There is not enough information in the question to answer.
y can be any value, unless it is linked to x. The mathematical term would be that y is a function of x. For example, y(x) := 2 + 5x. In this case, y is a dependent variable (it depends on x), and it has a defined value for x = 3: y(1) = 2 + 5x3 = 17.
Without a known relation between y and x, however, y and x are independent and can assume any value.
x+y=3 y=3-2x so x+3-2x=3 so x=0 0+y=3 so y=3
x5 You have x*x*x*x*x just add up the amount of x's you have and put it after as an exponent Example: 3*3*3*3=34 y*y*y*y*y*y=y6
-1
Follow this example. f(x) = (x+3)/5 To find its inverse, write y=f(x) y= (x+3)/5 Interchange x and y x = (y+3)/5 solve for y in terms of x 5x=y+3 y=5x-3 The inverse of f(x) is f^-1(x) = 5x-3
0
x - y = -3 -y = -x - 3 y = x + 3 y-int. = 3 x-int. = -3
Two integers (X & Y). X+Y=-3, X-Y=-11. x=-11+y --> (x)+y=-3 --> (-11+y)+y=-3 --> y=4 x=-11+4=-7 Hope that helps! Two integers (X & Y). X+Y=-3, X-Y=-11. x=-11+y --> (x)+y=-3 --> (-11+y)+y=-3 --> y=4 x=-11+4=-7 Hope that helps!
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
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
4
6
there are 4 possible answers. X= 1 , Y=2 Y=1 , X=2 X=0 , Y-3 Y=0 , X=3
1545(x-y)=3(x²-y²) (1545÷3)(x-y)=(x²-y²) 515(x-y)=(x²-y²) 515=(x²-y²)÷(x-y)
The problem to solve is: xy+x+3y+3 Multiply y and x Multiply the y and x Multiply y and x The y just gets copied along. The x just gets copied along. The answer is yx yx x*y evaluates to yx x*y+x evaluates to yx+x Multiply y and 3 Multiply y and 1 The y just gets copied along. The answer is y y 3*y evaluates to 3y The answer is yx+x+3y x*y+x+3*y evaluates to yx+x+3y The answer is yx+x+3y+3 x*y+x+3*y+3 evaluates to yx+x+3y+3 ---- The final answer isyx+x+3y+3----
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)
X = -Y + 3 add Y to each side Y + X = 3 subtract X from each side Y = -X + 3
x+y=3 y=3-2x so x+3-2x=3 so x=0 0+y=3 so y=3