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The equation you provided, 3x - y = 1, cannot be solved for any specific values of x and y without additional information. It is a linear equation in two variables, so it represents a straight line on a graph. To find a solution, you would need another equation or constraint.
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How do you solve y -x 1 if x -3?
4
What is the implicit differentiation of y equals sin x plus y?
y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]
What is the derivative of y equals square root of x when x is positive?
y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)
Does the graph x-y2 equals 1 represent x as a function of y?
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
How do you simplify x plus y all over x ALL OVER x to the negative 1 plus y to the negative 1?
(x + y) / (x-1 + y-1) = (x + y) divided by ( 1/x + 1/y)Multiply numerator and denominator by 'xy' :xy (x + y) / (y + x)Then the whole thing falls apart, and we're left with:xy
What is y x -y?
(y * x) - y = y * (x - 1)
What is the answer of x plus y equals 1?
x+y=1 x=1-y or y=1-x or if you fill that in 1-y+y=1 1 = 1 or 1-x+x = 1 1 = 1
How do you differentiate y equals x minus a over x?
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
Why does't the number get bigger with multiplication?
The assertion in the question is false. The result of a multiplication depends on the values. Given two numbers X and Y, if X < 0 and if Y < 0 then X*Y is greater than either; if X > 1 and if Y > 1 then X*Y is greater than either; if X < 0 and if Y > 1 then X*Y is smaller than either; if X > 1 and if Y < 0 then X*Y is smaller than either; if 0 < X < 1 and if 0 < Y < 1 then X*Y is smaller than either; If X < 0 and if 0 < Y < 1 then X*Y is greater than X but smaller than Y; If 0 < X < 1 and if Y > 1 then X*Y is greater than X but smaller than Y; If 0 < X < 1 and if Y < 0 then X*Y is smaller than X but greater than Y; If X > 1 and if 0 < Y < 1 then X*Y is smaller than X but greater than Y.
Is there a no solution in y equals x-1 and y equals -x plus 1?
In the equations Y=X-1 and Y=-X+1, the solution is (1,0)
How do you factor x-y plus xy-1?
(x-y) + (xy - 1) = (x - 1)(y + 1)
How do you simplify x plus y all over x ALL OVER x to the negative 1 plus y to the negative 1?
(x + y) / (x-1 + y-1) = (x + y) divided by ( 1/x + 1/y)Multiply numerator and denominator by 'xy' :xy (x + y) / (y + x)Then the whole thing falls apart, and we're left with:xy
How should I solve x+y?
x y = x y Transposing x to the Left Hand Side will give us x y − x = y x ⋅ ( y − 1 ) = y ( x was a common factor on the Left Hand Side) Dividing both sides by ( −1) (Assuming y≠1) , we get x=yy−1
X-y equals 5 and x squared-y squared equals 5 How do you work this out using simultaneous equations?
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Factor 2x squared plus 2x plus xy plus y?
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How do you solve3 times 97?
97 x 3 is the same as 97 + 97 + 97 if that helps you understand how multiplication works, 97 x 3 = 291
How do you prove the multiplicative inverse is unique?
You can multiply each side of the multiplicative inverse equation by the other inverse to show that any two multiplicative inverses are equal. Here it is more formally. Theorem: For all x in R, there exists y in R s.t. x * y = 1. If there is a y' in R such that x * y' = 1, then y = y'. Proof: - Start with x * y = 1. - y * x = 1 (commutative) - (y * x) * y' = 1 * y' = y' - y * (x * y') = y' (associative) - y * 1 = y' (because x*y' = 1) - y = y'
