That equation has no "solution" ... no single pair of numbers for 'x' and 'y' that makes it true.
In fact, it has an infinite number of them.
Here are a few:
x=1, y=-4
x=2, y=-9
x=-3, y=16
y = -5x + 1 is the equation of a line.
Every point on the line has the coordinates (x,y), and every point satisfies the equation.
5x+1=20 5x=20-1 5x=19 5x/5=19/5 x=19/5
x=8
f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
2x + 5x - 1 = 9x 7x - 1 = 9x -1 = 9x - 7x -1 = 2x -1/2=x
7x -3 = 5x + 1 solve for xadd three to both sides7x = 5x + 4subtract 5x from both sides2x = 4divide both sides by 2x = 2
-5x+1 = 6x -5x-6x = -1 -11x = -1 x = 1/11
5x+1=20 5x=20-1 5x=19 5x/5=19/5 x=19/5
If: 5x+1 = 21 Then: 5x = 21-1 => 5x = 20 Divide both sides by 5 and so x = 4
x=8
5x + 9 = 4Subtract 9 from each side:5x = -5Divide each side by 5:x = -1
f(x) = x2 + 5x + 1 The roots of this equation are x = -0.2087 and x = -4.7913 (approx).
To solve for two unknowns (x and y) it is necessary to have two independent equations.
2x + 5x - 1 = 9x 7x - 1 = 9x -1 = 9x - 7x -1 = 2x -1/2=x
7x -3 = 5x + 1 solve for xadd three to both sides7x = 5x + 4subtract 5x from both sides2x = 4divide both sides by 2x = 2
the question is to solve (x^2-5x+5)^(x^2-36)=1
7X-1=9+5X. add 1 to both sides 7X=10+5X. Subtract 5X and get 2X=10 divide by 2 and x=5
3x+5=4-2x+6 5x=4-1 5x=3 x=3/5