How do you find the solution of the system of equations #y=-x^2+2x-3# and #y=x-5#?

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Answer 1 · 2015-05-02T07:28:59

We can substitute # y = x - 5# in the first equation

We get # x - 5 = -x^2 + 2x - 3#

Transposing all the terms on the right to the left, we get

#-> x - 5 + x^2 - 2x + 3 = 0#

#-> x^2 - x - 2 = 0#

To solve for x, we need to factorise the expression on the left

We can split the middle term of this expression to factorise it

#-> x^2 - 2x + x - 2 = 0#

#-> x(x - 2) + 1(x - 2) = 0#

#(x-2)# is a common factor to both the terms

# -> (x - 2)(x+1) = 0#

This tells us that:

Either #(x - 2) = 0# or #(x + 1)= 0#

#color(green)( x = 2 or x = -1# is the Solution