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2 years ago

Please Help solve simultaneous equations , I never got hang of it.

Mathematics

2 answers:

tekilochka [14]2 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

(1)

a + 3b = 5 ( subtract 3b from both sides )

a = 5 - 3b → (1)

3a + 2b = 8 → (2)

Substitute a = 5 - 3b into (2)

3(5 - 3b) + 2b = 8 ← distribute and simplify left side

15 - 9b + 2b = 8

15 - 7b = 8 ( subtract 15 from both sides )

- 7b = - 7 ( divide both sides by - 7 )

b = 1

Substitute b = 1 into (1)

a = 5 - 3(1) = 5 - 3 = 2

Then a = 2 and b = 1

(2)

3m - 4n = - 14 → (1)

2m + 2n = 14 → (2)

Multiplying (2) by 2 and adding to (1) will eliminate the n- term

4m + 4n = 28 → (3)

Add (1) and (3) term by term to eliminate n

7m + 0 = 14

7m = 14 ( divide both sides by 7 )

m = 2

Substitute m = 2 into either of the 2 equations and solve for n

Substituting into (2)

2(2) + 2n = 14

4 + 2n = 14 ( subtract 4 from both sides )

2n = 10 ( divide both sides by 2 )

n = 5

Then m = 2 and n = 5

AnnyKZ [126]2 years ago

3 0

$\huge\underline\mathtt\colorbox{cyan}{ATTACHMENT}$

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Read 2 more answers

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$