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)
6
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
-6z 2 = 3z+ 4z+28 -6z+2 +3z = 3z+ 4z +28 +4z -6z+4z+2 = 3z+ 28 -10z+2 -3z= 28 - 2 -10z + -3z = 26 -13z/13 = 26/-13 z = -2
5x-2y+3z-2x-y-4z=3x-3y-z
its 10x +9y +1z
-6z + 2 = 3z + 4z + 28 Add 6z to both sides: 2 = 3z + 4z + 28 + 6z = 28 + 13z Subtract 28 from both sides: - 26 = 13z Divide both sides by 13: -2 = z
3x-3y+7z
To solve for h, first isolate 4h: T=4h+4z T-4z=4h Now divide both sides by 4. T/4-z=h
7z - 3 = 11 + 3z ie 7z -3z = 11 + 3 4z = 14 z = 3.5 Check: (7 x 3.5) - 3; 11 + (3 x 3.5) 24.5 - 3 = 11 + 10.5 21.5 = 21.5 QED
7x + y + z = 6 7x + y + 4z = 9 y + x = 6 Take the first two statements and subtract them. 7x + y + z = 6 7x + y + 4z = 9 _____________ 0 + 0 - 3z = -3 Therefore: z = 1 Then you are left with: 7x + y + 1 = 6 7x + y + 4 = 9 which is impossible, so this is an impossible equation.
-8z + 12 = 12 - 4z-8z +4z + 12 = 12 - 4z +4z-4z + 12 -12 = 12 -12-4z / -4 = 0 / -4z = 0:)
To solve for 3 unknown variables you need 3 independent equations. You have only 1.
9z
7 + 4z = 437 + 4z -7 = 43 - 7 4z = 43 - 7 4z = 36 z = 9