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Katyanochek1 [597]
3 years ago
12

Help with my algebra homework

Mathematics
1 answer:
Maru [420]3 years ago
3 0

Let's solve your equation step-by-step.

2x /5 + 3 = x/10 - 1

Step 1: Simplify both sides of the equation.

2x /5 + 3 =  x/10 - 1

2/5x + 3 = 1/10x - 1

Step 2: Subtract 1/10x from both sides.

2/5x + 3 - 1/10x = 1/10x - 1 - 1/10x

3/10x + 3 = -1

Step 3: Subtract 3 from both sides.

3 /10x + 3 - 3 = -1 - 3

3/10x = -4

Step 4: Multiply both sides by 10/3.

(10/3) * (3/10x) = (10/3) * (-4)

x = -40/3

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Use the intersect method to solve the equation. 14x^3-53x^2+41x-4=-4x^3-x^2+1x+4
UNO [17]

Answer:

x = (68 2^(1/3) + (27 i sqrt(591) + 445)^(2/3))/(27 (1/2 (27 i sqrt(591) + 445))^(1/3)) + 26/27 or x = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 26/27 or x = 1/27 ((-2)/(27 i sqrt(591) + 445))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3)) + 26/27

Step-by-step explanation:

Solve for x over the real numbers:

14 x^3 - 53 x^2 + 41 x - 4 = -4 x^3 - x^2 + x + 4

Subtract -4 x^3 - x^2 + x + 4 from both sides:

18 x^3 - 52 x^2 + 40 x - 8 = 0

Factor constant terms from the left hand side:

2 (9 x^3 - 26 x^2 + 20 x - 4) = 0

Divide both sides by 2:

9 x^3 - 26 x^2 + 20 x - 4 = 0

Eliminate the quadratic term by substituting y = x - 26/27:

-4 + 20 (y + 26/27) - 26 (y + 26/27)^2 + 9 (y + 26/27)^3 = 0

Expand out terms of the left hand side:

9 y^3 - (136 y)/27 - 1780/2187 = 0

Divide both sides by 9:

y^3 - (136 y)/243 - 1780/19683 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

-1780/19683 - 136/243 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ - 136/243) - (1780 z^3)/19683 + z^2 (3 λ^2 - (136 λ)/243) + λ^3 = 0

Substitute λ = 136/729 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 - (1780 u)/19683 + 2515456/387420489 = 0

Find the positive solution to the quadratic equation:

u = (2 (445 + 27 i sqrt(591)))/19683

Substitute back for u = z^3:

z^3 = (2 (445 + 27 i sqrt(591)))/19683

Taking cube roots gives 1/27 2^(1/3) (445 + 27 i sqrt(591))^(1/3) times the third roots of unity:

z = 1/27 2^(1/3) (445 + 27 i sqrt(591))^(1/3) or z = -1/27 (-2)^(1/3) (445 + 27 i sqrt(591))^(1/3) or z = 1/27 (-1)^(2/3) 2^(1/3) (445 + 27 i sqrt(591))^(1/3)

Substitute each value of z into y = z + 136/(729 z):

y = (68 2^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 1/27 (2 (27 i sqrt(591) + 445))^(1/3) or y = (68 (-2)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) - 1/27 (-2)^(1/3) (27 i sqrt(591) + 445)^(1/3) or y = 1/27 (-1)^(2/3) (2 (27 i sqrt(591) + 445))^(1/3) - (68 (-1)^(1/3) 2^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3))

Bring each solution to a common denominator and simplify:

y = (2^(1/3) ((27 i sqrt(591) + 445)^(2/3) + 68 2^(1/3)))/(27 (445 + 27 i sqrt(591))^(1/3)) or y = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (445 + 27 i sqrt(591))^(1/3)) or y = 1/27 2^(1/3) (-1/(445 + 27 i sqrt(591)))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3))

Substitute back for x = y + 26/27:

Answer:  x = (68 2^(1/3) + (27 i sqrt(591) + 445)^(2/3))/(27 (1/2 (27 i sqrt(591) + 445))^(1/3)) + 26/27 or x = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 26/27 or x = 1/27 ((-2)/(27 i sqrt(591) + 445))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3)) + 26/27

5 0
3 years ago
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