A Linear Equation is any equation that can written in the form \(ax + b = c\)
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.
Process for Solving Linear Equations
- If the equation contains fractions on either side, use the Lowest Common Denominator (LCD) to clear the fractions
- Simplify both sides of the equation by clearing out any parentheses using distributive property and combining like terms
- Rearrange the equation so that all variables are on one side and all constants are on the other side
- If there is a coefficient on the variable (\(a\) ≠ 1), divide or multiply both sides by the same term so that \(a\) = 1
- 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 \(12 = 3x\)
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 \(x\) = 4.
Example
Solve \(\cfrac{x}{11}=3\)
As the variable has a fraction, we need to multiply both sides by the denominator (11) to get the solution:
\((\textcolor{red}{11})(\cfrac{x}{11} = 3)\)
\((\textcolor{red}{11})(\cfrac{x}{11}) = (\textcolor{red}{11})(3)\)
\(x = 33\)
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 \(x\) = 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 \(3(x+6) = 2(x-1)\)
Show Answer
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 \(x = -20\).
Solve \(\cfrac{k}{4}-2=-\cfrac{4}{3}\)
Show Answer
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:
\(\textcolor{red}{12}(\cfrac{k}{4}-2=-\cfrac{4}{3}\))
\(\textcolor{red}{12}(\cfrac{k}{4}) - (\textcolor{red}{12})(2) = (\textcolor{red}{12}) (-\cfrac{4}{3})\)
\(3k - 24 = -16\)
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:
\(\cfrac{8}{3}/4 - 2 = -\cfrac{4}{3}\)
\((\cfrac{8}{3})(\cfrac{1}{4}) - 2 = -\cfrac{4}{3}\)
\(\cfrac{8}{12} - \cfrac{24}{12} = -\cfrac{4}{3}\)
\(-\cfrac{16}{12} = -\cfrac{4}{3}\)
\(-\cfrac{4}{3} = -\cfrac{4}{3}\)
Therefore, we can determine that \(x = \cfrac{8}{3}\).
Unique Types of Equations
These uncommon forms of linear equations are notable for not having a single solution. These include:
Contradictions
Equations with one variable that have no solutions.
Example
\(2x + 4 = 2x - 3\)
\(2x + 4 - (2x + 4) = 2x - 3 - (2x + 4)\)
\(0 = -7\)
Identities
Equations with one variable that have all real numbers as solutions.
Example
\(9x + 3x = 12x\)
\(12x = 12x\)