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Troyanec [42]
3 years ago
9

PLEASEEE HELP MEEEENO LINKSSS

Mathematics
1 answer:
Artemon [7]3 years ago
3 0

Answer:

\huge\boxed{(2,\ 3);\ (-1,\ 6)}

Step-by-step explanation:

f(x)=x^2-2x+3\to y=x^2-2x+3\\f(x)=-x+5\to y=-x+5\\\\\left\{\begin{array}{ccc}y=x^2-2x+3&(1)\\y=-x+5&(2)\end{array}\right

substitute (1) to (2):

x^2-2x+3=-x+5     add x to both sides

x^2-2x+x+3=-x+x+5

x^2-x+3=5        subtract 5 from both sides

x^2-x+3-5=5-5

x^2-x-2=0\\\\x^2+x-2x-2=0\\\\x(x+1)-2(x+1)=0\\\\(x+1)(x-2)=0\iff x+1=0\ \vee\ x-2=0\\\\\boxed{x=-1\ \vee\ x=2}

substitute the values of x to (2):

x=-1\\\\y=-(-1)+5=1+5=6\\\\\huge\boxed{(-1,\ 6)}\\\\x=2\\\\y=-2+5=3\\\\\huge\boxed{(2,\ 3)}

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