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Since the equation is in terms of x, you need to switch the x's and y's, and then solve it for y again. y=3x^3 --> x=3y^3 --> y=(x/3)^(1/3)
This is only half of an equation. If -x^2 + 3x +2y = 0, then x^2 -3x = 2y and y = (x^2-3x)/2 Solving for x is the inverse, but difficult without a second equation.
If you mean: 3x-y = 1 then it is a straight line equation
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If you mean: y-2 = 3x+2 then as a straight line equation it is y = 3x+4
So y = (3x - 4)/2. To find the find the inverse equation, y-1,(1) Rewrite the equation replacing all ys with xs and all xs with ys.x = (3y - 4)/2(2) Solve the new equation for yx = (3y - 4)/22x = 3y - 42x + 4 = 3y(2x + 4)/3 = yThis is your inverse equation!y-1 = (2x + 4)/3
3x+y=1 is the equation of a line with slope -3 and y intercpet 1.
y = 3x + 2 y-2 = 3x x = (y-2)/3 So the inverse is (x-2)/3
y - 3x = 8 standard form is y = mx + b m = slope y = 3x + 8 in this form slope m is 3 The perpendicular line slope is the negative inverse of m -1/m = -1/3
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Since the equation is in terms of x, you need to switch the x's and y's, and then solve it for y again. y=3x^3 --> x=3y^3 --> y=(x/3)^(1/3)
This is only half of an equation. If -x^2 + 3x +2y = 0, then x^2 -3x = 2y and y = (x^2-3x)/2 Solving for x is the inverse, but difficult without a second equation.
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