What is the solution to Y=x-5 and x^2+y^2=30^?
Hi @trug
@trug first substitute the linear equation into the quadratic equation and solve for a variable. Then find the other variable. You may have more than one solution since there is a quadratic equation. Good Luck!
Ok draw the graphs of both the ellipse and the straight line. the points of intersection are the required solution. @trug
Sketching may not always yield accurate or exact results unless performed by a graphing calculator. Although I believe you are supposed to find a solution algebraically.
I don't really know how to go about any of that, since there is a ? for a power. i can get it into a quadratic equation but I'm lost after that @calculusfunctions @Princer_Jones
y=x-5 putting this value of y in the other equation we have x^2+(x-5)^2=30 what did you get after simplifying this equation @trug ?
the question is incomplete then???
after simplifying you get 2x^2-25=30^? right?
what is 30^ ???
@Princer_Jones I have no idea it's just what it says on the worksheet 30 raised to the ? power
how would you go about solving it if you just completely disregarded the ?
see ^ means raised to the power. here a term is missing after ^ in the question
the term is the ?
see the question... how can i say what is written in the book/?.:)
Usually questions end in question marks. Maybe the ^ is a type-o.
ok lets just completely blow off the ? how would you solve it if it were just Y=x-5 and X^2+Y^2=30
that could be it @freckles
Ohkay substitute y=x-5 in the second equation and we have x^2+(x-5)^2=30 this implies x^2+x^2-10x+25=30 this implies 2x^2-10x-5=0 from here you get two values of x and putting these two values of x in y=x-5, you get the value of y
do you solve for the two x's individually?
just the quadratic formula to compute the values of x
so the solutions are 2 and -4?
Wait, or is it X= 5-√ 5 and X= 5+√ 5 ----- ----- 2 2
check it
@trug
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