To solve for y in terms of x, divide both sides of the equation by 2: y = x/2.If x=2y then you have already solved for x.
You have 2 unknowns (x,y) and only one equation, so it is unsolvable if you are trying to solve for values for x and y. However, if you are just trying to solve for x in terms of y, then the answer is x=-4+2y.
No. There is not enough information in the equation x + 2y = 2, by itself, to solve it. There are an infinite number of solutions. A second equation, or information to allow a second equation to be derived, must be given to find a solution.
If the two equations are (i) 4x + 2y = 12 and (ii) x + 2y = 6, proceed as follows: Rewrite (ii) as x = 6 - 2y; substitute this expression for x into (i) to result in 4(6 - 2y) + 2y = 12, or 24 - 8 y + 2y = 12, or - 6 y = 12 -24 = -12, or y = -12/-6 = 2. Then, back to (ii) again, x = 6 - 2(2) = 2.
x2 - 4y2 = 16∴ (x - 2y)(x + 2y) = 162y - x = 2∴ x = 2y - 2∴ ([2y - 2] - 2y)([2y - 2] + 2y) = 16∴ (y - 1 - y)(y - 1 + y) = 16∴ -1(2y - 1) = 16∴ 1 - 2y = 16∴ -2y = 15∴ y = -7.52y - x = -2∴ -15 - x = -2∴ x = -13So the point of intersection is (-13, -7.5)
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Your answers are x = 2 & y = 1Original Equationsx - 2y = 03x + 2y = 8Solve for a variable (x)x = 2ySubstitute x with its equation3*(2y) + 2y = 86y + 2y = 88y = 8y = 1Solve for second variablex = 2*(1)x = 2Check your answer(2) - 2*(1) = 03*(2) + 2*(1) = 8
x = 2y y = x - 2 or x = 2y subtract 2y to both sides x - 2 = y subtract y and add 2 to both sides x - 2y = 0 multiply by -1 to both sides x - y = 2 -x + 2y = 0 x - y = 2 add these two equations together y = 2 substitute 2 for y into the first equation, x = 2y x = 2(2) = 4 Thus the solution of these linear system of equations is the point (4, 2). Check:
The answer to x2 - 2x - 4y2 - 4y =(x - 2y)(x - 2y - 2)
3x - 2y = 20 Substitute x = 4y into this equation: 3(4y) - 2y = 20 ie 12y - 2y = 20 10y = 20 so that y = 2 and then x = 4y gives x = 2*4 = 8
Given: 2y = x + 2 x - 3y = -5 ∴ y = x/2 + 1 ∴ x - 3(x/2 + 1) = -5 ∴ x - 3x/2 - 1 = - 5 ∴ -x/2 = -4 ∴ x = 2 ∴ 2y = 2 + 2 ∴ y = 2 ∴ {x, y} = {2, 2}
Given: 2x - 3y = 2 3x + 2y = 3 Take the first equation, and solve for x: x = (2 + 3y) / 2 Now plug it into the second equation: ∴ 3(2 + 3y) / 2 + 2y = 3 ∴ 3 + 9y/2 + 2y = 3 ∴ 9y/2 + 2y = 0 ∴ 22y = 0 ∴ y = 0 Then you can take that value of y, and plug it into either of our first equations to find x; 2x - 3y = 2 ∴ 2x - 3(0) = 2 ∴ 2x = 2 ∴ x = 1 So x is equal to one, and y is equal to zero.
X 2y-1 4x-3x=18 X 2y-1 7x=18 X 1y 7x=18 X Y 7x=18 X Y X = 2 4/7 and there is ur anser
-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x
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.
Sum = addition Difference = subtraction Say your first number is x and your second number is y. So: x+y = 10 and x-y = 20 Solve for one variable for one of the above equations and substitute that variable into the other equation: x = 10-y (10-y)-y = 20 10-2y = 20 Get y by itself: -2y = 20-10 -2y = 10 Solve for y: -2y/-2 = 10/-2 y = -5 To solve for x: x = 10-y x = 10-(-5) x = 10+5 x = 15 Therefore, the two numbers you are looking for are -5 and 15.
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