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Solving Two Step Linear Equations


The solving one step equations page showed examples of solving linear equations where you could solve them in one step or operation, and then you were done.

We can also encounter equations that can be classed as two step equations, where you have to perform more than one step or operation to establish a solution to the equation.
This page will introduce some examples of solving two step linear equations and show how to approach some standard and common types of such equations.




Order of Steps:

The order of which to do certain steps in two step equation examples can differ, but some general approaches can be used to make things a bit quicker and easier depending on the set up of the equation.

If there is a division required, it can often be better to not perform this step first, as it can create a fraction or fractions that may make things a bit more complicated.

While there is no set order one should perform the steps in solving two step equation examples, with enough practice it can become easier to identify the best order.





Solving Two Step Linear Equations,
Examples



(1.1) 

Solve for x,     3x \space {\text{--}} \space 4 \space = \space 14.

Solution   

First, we can add  4  to both sides.

3x \space {\text{--}} \space 4 + 4 \space = \space 14 + 4     ,     3x \space = \space 18

Now a division by  3  will give a solution.

{\frac{3x}{3}} \space = \space {\frac{18}{3}}     ,     x \space = \space 6



(1.2) 

Solve for a,     {\text{-}}4a + 5 \space = \space 21.

Solution   

Firstly we can take away  5  from both sides.

{\text{-}}4a + 5 \space {\text{--}} \space 5 \space = \space 21 \space {\text{--}} \space 5     ,     {\text{-}}4a \space = \space 16

We can now divide by  –4  as the second step.

{\frac{{\text{-}}4a}{{\text{-}}4}} \space = \space {\frac{16}{{\text{-}}4}}     ,     a \space = \space {\text{-}}4




(1.3) 

a)    Solve for x,     6x + x \space = \space 28.

Solution   

When there are separate like terms in a linear equation, they should be combined before proceeding for a solution.

6x + x \space = \space 28     ,     7x \space = \space 28

{\frac{7x}{7}} \space = \space {\frac{28}{7}}     ,     x \space = \space 4


b)    Solve for x,     8x \space {\text{--}} \space 12 \space = \space 2x.

Solution   

We again want the like terms to be combined, though initially here they are on opposite sides of the  =  sign.
We can easily move terms and numbers across  =  signs, but just remember to change the nature of them when we do.

Say you had  5 + 3 = 8.
Moving the  3  across as is would result in   5 = 8 + 3,   which is INCORRECT.

But moving the  3  across and changing the nature results in   5 = 8 \space {\text{--}} \space 3,   which is CORRECT.

Changing the nature when moving across is something that is important to remember.


So here, moving the  2x  and the  12:    8x \space {\text{--}} \space 2x \space = \space 12

Now:    6x \space = \space 12

{\frac{6x}{6}} \space = \space {\frac{12}{6}}     ,     x \space = \space 2




(1.4) 

Solve for b,     4b \space {\text{--}} \space 7\space = \space {\text{-}}8b + 29.

Solution   

4b + 8b \space = \space 29 + 7

12b \space = \space 36

{\frac{12b}{12}} \space = \space {\frac{36}{12}}     ,     b \space = \space 3





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