Questions on Algebra: Polynomials, rational expressions and equations answered by real tutors!

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Question 152559: Give 3 sets of points that satisfy the equations:
2x+2y=4 and y=x^2-4: Give 3 sets of points that satisfy the equations:
2x+2y=4 and y=x^2-4
Answer by jim_thompson5910(9168) About Me  (Show Source):
You can put this solution on YOUR website!
Are you trying to solve this as a system? If you are, then there are two points that satisfy this system

2x+2y=4 Start with the first equation.


2x+2(x^2-4)=4 Plug in y=x^2-4


2x+2x^2-8=4 Distribute


2x+2x^2-8-4=0 Get all terms to the left side.


2x^2+2x-12=0 Combine like terms.


Notice we have a quadratic equation in the form of ax^2+bx+c where a=2, b=2, and c=-12


Let's use the quadratic formula to solve for x


x = (-b +- sqrt( b^2-4ac ))/(2a) Start with the quadratic formula


x = (-(2) +- sqrt( (2)^2-4(2)(-12) ))/(2(2)) Plug in a=2, b=2, and c=-12


x = (-2 +- sqrt( 4-4(2)(-12) ))/(2(2)) Square 2 to get 4.


x = (-2 +- sqrt( 4--96 ))/(2(2)) Multiply 4(2)(-12) to get -96


x = (-2 +- sqrt( 4+96 ))/(2(2)) Rewrite sqrt(4--96) as sqrt(4+96)


x = (-2 +- sqrt( 100 ))/(2(2)) Add 4 to 96 to get 100


x = (-2 +- sqrt( 100 ))/(4) Multiply 2 and 2 to get 4.


x = (-2 +- 10)/(4) Take the square root of 100 to get 10.


x = (-2 + 10)/(4) or x = (-2 - 10)/(4) Break up the expression.


x = (8)/(4) or x = (-12)/(4) Combine like terms.


x = 2 or x = -3 Simplify.


So the answers are x = 2 or x = -3


So the two solutions are (2,0) and (-3,5)