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2 years ago

10

Solve the equation -6(2a + 7) = -22- 8a

2 answers:

Andru [333]2 years ago

3 0

Answer: a = -5

Step-by-step explanation:

First, open up the brackets, and solve it

-12a -42 = -22 -8a. (-6 x 2a, and -6 x 7)

bring the -8a to the left side, so that all the a’s are together

-12a -42 + 8a = -22 (add -42 on both sides like we did with -8a)

-12a + 8a = -22 + 42

solve

-4a = 20

divide -4 on both sides

a = -5

katen-ka-za [31]2 years ago

3 0

Answer:

a=-5

Step-by-step explanation:

Here you go mate

PEMDAS: Parenthesis,Exponent,Multiplication,Division,Addition,Subtraction

Step 1

-6(2a + 7) = -22- 8a Equation/Question

Step 2

-6(2a + 7) = -22- 8a Remove parenthesis

-12a-42=-8a-22

Step 3

-12a-42=-8a-22 Add 8a to the sides

-4a-42=-22

Step 4

-4a-42=-22 Subtract 42

4a=20

Step 5

4a=20 Divide both sides by 4

answer

-5

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The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

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${y_p}'=a\cos x-b\sin x$

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Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

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$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

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$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

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2 years ago