All categories

3 years ago

Solve the system using combination method to get a solution (x,y) 4x+7y=10 -4x-6y=28

Mathematics

2 answers:

lilavasa [31]3 years ago

8 0

$4x+7y=10\\
-4x-6y=28\\
-------\\
y=38\\\\
4x+7\cdot38=10\\
4x+266=10\\
4x=-256\\
x=-64$

PIT_PIT [208]3 years ago

3 0

$\left \{ {{4x+7y=10} \atop {-4x-6y=28}} \right. \\\\add\ those\ two\ equations\\\\
y=38\\\\
4x+7y=10\ \ \ | subtract\ 7y\\\\
4x=10-7x\ \ \ \ | divide\ by\ 4\\\\
x=\frac{10-7x}{4}=\frac{10-7*38}{4}=\frac{10-266}{4}=\frac{-256}{4}=-64\\\\
solution:\\ \left \{ {{y=38} \atop {x=-64}} \right.$

You might be interested in

2x-5y=17 ..solve for y

AlladinOne [14]

$2x-5y=17\ \ \ \ | subtract\ 2x\\\\
-5y=17-2x\ \ \ | divide\ by\ -5\\\\
y=\frac{-17+2x}{5}\\\\Solution\ is\ y=\frac{-17+2x}{5}.$

7 0

2 years ago

Two telephone calls come into a switchboard at random times in a fixed one-hour period. Assume that the calls are made independe

svp [43]

Answer:

0.25

Step-by-step explanation:

From the given information:

Since the times are uniform distributed over the one hour period (0,1):

Then;

$f_1(y_1) = 1$

$f_2(y_2) = 1$

So, $Y_1$ and $Y_2$ are independent; then:

$f(y_1|y_2) = f_1(y_1)f_2(y_2) \\ \\ = 1(1) = 1$

$P(Y_1\le 0.5 Y_2 \le 0.5) = \int ^{0.5}_{0} \int ^{0.5}_{0} f(y_1,y_2) dy_{2}dy_{1}$

$\implies \int ^{0.5}_{0} \int ^{0.5}_{0} (1) dy_{2}dy_{1}$

$= \dfrac{1}{4}$

= 0.25

3 0

2 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago

If AD = 5x + 3, TD = x + 27, what is AD?

Eddi Din [679]

The question is incomplete, but if we assume that you mean AD = TD , we have :

AD = TD ===》 5x + 3 = x + 27

===》 4x = 24

===》 x = 6

_________________________________

AD = 5 × ( 6 ) + 3

AD = 30 + 3

AD = 33

7 0

2 years ago

Help me with #19, please

densk [106]

You would pay 6 dollars

4 0

2 years ago