Solving Equations

A Linear Equation is any equation that can written in the form \(\boldsymbol{ax + b = c}\) where \(a\), \(b\), and \(c\) are all constant values. Make note that the variable can be expressed using different letters aside from \(x\) such as \(n\) or \(m\).
To solve the linear equation, we must calculate the Solution, a value that makes the equation come true when put in place of the variable.

Equations can also be solved graphically. We can do so by identifying the value at the corresponding \(x\)-coordinate.

Process for Solving Linear Equations

  1. If the equation contains fractions on either side, use the Lowest Common Denominator (LCD) to clear the fractions
  2. Simplify both sides of the equation by clearing out any parentheses using distributive property and combining like terms
  3. Rearrange the equation so that all variables are on one side and all constants are on the other side
  4. If there is a coefficient on the variable \((a \neq 1)\), divide or multiply both sides by the same term so that \(a = 1\)
  5. Check your work! Do this by inputting the solution into the variable of the original equation to verify that both sides are the same

Example

Solve \(12 = 3x\).

First, as there aren't any fractions or parantheses, we can skip Steps 1 and 2. Likewise, as variables and constants are already on seperate sides, we can skip Step 3.

As we can identify the variables coefficient (or \(a\)) as \(3\), we need to divide both sides by \(3\) to get the solution:

\(\cfrac{12}{\textcolor{red}{3}} = \cfrac{\cancel3x}{\cancel{\textcolor{red}{3}}}\)

\(x = 4\)

To verify the result, we can plug the solution back into the original equation:

\(12 = 3(4)\)

\(12 = 12\)

Therefore, we can determine that \(\boldsymbol{x = 4}\).


Example

Solve \(\cfrac{x}{11}=3\)

First, as the variable contains a fraction, we need to multiply both sides by the denominator, \(11\), to get the solution:

\(\textcolor{red}{11}\left(\cfrac{x}{11} = 3\right)\)

\(\textcolor{red}{11}\left(\cfrac{x}{11}\right) = \textcolor{red}{11}(3)\)

\(x = 33\)

As we have already solved the \(x\)-value, we can skip steps 2, 3, and 4.

To verify the result, we can plug the solution back into the original equation:

\(\cfrac{33}{11}=3\)

\(11 = 11\)

Therefore, we can determine that \(\boldsymbol{x = 33}\).


Strategies for Solving Linear Equations

Addition and Subtraction Properties of Equality
  • If \(a = b\), then \(a + c = b + c\)
  • If \(a = b\), then \(a - c = b - c\)

These rules essentially state that when 2 expressions are equal to each other, we can add or subtract a value, \(c\), to both expressions while they remain equal.


Multiplication and Division Properties of Equality
  • If \(a = b\), then \(a(c) = b(c)\)
  • If \(a = b\), then \(a/c = b/c\)

These rules essentially state that when 2 expressions are equal to each other, we can multiply or divide a value, \(c\), to both expressions while they remain equal.


Solve \(3(x+6) = 2(x-1)\)

First, we can simplify the equation by expanding the expressions on both sides and collecting like terms:

\(3(x) + 3(6) = 2(x) -2(1)\)

\(3x + 18 = 2x -2\)

Next, we can shift all like terms onto separate sides by using the Subtraction Property of Equality. In this case, we can subtract \(2x\) and \(18\) from both sides to keep variables on the left side and constants on the right side:

\(3x + 18 \textcolor{red}{- 2x - 18} = 2x -2 \textcolor{red}{- 2x - 18}\)

\(x = -20\)

To verify the result, we can plug the solution back into the original equation:

\(3(-20 + 6) = 2(-20 - 1)\)

\(3(-14) = 2(-21)\)

\(-42 = -42\)

Therefore, we can determine that \(\boldsymbol{x = -20}\).


Solve \(\cfrac{k}{4}-2=-\cfrac{4}{3}\)

First, as both sides contain fractions, we need to multiply the equation by the lowest common denominator. As the denominators are \(3\) and \(4\), we can determine the LCD to be \(12\):

\(\textcolor{red}{12}\left(\cfrac{k}{4}-2=-\cfrac{4}{3}\right)\)

\(\textcolor{red}{12}\left(\cfrac{k}{4}\right) - \textcolor{red}{12}(2) = \textcolor{red}{12} \left(-\cfrac{4}{3}\right)\)

\(3k - 24 = -16\)

Next, we can shift all like terms onto separate sides by using the Addition Property of Equality. In this case, we can add \(24\) to both sides to keep variables on the left side and constants on the right side. Then, solve for \(k\):

\(3k - 24 \textcolor{red}{+ 24} = -16 \textcolor{red}{+ 24}\)

\(\cfrac{3k}{\textcolor{red}{3}} = \cfrac{8}{\textcolor{red}{3}}\)

\(k = \cfrac{8}{3}\)

To verify the result, we can plug the solution back into the original equation:

\(\cfrac{8/3}{4} - 2 = -\cfrac{4}{3}\)

\(\left(\cfrac{8}{3}\right)\left(\cfrac{1}{4}\right) - 2 = -\cfrac{4}{3}\)

\(\cfrac{8}{12} - \cfrac{24}{12} = -\cfrac{4}{3}\)

\(-\cfrac{16}{12} = -\cfrac{4}{3}\)

\(-\cfrac{4}{3} = -\cfrac{4}{3}\)

Therefore, we can determine that \(\boldsymbol{x = \cfrac{8}{3}}\).


Unique Types of Equations

These uncommon forms of linear equations are notable for not having a single solution. These include:

Contradictions

Contradictions are equations with one variable that have no solutions.

Example
\(2x + 4 = 2x - 3\)
\(2x + 4 - (2x + 4) = 2x - 3 - (2x + 4)\)
\(0 = -7\)

Identities

Identities are equations with one variable that have all real numbers as solutions.

Example
\(9x + 3x = 12x\)
\(12x = 12x\)
Determine if the following linear equations contain \(1\), \(0\), or multiple solutions.

\(x + 4 = 3x - 2(x - 2)\)

First, we can expand the expression on the right side of the equation:

\(x + 4 = 3x - 2(x) - (2)(-2)\)

\(x + 4 = 3x - 2x + 4\)

Next, we can simplify the expression on the right of the equation by collecting like terms:

\(x + 4 = x + 4\)

Therefore, as this equation represents an identity, we can determine that its contains multiple solutions.


\(\cfrac{3}{2}x + 5 = \cfrac{15}{5}x + 7\)

First, as both sides contain fractions, we need to multiply the equation by the lowest common denominator. As the denominators are \(2\) and \(5\), we can determine the LCD to be \(10\):

\(\textcolor{red}{10}\left(\cfrac{3}{2}x + 5 = \cfrac{15}{5}x + 7\right)\)

\(\textcolor{red}{10}\left(\cfrac{3}{2}x\right) + \textcolor{red}{10}(5) = \textcolor{red}{10}\left(\cfrac{15}{5}x\right) + \textcolor{red}{10}(7)\)

\(15x + 50 = 15x + 70\)

Next, we can shift all like terms onto separate sides by using the Subtraction Property of Equality. In this case, we can subtract \(15x + 50\) on both sides:

\(15x + 50 \textcolor{red}{- (15x + 50)} = 15x + 70 \textcolor{red}{- (15x + 50)}\)

\(15x + 50 - 15x - 50 = 15x + 70 - 15x - 50\)

\(0 = 20\)

Therefore, as this equation represents a contradiction, we can determine that it contains \(\boldsymbol{0}\) solutions.


\(4x + 9 = -x - 6\)

First, we can shift all like terms onto separate sides by using the Addition Property of Equality. In this case, we can add \(2x - 8\) to both sides:

\(4x + 9 \textcolor{red}{+ (x - 9)} = -x - 6 \textcolor{red}{+ (x - 9)}\)

\(4x + 9 + x - 9 = -x - 6 + x - 9 \)

\(5x = -15\)

Next, we can use the Divison Property of Equality to determine \(x\):

\(\cfrac{\cancel{5}x}{\textcolor{red}{\cancel{5}}} = \cfrac{-15}{\textcolor{red}{5}}\)

\(x = -3\)

Therefore, as \(x = -3\), we can determine that this expression contains only \(\boldsymbol{1}\) solution.




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