Solving Systems of Equations with Row Operations

Question

x + y + z = 3

3x - y - 2z = -4

3x + 4y + 3z = 11

Answer 1

You have a 3 x 3 system in x, y, and z the coefficients are inside the brackets the constants are outside

⌈1 1 1⌉ 3

¦ 3 -1 -2¦-4

⌊3 4 3⌋ 1

x = 0

y = 2

z = 1

First lets number the Equations

Equation 1

x + y + z = 3

Equation 2

3x - y - 2z = -4

Equation 3

3x + 4y + 3z = 11

Both x and z can be eliminated when Equation 1 is multiplied by -3 and combined with Equation 3

-3x - 3y - 3z = -9 (this is Equation 1 multiplied by -3)

3x + 4y - 3z = 11

Combining leaves

y = 2

Substitute y = 2 in the other two equations to get them in terms of x and z then combine them to eliminate another variable,

Equation 2

3x - y - 2z = -4

3x - 2 - 2z = -4

Leaves

3x - 2z = -2

Equation 1

x + y + z = 3

x + 2 + z = 3

Subtract 2 from both sides

Leaves

x + z = 1

Combine equations in x and z to solve for those variable

x + z = 1

3x - 2z = -2

Eliminate z by multiplying (x + z ) by 2

2x + 2z = 1

3x - 2z = -2

Leaves

5x = 0

x = 0

Plug y = 2 and x = 0 into Equation 1 to find z

x + y + z =3

0 + 2 + z = 3

Subtract 2 from both sides of the equation

z = 1

Check the values in the original equations

Equation 1

x + y + z = 3

0 + 2 + 1 = 3

3 = 3

Equation 2

3x - y - 2z = -4

3(0) - 2 -2(1)

0 -2 - 2 = -4

-4 = -4

Equation 3

3x + 4y + 3z = 11

3(0) + 4(2) + 3(1) = 11

0 + 8 + 3 = 11

11 = 11

Answer 2

I'll tell ya, these confused me greatly when I first started learning them.

If we look at the first two equations, we have y and - y, so adding the equations cancels out y.

4x - z = -1

z = 4x + 1

What if we subtract the 3rd from the 2nd?

-5y - 5z = -15

y + z = 3

Keep playing with it until you solve for all three variables.