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Solve x^2+y^2-4x-6y+4=0 intersects the straightline y=x​

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Answer: the points of intersection are:

((5 + sqrt(17)) / 2, (5 + sqrt(17)) / 2) and ((5 - sqrt(17)) / 2, (5 - sqrt(17)) / 2)

Step-by-step explanation: A substitution of y=x in the equation can assist in determining the point of intersection with the line y=x.

2x^2-10x+4=0

A quadratic equation 2x squared minus 10x plus 4 equals zero.

The equation is x squared minus 5x plus 2 equals zero.

By applying the quadratic equation, we obtain:

One possible paraphrasing of this mathematical expression could be: The value of x can be found by adding or subtracting the square root of the difference between 5 squared and 412, and then dividing the result by 2.

One possible paraphrase: The value of x can be found by taking the solutions of the equation x^2 - 5x + 17/4 = 0, which are x = (5 + sqrt(17))/2 and x = (5- sqrt(17))/2.

Thus, the values on the x-axis where the points intersect can be identified as:

Two expressions that give the value of x are: x equals half the sum of 5 and the square root of 17, and x equals half the difference between 5 and the square root of 17.

By inserting these x-values into the equation y=x, we can obtain the corresponding y-coordinates.

Two equations yield the value of y. The first equation involves adding the square root of 17 to 5 and dividing the sum by 2. The second equation involves subtracting the square root of 17 from 5 and dividing the difference by 2 to obtain y.

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