Given: 3x + y + 2z = 1 2x - y + z = -3 x + y - 4z = -3 Take any one of the equations (we'll use the first one), and solve for any one of the variables (we'll use y): y = 1 - 2z - 3x Now plug that value of y into the latter two equations: 2x - (1 - 2z - 3x) + z = -3 x + (1 - 2z - 3x) - 4z = -3 Now take either of those (again, we'll use the first one), and solve it for either of the remaining variables (we'll go for x): 2x - 1 + 2z + 3x + z = -3 ∴ 5x = -2 - 3z ∴ x = (3z + 2) / -5 Now take that value, and plug it into our other equation that uses x and z: (3z + 2) / -5 + 1 - 2z - 3(3z + 2) / -5 - 4z = -3 Then solve for z: ∴ 2(3z + 2) / 5 - 6z = -4 ∴ (6z + 4 - 30z) / 5 = -4 ∴ 4 - 24z = -20 ∴ 24z = 24 ∴ z = 1 Now we can take that value for z, and plug it back into our previous equation for x and z: x = (3z + 2) / -5 ∴ x = (3 + 2) / -5 ∴ x = -1 Finally, we can take those two values, and plug them into our equation for y: y = 1 - 2z - 3x ∴ y = 1 - 2 + 3 ∴ y = 2 So x = -1, y = 2, and z = 1 You can test these values by plugging them into each of the original three equations, and seeing if they solve correctly: 3x + y + 2z = 1 ∴ 3(-1) + 2 + 2(1) = 1 ∴ 2 + 2 - 3 = 1 ∴ 1 = 1 2x - y + z = -3 ∴ 2(-1) - 2 + 1 = -3 ∴ -2 - 2 + 1 = -3 ∴ -3 = -3 x + y - 4z = -3 ∴ -1 + 2 - 4 = -3 ∴ -3 = -3 which shows our answer to be correct.
4z - 1 = 9 4z = 10 (add 1 to both sides) z = 2.5 (divide both sides by 4)
It is: 4(x+y+z)
(x+y)^2+z^2=x^2+y^2+z^2+2xy or ((x+y)^2+z)^2= (x^2+y^2+2xy+z)^2= x^4+y^4+z^2+6x^2y^2+4x^3y+2x^2z^2+4xy^3+4xyz^2+2z^2y^2
Start with the equation z = 1, then do any valid manipulation on the left side, and the same manipulation on the right side. The manipulations that won't change the solution set are: adding the same constant to both side; multiplying both sides by the same non-zero constant. Squaring and taking roots will sometimes add additional solutions, or eliminate solutions. For example, and this is just an example of what you can do: z = 1 Multiply both sides by 2: 2z = 2 Add 5 to both sides: 2z + 5 = 7 Add 1 to each side: 2z + 6 = 8 Multiply both sides by 2: 4z + 12 = 16 Add z to both sides: 5z + 12 = z + 16
Since 2z^2 is a factor fo 4z^3, it is automatically the GCF.
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4z + 3 = 6 + 2zSubtract 2z from each side of the equation:2z + 3 = 6Subtract 3 from each side:2z = 3Divide each side by 2:z = 3/2 = 1.5
The GCF is 2z
There are a few ways to isolate a variable. First, add or subtract anything to get a monomial on one side containing the variable. Then divide by anything that is being multiplied to the variable. Here is an example: xy-4z=2z-4y solving for x xy=(2z+4z)-4y xy=6z-4y x=(6z-4y)/y
6xy^2z^2 over 6xy^4z^4 or 1 over y^2z^2
I used the matrix method to find the answer: x=4, y=-7, z=-5.
(-3) x (-2z - 7) = 6z + 21 = 3 (2z + 7)
Well, you combine the like terms (-4z and 4z). -4z - 4z = -8z Now I'm not sure what to do after this because you haven't written the operation that comes before the 9, whether it's +, -, x, or /
z = 18.5 6z - 24 = 2z + 50 Take the -24 to the other side of the equals and change its sign; take the 2z to the other side of the equals and change its sign: => 6z - 2z = 50 + 24 => 4z = 74 Divide both sides by 4: => z = 18.5
Given: 3x + y + 2z = 1 2x - y + z = -3 x + y - 4z = -3 Take any one of the equations (we'll use the first one), and solve for any one of the variables (we'll use y): y = 1 - 2z - 3x Now plug that value of y into the latter two equations: 2x - (1 - 2z - 3x) + z = -3 x + (1 - 2z - 3x) - 4z = -3 Now take either of those (again, we'll use the first one), and solve it for either of the remaining variables (we'll go for x): 2x - 1 + 2z + 3x + z = -3 ∴ 5x = -2 - 3z ∴ x = (3z + 2) / -5 Now take that value, and plug it into our other equation that uses x and z: (3z + 2) / -5 + 1 - 2z - 3(3z + 2) / -5 - 4z = -3 Then solve for z: ∴ 2(3z + 2) / 5 - 6z = -4 ∴ (6z + 4 - 30z) / 5 = -4 ∴ 4 - 24z = -20 ∴ 24z = 24 ∴ z = 1 Now we can take that value for z, and plug it back into our previous equation for x and z: x = (3z + 2) / -5 ∴ x = (3 + 2) / -5 ∴ x = -1 Finally, we can take those two values, and plug them into our equation for y: y = 1 - 2z - 3x ∴ y = 1 - 2 + 3 ∴ y = 2 So x = -1, y = 2, and z = 1 You can test these values by plugging them into each of the original three equations, and seeing if they solve correctly: 3x + y + 2z = 1 ∴ 3(-1) + 2 + 2(1) = 1 ∴ 2 + 2 - 3 = 1 ∴ 1 = 1 2x - y + z = -3 ∴ 2(-1) - 2 + 1 = -3 ∴ -2 - 2 + 1 = -3 ∴ -3 = -3 x + y - 4z = -3 ∴ -1 + 2 - 4 = -3 ∴ -3 = -3 which shows our answer to be correct.
post an equation first.