Find the points of intersection of the line X+Y=5, and the circle x^2+y^2=25 by solving the system of equations.
well, y = 5-x, so
x^2 + (5-x)^2 = 25
now just expand and solve the quadratic.
x+5/30=4x-10/75
2*3*5*5*(x+5)/2*3*5=2*3*5*5(4x-10)/3*5*5
5*(x+5)=2*(4x-10)
5x+25=8x-20
25+20=3x
3x=45
x=15
x=0, y=0
x=5,y=25
no other solutions
The x=15 is ok
Not sure what the last set of values is for, but if they are for the 1st problem, no good.
x^2 + (5-x)^2 = 25
x^2 + x^2-10x+25 = 25
2x^2-10x = 0
2x(x-5) = 0
x = 0 or x=5
So, the points of intersection are (0,5) and (5,0)
To find the points of intersection of the line X+Y=5 and the circle x^2+y^2=25, we can solve the system of equations by substitution or elimination.
Let's solve the system using substitution:
1. Solve one of the equations for one variable. In this case, let's solve the line equation X+Y=5 for X.
X = 5 - Y
2. Substitute the value of X in the second equation.
(5 - Y)^2 + Y^2 = 25
3. Simplify the equation.
25 - 10Y + Y^2 + Y^2 = 25
2Y^2 - 10Y = 0
4. Factor out 2Y from the equation.
2Y(Y - 5) = 0
5. Solve for Y by setting each factor equal to zero.
2Y = 0 or Y - 5 = 0
Y = 0 or Y = 5
6. Substitute the values of Y back into the equation for X.
X = 5 - 0 = 5 or X = 5 - 5 = 0
So we have two points of intersection:
Point 1: (X, Y) = (5, 0)
Point 2: (X, Y) = (0, 5)
Therefore, the points of intersection of the line X+Y=5 and the circle x^2+y^2=25 are (5, 0) and (0, 5).