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Novosadov [1.4K]
3 years ago
10

I need working out and answers for the following:

Mathematics
1 answer:
LenaWriter [7]3 years ago
4 0
A. 4x - 3 = 2x + 7
4x - 2x = 7 + 3
2x = 10
x = 5

B. 2x + 6 = 7x - 14
2x - 7x = - 14 - 6
-5x = -20
x = 4

C. 2( x + 3) = x - 4
2x + 6 = x - 4
2x - x = -4 - 6
x = -10

D.4 ( 5x - 2) = 2( 9x + 3)
20x - 8 = 18x + 6
20x - 18x = 6 + 8
2x = 14
x = 7

E.4x - 1/2 = x - 7
4x - x = -7 + 1/2
3x = (-14 + 1)/2
3x = -13/2
x = -13/6

F. 3x + 2 = 2x + 13/3
3x - 2x = 13/3 - 2
x = (13 - 6)/3
x = 7/3


Hope This Helps You!
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PLEASE ANSWER QUICKLY
Olegator [25]

Answer:

x=3

Step-by-step explanation:

Set equations equal to each other:

x^3-3x^2+2=x^2-6x+11

Subtract two from both sides

x^3-3x^2=x^2-6x+9
Factor out x^2 from the left side

x^2(x-3)=x^2-6x+9

Factor the right side, look for two factors of 9, which add to -6, which is just -3 and-3

x^2(x-3) = (x-3)(x-3)

So as you can see here, we have both sides in factored form, and both have a factor of (x-3) on both sides, and if one of the factors is zero, then the entire thing becomes zero, since 0 * anything= zero.

this means a solution would be at x=3, since it makes the right side 0 * 0 = 0, and the left side (3)^2 * 0 = 0

It's important to notice this, since had we divided by x-3 to make the equation a quadratic, we would've excluded this real solution, because when dividing by x-3 the solution x=3 makes it 0, so we would've been dividing by zero.

So one of the real solutions is at x=3

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