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3 years ago

Does the system have one solution, no solution, or infinitely many? Use your substitution to justify your answer

Mathematics

1 answer:

Taya2010 [7]3 years ago

4 0

Answer: It has one solution. The solution is (x,y) = (-4,-3)

Add up the equations doing so straight down

x + -x = 0x = 0 so the x terms go away

2y + 2y = 4y

-10 + (-2) = -12

We end up with 4y = -12 so y = -3 after you divide both sides by 4. Use this y value to find the value of x

x+2y = -10

x + 2(-3) = -10

x - 6 = -10

x = -10+6

x = -4

The single solution is (x,y) = (-4,-3)

As a check, plug this solution into each equation to see if you get a true statement or not. Let's do so with the first equation

x+2y = -10

-4 + 2(-3) = -10

-4 - 6 = -10

-10 = -10 .... true

and then the second equation

-x+2y = -2

-(-4) + 2(-3) = -2

4 - 6 = -2

-2 = -2 .... true

both equations are true, so the solution is confirmed

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3 years ago

Read 2 more answers

15. If x=a Sin2t (I+Cos2t) and y=b Cos 2t (1-Cos2t) then find dy/dx at =22/7*4

Sav [38]

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

It looks like we're given

$\begin{cases}x=a\sin(2t)(1+\cos(2t))\\y=b\cos(2t)(1-\cos(2t))\end{cases}$

where a and b are presumably constant.

Recall that

$\cos^2t=\dfrac{1+\cos(2t)}2$

$\sin^2t=\dfrac{1-\cos(2t)}2$

so that

$\begin{cases}x=2a\sin(2t)\cos^2t\\y=2b\cos(2t)\sin^2t\end{cases}$

Then we have

$\dfrac{\mathrm dx}{\mathrm dt}=4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t$

$\dfrac{\mathrm dy}{\mathrm dt}=-4b\sin(2t)\sin^2t+4b\cos(2t)\sin t\cos t$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4b\cos(2t)\sin t\cos t-4b\sin(2t)\sin^2}{4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t}$

$\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac ba\tan t}$

where the last reduction follows from dividing through everything by $\cos(2t)\cos^2t$ and simplifying.

I'm not sure at which point you're supposed to evaluate the derivative (22/7*4, as in 88/7? or something else?), so I'll leave that to you.

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3 years ago