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3 years ago

Solve the system of linear equations by elimination. x + 5y = 15 and -x - 2y = 0

Mathematics

1 answer:

Zolol [24]3 years ago

4 0

Answer:

X= - 10

Y=5

Step-by-step explanation:

First add the equations and x will be eliminated, then solve for y as a normal linear function, and the answer is 5.

Substitute thw y in any function, I substituted in the first one, and solved for x and got negative 10 as an answer.

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Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

Evaluate f(x)=2x- 1 at the given value of x. f(1) f(1)=

USPshnik [31]

F(1)=2(1)-1
f(1)=2-1
f(1)=1

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3 years ago

A) Factorise 2x² - 9x - 5 (x-5)(2x+1) (2) b) Hence, simplify fully 2x^2- 9x - 5 3x – 15x (2)

Lyrx [107]

Answer:

Step-by-step explanation:

b).

2x² - 9x - 5 = ( x - 5 )( 2x + 1 )

3x² - 15x = 3x ( x - 5 )

$\frac{(x-5)(2x+1)}{3x(x-5)}$ = $\frac{2x+1}{3x}$

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How does understanding place value help you to subtract across zeros?

Alex_Xolod [135]

Oh it helps because a zero could count as a number when you write it down but it doesn't seem that way when u say it

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What is the tangent ratio for ∠A?

tiny-mole [99]

for this case we have to define trigonometric relations of rectangular triangles, that the tangent of an angle is given by the leg opposite the angle on the leg adjacent to the angle. So:

$tg (A) = \frac {6} {8} = \frac {3} {4}$

Answer:

$tg (A) = \frac {3} {4}$

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