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x - y + z = 1, -x + z = 1, x + y + z = 0. Solve for x, y and z.

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First add the first and the third equations together: 2x + 2z = 1. Then 2x = 1 - 2z, and x = (1 - 2z) / 2. Now express the second equation as z - x = 1, and plug the x value into it: z - (1 - 2z)/2 = z - 1/2 + z = 2z - 1/2 = 1. That means 2z = 1½, and z = (3/2) / 2 = 3/4. Now plug that z value into the second equation to get x = -1/4. Finally, plug our x and z values into the first or third equation to get y = -1/2.

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Solve the system of linear equation by using Cramer's rule x+y+z=-2, x-y+z=0, 4x+2y+z=3

I_l1ke_gam3s

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Multiply the first equation by -1. -x -y -z = 2. Add that to the second equation to get: -2y = 2. y = -1. Substitute y in equation 1 and equation 2. -x + 1 - z = 2, 4x + 2 + z = 3. Add these to eliminate z. 3x + 3 = 5, 3x = 2, x = 2/3. -2/3 + 1 - z = 2. 1/3 - z = 2. -z = 5/3. z = -5/3. It’s fast and simple without matrices, and it’s best to use elimination or substitution.

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How do I solve this: 3x 4y - 2z =5 & x-5y+6z=2 & x+y+z=3 ?

I_l1ke_gam3s

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Let’s take the second equation, and the third equation, x - 5y + 6z = 2 and x + y + z = 3. Subtract these two equations. -6y + 5z = -1, Then we take the first equation and the third equation. x + y + z = 3, and 3x + 4y - 2z = 5. Let’s multiply the first equation by -3 so we get -3x -3y -3z = -9. Then when we add, we get y - 5z = -3. Then let’s take the -6y + 5z = -1. Using elimination, we can eliminate the z’s by adding up and down, so -5y = -4, and y = 4/5. Then -6(4/5) + 5z = -1. -24/5 + 5z = -1. 5z = 19/5. Then multiply both sides by 1/5 to find z = 19/25. Then x + 4/5 + 19/25 = 3, x + 29/25 = 3, x = 46/25,

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How do I solve 3x3 matrix linear equations?

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You can use Cramer's rule if applicable, or try methods like elimination, substitution, and graphing.

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How do I solve 3x3 matrix linear equations?

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Follow the instructions above.

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