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Solve the system of equations 3x plus y plus 2z equal 1 also 2x minus y plus z equals negative 3 also x plus y minus 4z equals negative 3?
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.
If 4z-1 equals 9 what is z?
4z - 1 = 9 4z = 10 (add 1 to both sides) z = 2.5 (divide both sides by 4)
How do you write an equivalent expression for 4x plus 4y plus 4z using the distributive property?
It is: 4(x+y+z)
What is 4z x -2z?
6
What equations will give you the answer to z equals 1?
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
