The question software does not support mathematical symbols, so you must use as much English as possible. If you meant y = 3x - 2, then you have the equation of a line. There is no definite solution to it. There are pairs of numbers (x,y) that make the equation true. Those pairs define points on a graph. All those points will fall on a straight line. All the points on the line will make the equation true. For your equation, the line crosses the y-axis at -2. It has a positive slope of three, which means for every increment of one in the positive x direction, there is an increment of three in the positive y direction.
There is a missing equals. When the equation of a line is written in the form: y = mx + c m is the gradient/slope of the line; and c is the y intercept. If your line is ``y = 3/2 x'' then y = 3/2 x → y = 3/2 x + 0 → slope is 3/2 & y intercept is 0.
y = X^2 Is this what you mean? functions are y values when x = 1; y =1 when x = -1; y = 1 when x = 2; y = 4 when x = -2; y = 4 when x = 3; y = 9 when x = -3; y = 9 get it? A upward opening parabola with its vertex at the origin.
2X-Y=3 X+Y=3 ---------- 3X = 6 X=2 2(2)-Y=3 4-Y=3 Y=1 Point of interesection: (2,1).
2xy substitute 3 for y and 5 for x 2(5)(3) = 30
if y = 2- x then x^2 + 2x(2 - x) = 3 ie x^2 + 4 x -2x^2 = 3 ie 4x - 3 = x^2 x^2 - 4x +3 = 0 = (x - 1)(x - 3) so x = 1 or 3 Check: y= 2 - 1 =1; x^2 = 1, 2xy = 2 x 1 x 1 = 2, 1 +2 = 3. y = 2 - 3 = -1; x^2 = 9, 2xy = 2 x 3 x -1 = -6, 9 + -6 = 3. QED.
To find the inverse you switch the x and the y and then solve for y. x=2 radical( y + 3) radical(y + 3) = x/2 y+3= (x/2)² y = (x/2)² -3 So the answer is y = (x/2)² -3
X - Y = 2 When X = 3 you have 3 - Y = 2 Moving Y to RHS: 3 = Y + 2 Moving 2 to LHS: 1 = Y
y = 3√x y = 3x^(1/2) y' = 3(1/2)x^(1/2 -1) y'= (3/2)x^(-1/2) y' = 3/[2x^(1/2)] y' = 3/(2√x)
y^2 X y^3 = y^(2 + 3) = y^5 You can only do this if the coefficient 'y' is the same for both terms. Remember y^2 = y X y y^3 = y X y X y Hence y^2 X y^3 = y X y X y X y X y = y^5 Similarly for division/subtraction y^3 / y^2 = y^(3 - 2 ) = y^1 = y The power of '1' is trivial and not normally shown. NB You CANNOT do z^2 X y^3 by adding the indices. z^2 X y^3 is (z^2)*(y^3)
let y= xsqrt(x) -1 y= x^(3/2) -1 ---- since xsqrt(x) is the same as x^(3/2) y' = (3/2) x^(3/2-1) y' = (3/2) x^(1/2) y'' = (3/2) (1/2) x^(1/2-1) y'' =(3/2)(1/2) x^(-1/2) y'' = 3/4x^(1/2) y'' = 3 / 4sqrt(x)
Explanation: The difference of squares identity can be written: a 2 β b 2 = ( a β b ) ( a b ) The difference of cubes identity can be written: a 3 β b 3 = ( a β b ) ( a 2 a b b 2 ) The sum of cubes identity can be written: a 3 b 3 = ( a b ) ( a 2 β a b b 2 ) So: x 6 β y 6 = ( x 3 ) 2 β ( y 3 ) 2 = ( x 3 β y 3 ) ( x 3 y 3 ) = ( x β y ) ( x 2 x y y 2 ) ( x y ) ( x 2 β x y y 2 ) If we allow Complex coefficients, then this reduces into linear factors: = ( x β y ) ( x β Ο y ) ( x β Ο 2 y ) ( x y ) ( x Ο y ) ( x Ο 2 y ) where Ο = β 1 2 β 3 2 i = cos ( 2 Ο 3 ) sin ( 2 Ο 3 ) i is the primitive Complex cube root of 1 .
There are an infinite number of them. Here are a few: x= -2, y= -5/3 x= -1, y= -4/3 x= 0, y= -1 x= 1, y= -2/3 x= 2, y= -1/3 x= 3, y= 0 x= 12, y= 3
(-x) + (-y) = -(x + y) Example: x = 2, y = 3 -2 + -3 = -5
there are 4 possible answers. X= 1 , Y=2 Y=1 , X=2 X=0 , Y-3 Y=0 , X=3
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
2x=3-y 2y=12-x y=3-2x 2(3-2x)=12-x 6-4x=12-x -3x=6 x=-2 2(-2)=3-y -4=3-y y=7 x=-2 y=7
y = x + 2 y = -x + 4 x + 2 = -x + 4 2x + 2 = 4 2x = 2 x = 1 y = x + 2 y = 1 + 2 y = 3 (1, 3)