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Y=-x^2+2x+10
y=x+2

Substitution
Need ASAP

User Hackman
by
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1 Answer

4 votes

Answer:


x_1=(1+√(33) )/(2),y_1=(5+√(33) )/(2)\\ \\ x_2=(1-√(33) )/(2),y_2=(5-√(33) )/(2)

Step-by-step explanation:

You must find the solution of the system using the substitution method.

System:


y=-x^2+2x+10\\\\ y=x+2

Substitute x+2 for y:


x+2=y=-x^2+2x+10

Pass all the terms to one side:


x^2-2x+x+2-10\\ \\ x^2-x-8=0

Use the quadratic formula to solve:


x=(-b+/-√(b^2-4ac) )/(2a)\\ \\ x=(-(-1)+/-√((-1)^2-4(1)(-8)) )/(2(1))\\ \\ x=(1+/-√(1+32) )/(2)


x_1=(1+√(33) )/(2)\\ \\x_2=(1-√(33) )/(2)

Use y = x + 2 to find the value of y:


y_1=2+x_1=2+(1+√(33) )/(2)\\ \\y_1=(5+√(33) )/(2) \\\\y_2=2+x_2=2+(1-√(33) )/(2)\\ \\ y_2=(5-√(33) )/(2)

User Pritish
by
6.6k points