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.
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 \(\boldsymbol{x = 4}\).
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:
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:
Therefore, we can determine that \(\boldsymbol{x = 33}\).
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.
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.
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 \(\boldsymbol{x = -20}\).
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 \(12\):
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:
Therefore, we can determine that \(\boldsymbol{x = \cfrac{8}{3}}\).
These uncommon forms of linear equations are notable for not having a single solution. These include:
Contradictions are equations with one variable that have no solutions.
ExampleIdentities are equations with one variable that have all real numbers as solutions.
Example\(x + 4 = 3x - 2(x - 2)\)
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:
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 \(10\):
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.