y + x = 9
10y + 20x = 100
This is a basic linear system of equalities. We can start by rewriting the first equation as:
-10y - 10x = -90 by multiplying the equation by -10, from here we can eliminate y and solve for x
(-10y - 10x = -90) + (10y + 20x = 100) = (10x = 10)
This yields x = 1, and by plugging x into the first equation y = 8
They touch each other at (0, 100) on the x and y axis.
x2 + 20x + 4 = 0 x2 + 20x + 100 = 96 (x + 10)2 = 96 x + 10 = ±961/2 x = -10 ±961/2 x = -10 ±4√6
2x + 20x = 111,45222x = 111,452x = 5,066
Improved Answer:-If: 20x+27 = 143Then: x = 5.8
11x = 20x +1811x-20x=18x(11-20)=18-9x=18x=-2
2
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)
The graphs of the two equations will intersect when x² + 20x + 100 = y = x² - 20x + 100 Subtracting x² +100 from both sides you get 20x = -20x that will only be true when x = 0. At x = 0, y = 100 for both equations - so the point of contact would be (0,100)
They touch each other at (0, 100) on the x and y axis.
x2 + 20x - 39 = 0: Add 61 giving: x2 + 20x + 100 = (x + 10)2
x2 + 20x + 4 = 0 x2 + 20x + 100 = 96 (x + 10)2 = 96 x + 10 = ±961/2 x = -10 ±961/2 x = -10 ±4√6
x2 + 20x +0 =30 [(20/2)2 =100] x2 + 20x + 100 =30 +100 √(x+10)2=√130 x+10=√130 x= −10+√130 √ means square root
2x + 20x = 111,45222x = 111,452x = 5,066
x2-20x+100 = (x-10)(x-10) when factored
Improved Answer:-If: 20x+27 = 143Then: x = 5.8
11x = 20x +1811x-20x=18x(11-20)=18-9x=18x=-2
20x + 80y = 0Subtract 20x from each side of the equation:80y = -20xDivide each side by 80:y = -1/4 xm = -1/4b = 0