7370
0.12y + 0.2y = 7370
0.32y = 7370
y = 7370/0.32
y = $23031.25
x = 2 × 23031.25
x = 46062.5