10. Solve using the substitution method.

2x + 3y + z = 10

x - y + z = 4

4x - y - 5z = -8

Solve one of the equations for one of the unknowns. Which equation you solve for which unknown will not make a difference, so we look for the easiest unknown in the easiest equation. In this case, the y in the second equation looks very reasonable. Solve the second equation for y.

y = x + z - 4

Now substitute this formula in parentheses for y in the other two equations.

2x + 3(x + z - 4) + z = 10

4x - (x + z - 4) - 5z = - 8

This gives us two equations in x and z. Remove parentheses

2x + 3x + 3z - 12 + z = 10

4x - x - z + 4 - 5z = - 8

Combine like terms.

5x + 4z - 12 = 10

3x - 6z + 4 = -8

Transpose the numbers to the right

5x + 4z = 12 + 10

3x - 6z = -4 - 8

 

5x + 4z = 22

3x - 6z = -12

and we have reduced our problem down to a system of two equations in two unknowns.

At this point we could use any method at our disposal to solve this system, but for consistency's sake let us continue with the substitution method.

Divide the bottom equation by 3

x - 2z = -4

we can solve for x.

x = 2z - 4

If we substitute this into the other equation we get

5(2z - 4) + 4z = 22

Remove parentheses

10z - 20 + 4z = 22

Combine like terms

14z - 20 = 22

Transpose

14z = 20 + 22

14z = 42

Divide,

z = 42/14

z = 3

Now that we know what z is, we can substitute it into any equation which has only x's and z's and solve for x. The simplest equation would be the one where we expressed x as a function of z.

x = 2z - 4

This now becomes

x = 2(3) - 4

or

x = 6 - 4

x = 2

Now that we know both x and z, we can substitute them into any equation which also has a y and solve. The best one would be the equation we got by solving the second original equation for y.

y = x + z - 4

This now becomes.

y = (2) + (3) - 4

y = 1

leaving us with the solution of

x = 2

y = 1

z = 3

Which will check

Compare to the addition method, Cramer's rule, row operations and inverse matrices.

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