# Solving Multi Step Linear Equations

The two step linear equations page showed a range of examples of solving linear two step equations.

It’s also possible to encounter what can be classed as multi step linear equations, where a larger number of steps need to be performed to solve the equation.

Though it is true that solving two step equations is by definition solving multi step linear equations, seeing as more than one step is required for solving. Multi step does often refer to equations needing more than 2 steps to solve.

For a brief recap, we can look at a couple of cases of two step equations.

Examples

(1.1)

Solve for  x:     4x + 2 = \space 14.

Solution

1)  Subtract 2 from both sides.
4x + 2 \space {\text{–}} \space 2 \space = \space 14 \space {\text{–}} \space 2    =>    4x = 12

2)  \frac{4x}{4} = \space \frac{12}{4}     ,     x = 3

(1.2)

Solve for  t:     3t + 3t = \space 24.

Solution

1)  Combine the like terms.
3t + 3t = \space 24    =>    6t = \space 24

2)  \frac{6t}{6} = \space \frac{24}{6}     ,     t = 4

## Solving Multi Step Linear Equations,Examples

(2.1)

Solve for  x:     2x = \space 20 \space {\text{–}} \space 3x.

Solution

1) Move 3x across.    2x + 3x = \space 20

2) Combine like terms.    5x = \space 20

3)    \frac{5x}{5} = \space \frac{20}{5}     ,     x = 4

(2.2)

Solve for  a:     2a + 8 = \space 4a \space {\text{–}} \space 16.

Solution

1)   2a \space {\text{–}} \space 2a + 8 = \space 4a \space {\text{–}} \space 2a \space {\text{–}} \space 16
=>   8 = \space 2a \space {\text{–}} \space 16

2)   Move 16 across.    8 + 16 = \space 2a
=>   24 = \space 2a

3)    \frac{24}{2} = \space \frac{2a}{2}     ,     12 = a

(2.3)

Solve for  x:     30 = \space 3(4x \space {\text{–}} \space 6).

Solution

1)   Eliminate the brackets present.    30 = \space 3(4x \space {\text{–}} \space 6)     =>     30 = \space 12x \space {\text{–}} \space 18

2)   Move 18 across.    30 + 18 = \space 12x     =>     48 = \space 12x

3)    \frac{48}{12} = \space \frac{12x}{12}     ,     4 = x

(2.4)

Solve for  r:     13 + 2r = \space 3r + 10 \space {\text{–}} \space r.

Solution

1)   13 + 2r = \space 3r + 10 \space {\text{–}} \space r
=>   13 + 2r = \space 2r + 10

2)   Subtract 2r from both sides.

13 + 2r \space {\text{–}} \space 2r = \space 2r \space {\text{–}} \space 2r + 10 &nbsp =>   13 = 10

Here after 2 steps it can be seen that there is NO solution for r, as we are left with 13 equals 10 which is untrue.

This example shows that although we start off a linear equation assuming that there will be a solution for a variable, there are times where there isn’t a solution.

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