Question #59984
1 Answer
Explanation:
Solve using
color(blue)"substitution"substitution Rearrange the linear equation to give x in terms of y, then substitute into the equation of the circle.
x+3y=-3rArrx=color(blue)(-3-3y)to(1)x+3y=−3⇒x=−3−3y→(1)
rArr(color(blue)(-3-3y))^2+y^2=13⇒(−3−3y)2+y2=13 distributing the bracket using the FOIL method.
rArr(9+18y+9y^2)+y^2=13⇒(9+18y+9y2)+y2=13 simplify and equate to zero.
rArr10y^2+18y-4=0⇒10y2+18y−4=0 factorising the left side.
2(5y-1)(y+2)=02(5y−1)(y+2)=0 solving for y gives.
•5y-1=0rArry=1/5∙5y−1=0⇒y=15
•y+2=0rArry=-2∙y+2=0⇒y=−2 Substitute these values for y into (1) to obtain corresponding
x-coordinates.
• y=1/5to x=-3-3/5=-18/5to(-18/5,1/5)∙y=15→x=−3−35=−185→(−185,15)
• y=-2 to x=-3+6=3 to(3,-2)∙y=−2→x=−3+6=3→(3,−2)
The graph illustrates, geometrically, the intersections of the linear equation with the circle.
graph{(x^2+y^2-13)(y+1/3x+1)=0 [-10, 10, -5, 5]}
