MHB Determine if x = -2 is a solution to the equation 2x – 2 = 3x

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To determine if x = -2 is a solution to the equation 2x – 2 = 3x, substituting -2 into the equation shows that both sides equal -6. The left side calculates as 2(-2) - 2 = -4 - 2 = -6, while the right side computes as 3(-2) = -6. Since both sides are equal, x = -2 is indeed a solution to the equation. The verification process confirms the solution without needing to solve the equation algebraically. Thus, x = -2 satisfies the original equation.
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Showing full working determine if x = -2 is a solution to the equation 2x – 2 = 3x
 
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lmae said:
Showing full working determine if x = -2 is a solution to the equation 2x – 2 = 3x

$$2x-2=3x \Leftrightarrow 2x-2-3x=3x-3x \Leftrightarrow -x-2=0 \Leftrightarrow -x-2+2=2 \Leftrightarrow -x=2 \\ \Leftrightarrow -(-x)=-2 \Leftrightarrow x=-2$$
 
The problem was to "show that x= -2 is a solution to the equation 2x- 2= 3x. It is not necessary to solve the equation! Just replace x with -2 in the equation and show that it is true. If x= -2, the 2x= -2(-2)= 4 and 2x- 2= -4- 2= -6. On the right side, 3x= 3(-2)= -6. Yes! They are equal!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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