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The solution to the equation y = 1/2x will be x= 2y.
To solve this system of equations using substitution, we can isolate one variable in one equation and substitute it into the other equation. From the second equation, we can express x in terms of y as x = 4 + 2y. Then, substitute this expression for x into the first equation: 4(4 + 2y) - 3y = 1. Simplify this equation to solve for y. Once you find the value of y, substitute it back into x = 4 + 2y to find the corresponding value of x.
As an equation, x = -2y cannot be solved without further information. To solve an equation (or multiple equations), the number of unknowns (x, y etc.) must be equal to or less than the number of equations. In your example you have 2 unknowns and 1 equation, so it does not work.
This is a system of two linear equations: x+2y=11 3x-4y=-17 One way to solve this system is to solve for x (or y) in the first equation and then plug it into the second equation to solve for the other variable, which then allows us to solve for our original variable. (1) Take the first equation and solve for x: x=11-2y (2) Plug this value into the second equation: 3x-4y=3(11-2y)-4y=33-10y=-17. We see that y=5. (3) We now plug this into either equation to solve for x: From (1): x=11-2y=11-2(5)=1 We get x=1 and y=5. It's also always a good idea to plug the numbers we found into our original two equations to verify they work. Indeed, x+2y=1+2(5)=1+10=11, and 3x-4y=3(1)-4(5)=3-20=-17.
If you mean: 80 = 3y+2y+4+1 then 80 =5y+5 and so y = 15
To solve the equation ( xy^0 + 5x + 2y - 3 = 0 ), note that ( y^0 = 1 ), simplifying the equation to ( x + 5x + 2y - 3 = 0 ) or ( 6x + 2y - 3 = 0 ). Rearranging gives ( 2y = 3 - 6x ), leading to ( y = \frac{3 - 6x}{2} ). This represents a linear relationship between ( x ) and ( y ).
x - 2y = -4 2x - y = 1 To solve, multiply both sides of 2nd equation by -2 (remember if you do the same thing to both sides you do not change the equation), then add equations. x - 2y = -4 -4x + 2y = -2 adding, -3x = -6 divide by -3 both sides x = 2 substitute x = 2 into 2nd equation 2(2) - y = 1 4 -y = 1 -y = -3 y = 3 Ans: x = 2, y = 3
It is not possible to do so because the question contains one equation (5x + y = 1) and one expression (3x + 2y + 2). An expression cannot be solved.
If: x-2y = 1 and 3xy -y^2 = 8 Then: x = 1+2y and 3(1+2y)y -y^2 = 8 Hence: 3y+6y^2 -y^2 = 8 => 5y^2 +3y -8 = 0 Solving the above quadratic equation: y =1 or y = -8/5 Solutions by substitution are: when y=1 then x=3 and when y=-8/5 then x=-11/5
get it to y= -2y = 1 - 2x y = x - 0.5
2y + 7x + y-1?
x=1, y=1