Introductory Algebra

1.4 – Solving Linear Equations

Solving Linear Equations

A Linear Equation is an equation that involves a single variable to the first power. All of the equations we dealt with in the previous sections were linear equations. Here, we will devise a general strategy for solving linear equations.

 

Solving Linear Equations
  1. Multiply both sides of the equation by the Lowest Common Denominator (LCD) to clear fractions, if there are any.
  2. Use the distributive property if necessary to remove parentheses.
  3. Simplify each side of the equation by combining like terms.
  4. Get all variable terms on one side of the equation, and all non-variable terms on the other by using the addition principle of equalities.
  5. Completely isolate the variable by using the multiplication principle of equalities.
  6. Check your solution by substituting it into the original equation.

Let’s view this process through a few examples:

Solve: 2(3x+5)-4=5(x-1)
Because there are no fractions here, we can skip step 1 and go directly to step 2. We distribute:
2(3x+5)-4=5(x-1)

6x+10-4=5x-5

Then we simplify each side of the equation (step 3):
6x+\underbrace{10-4}_{6}=5x-5

6x+6=5x-5

And then get all x values onto one side. here it is easy to get them to the left by subtracting 5x from each side.
6x+6 {\color{red}-5x}=5x-5 {\color{red}-5x}

x+6=-5

We will also move the non x term to the other side by subtracting the 6:
x+6{\color{red}-6}=-5 {\color{red}-6}

x=-11

We can now also skip step 5, since x has been completely isolated. We will then check our solution in the original equation (step 6):
2(3x+5)-4=5(x-1)

2({\color{red}3(-11)}+5)-4=5((-11)-1)

2({\color{red}-33+5})-4=5({\color{red}-11-1})

{\color{red}2(-28)}-4={\color{red}5(-12)}

-56-4=-60

-60=-60 \checkmark

 

Solve: 4(n+3)+8=-2(n-7)-3
Again, there are no fractions, so we skip to step 2, distributing gives:
4(n+3)+8=-2(n-7)-3

4n+12+8=-2n+14-3

Combining like terms yields:
4n+20=-2n+11

it is easy to isolate the variable terms onto the left side of the equation, and the non-variable terms onto the right. Doing so yields:
6n=-9

And finally, solving the equation for x by dividing each side by 6 yields:
n=-\frac{9}{6}

Which can be simplified to:
x=-\frac{3}{2}$

We can check this by putting it into the original equation:
4(n+3)+8=-2(n-7)-3$

4\left(\left(-\frac{3}{2}\right)+3\right)+8=-2\left(\left(-\frac{3}{2}\right)-7\right)-3$

4\left(-\frac{3}{2}+\frac{3}{1}\right)+8=-2\left(-\frac{3}{2}-\frac{7}{1}\right)-3$

4\left(-\frac{3}{2}+\frac{6}{2}\right)+8=-2\left(-\frac{3}{2}-\frac{14}{2}\right)-3$

4\left(\frac{3}{2}\right)+8=-2\left(-\frac{17}{2}\right)-3$

6+8=17-3

14=14 \checkmark

Some problems with fractions will require us to deal with them first (Step 1):

Solve: \dfrac{x}{4}-\dfrac{1}{3}=\dfrac{2}{3}x+\dfrac{2}{3}
In this example we have fractions to deal with, so we use step 1, and multiply each side by the LCD which is 12:
{\color{red}12}\cdot\left(\dfrac{x}{4}-\dfrac{1}{3}\right)={\color{red}12}\cdot\left(\dfrac{2}{3}x+\dfrac{2}{3}\right)

The twelve gets distributed through on each side, and we can write it as an improper fraction if we wish:
\dfrac{12}{1}\cdot\dfrac{x}{4}-\dfrac{12}{1}\cdot\dfrac{1}{3}=\dfrac{12}{1}\cdot\dfrac{2}{3}x+\dfrac{12}{1}\cdot\dfrac{2}{3}

And simplifying:
\dfrac{\cancel{12}^{\ 3}}{1}\cdot\dfrac{x}{\cancel{4}_{\ 1}}-\dfrac{\cancel{12}^{\ 4}}{1}\cdot\dfrac{1}{\cancel{3}_{\ 1}}=\dfrac{\cancel{12}^{\ 4}}{1}\cdot\dfrac{2}{\cancel{3}_{\ 1}}x+\dfrac{\cancel{12}^{\ 4}}{1}\cdot\dfrac{2}{\cancel{3}_{\ 1}}

yields:
3x-4=8x+8

There is no need to remove parentheses, so we skip step 2, and there is also no need to simply each side, so we can skip step 3. We will, however, move all x terms to one side of the equation. If we move them to the right side, we can avoid having a negative x term:
3x-4-3x=8x+8-3x

-12=5x

And then using the multiplication principle:
-\frac{12}{5}=x

Checking this answer by substituting into the original equation will be left to the student.

When there is a fraction on the outside of a parentheses, it might make the problem easier to distribute first:

Solve: \frac{1}{2}(x+3)=\frac{3}{5}x-\frac{1}{10}
{\color{red}\frac{1}{2}}(x+3)=\frac{3}{5}x-\frac{1}{10}

\frac{1}{2}x+\frac{3}{2}=\frac{3}{5}x-\frac{1}{10}

Our LCD here is 10, so we multiply every term by 10:
{\color{red}10\cdot}\frac{1}{2}x+{\color{red}10\cdot}\frac{3}{2}={\color{red}10\cdot}\frac{3}{5}x-{\color{red}10\cdot}\frac{1}{10}

and simplify:
^5 \cancel{10}\frac{1}{\cancel{2}}x+^5\cancel{10}\cdot\frac{3}{\cancel{2}}=^2\cancel{10}\cdot\frac{3}{\cancel{5}_2}x-\cancel{10}\cdot\frac{1}{\cancel{10}}

5x+15=6x-1
And we solve:
5x+15=6x-1

{\color{red}\underline{\ \ \ -5x+1 \phantom{.} -5x +1\ \ \ \ }}

16=x

There are, on occasion, equations without solutions:

Solve: 4(2+x)+1=7x-3(x-2)
Going through the general strategy, we have:
4(2+x)+1=7x-3(x-2)

8+4x+1=7x-3x+6

4x+9=4x+6

At this point, when we try to isolate our x term, we find that both x terms will cancel out, and we are left with:
9=6

this is a blatantly false statement. When our algebra results in a false statement such as this, then the statement will be false no matter what value of x we substitute, so there are no solutions. We might note this with the symbol \emptyset

Then there are solutions that have an infinite number of solutions.

Solve: 3x+2=3(x-4)+14
Again, through the general strategy, we have:
3x+2=3(x-4)+14

3x+2=3x-12+14

3x+2=3x+2

And we can see at this point that the expression on the left is the same as the one on the right. If we try to further isolate x terms, we will again eliminate them completely:
2=2

In this case, however, we come to a completely true statement, and no value of x will ever make this untrue. This is called an identity. The solution to this equation is all real numbers. This is often indicated by the use of the symbol \mathbb{R}

 

The last type of equation in this section is those with a variable in the denominator. These problems will be treated with the same process from above. Eliminate the fractions by multiplying by the lowest common denominator. Let us view an example of this:

Solve the equation for x:
\frac{x}{x+2}=3

The start, the lowest common denominator is multiplied. In this case, the lowest common denominator is x+2.

\frac{x}{x+2}=3

{\color{red}(x+2)\cdot}\frac{x}{x+2}=3{\color{red}\cdot(x+2)}

On the left side of the equation, (x+2) cancels out. On the right hand side, the 3 can distribute:
{\color{red}\cancel{(x+2)}\cdot}\frac{x}{\cancel{x+2}}=3x+6}

x=3x+6

Subtract x from each side to get the x terms to one side:
0=2x+6

and then subtract 6 from each side to isolate the x term:
-6=2x

And finally, divide each side by 2:
-3=x

 

Solve the equation for x
\frac{1}{x+2}=\frac{3}{x+4}

The lowest common denominator in this equation is the product of both denominators: (x+2)(x+4). This is multiplied to both sides of the equation:

{\color{red}(x+2)(x+4)}\frac{1}{x+2}=\frac{3}{x+4}{\color{red}(x+2)(x+4)}

The like-factors cancel:
{\color{red}\cancel{(x+2)}(x+4)}\frac{1}{\cancel{x+2}}=\frac{3}{\cancel{x+4}}{\color{red}(x+2)\cancel{(x+4)}}

Leaving a factor on each side:
1(x+4)=3(x+2)

Distributing on both sides:
x+4=3x+6

Moving x terms to the right by subtracting x:
4=2x+6

And moving the non-variable terms to the left by subtracting 6:
-2=2x

And finally dividing each side by 2 gives
-1=x