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Solve the system by substitution. y equals 2 x squared minus 3 x minus 1 y equals x minus 3

User Dally
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y = 2x ^2 - 3x - 1
y = x-3

First we put the equation in terms of x, we replace y = x-3 in the other equation
y = 2x ^2 - 3x - 1, so we get:

x-3 = 2x^2 - 3x - 1
0 = 2x^2 - 3x - x - 1 + 3
0 = 2x^2 - 4x + 2

We look for a common factor in the equation which is 2:

2 ( x^2 - 2x + 1) = 0

We send the 2 to the other side as a division of 0 so it becomes 0, because 0 divided by anything is 0:

x^2 - 2x + 1 = 0/2
x^2 - 2x + 1 = 0

In ax^2 + bx + c find two numbers that added are b and that multiplied are c, in this case is -1 and -1, because b = -2 and c = 1 and b = (-1) + (-1) b = -2 and c = (-1) • (-1) c = 1, so we get:

(x-1) (x-1) = 0

So we get that:

x - 1 = 0
x = 1

I hope I was helpful.
User Martin Sustrik
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