If: x+y = 4 and y = 2x+1 Then: 4-x = 2x+1 => 3 = 3x => 1 = x So by substitution: x = 1 and y = 3
x = -2 and y = 1
-2
x2 + y2 = 1x2- x2+ y2= 1 - x2y2 = 1 - x2y =± √(1 - x2)
In that instance, it means that the lines never touch.
If: x+y = 4 and y = 2x+1 Then: 4-x = 2x+1 => 3 = 3x => 1 = x So by substitution: x = 1 and y = 3
x = -2 and y = 1
By a process of elimination and substitution the lines intersect at: (1/4, 0)
-2
Substitution method: y = 1 + x so 2x + (1 + x) = 4, ie 3x = 3 so x = 1 and y = 2
4x+4y=16 y=5 4x+ 20=16 4x=-4 x=-1
x2 + y2 = 1x2- x2+ y2= 1 - x2y2 = 1 - x2y =± √(1 - x2)
x - 2y=0 so x = 2y 9y - 4x=1 Substituting for x, 9y-4*(2y) = 1 or 9y-8y = 1 so that y = 1 then x = 2y = 2*1 = 2
y= -4x-7 y=3x 3x= -4x-7 3x+4x=7 7x=7 x=1
The point of intersection of the given simultaneous equations of y = 4x-1 and 3y-8x+2 = 0 is at (0.25, 0) solved by means of elimination and substitution.
In that instance, it means that the lines never touch.
5x+8 = 10x+35x-10x = 3-8-5x = -5x = 1 and by substituting the value of x into the equations y = 13