An equations that can be written of the form \[ax+b=0,\qquad(a\neq0)\] where \(a\) and \(b\) are (fixed) real numbers and \(x\) is the variable (or the unknown) is called a linear equation. To solve the equation, subtract \(b\) from both sides \[ax+\cancel{b}-\cancel{b} = -b\] and then divide both sides of the resulting equation, \(ax=-b\) , by \(a\) : \[x=-\frac{b}{a}\] (Sometimes, this process is described as "transferring \(b\) to the other side of the other side of the equation and then "moving the coefficient \(a\) to the denominator of \(b\) ".)
- The solution of \(ax+b=0\) is the \(x\) -intercept of the straight line \(y=ax+b\) (the \(x\) -intercept is where the graph intersects the \(x\) -axis).
Example 1. Solve: \[5x+9=8x+12.\]
Solution
We first simplify and then rewrite the equations of the form that all terms containing the variable \(x\) are on one side and the constant terms are on the other side.
\[\begin{align} 5x+9-8x & =8x-8x+12 &&(\text{subtract } 8x \text{ from both sides})\\ -3x+9 & =\cancel{0x}+12\\ -3x+9-9 & =12-9 &&(\text{subtract } 9 \text{ from both sides})\\ -3x & =3\\ x & =-1&&\text{divide both sides by $-3$} \end{align}\]
Example 2. Solve \[\frac{2 x}{3}-\frac{x-2}{2}=\frac{x}{6}-(4-x).\]
Solution
To clear of fractions, multiply both members by the LCD, 6 . Then \[4 x-3(x-2) =x-6(4-x)\] or \[4 x-3 x+6 =x-24+6 x\]
Transpose and collect terms, \(-6 x=-30\) . Therefore \[x=5 .\] Verfication: \[\frac{2 \cdot 5}{8}-\frac{5-2}{2} \equiv \frac{5}{8}-(4-5).\]
Example 3. Solve \(\dfrac{3x+7}{4}=1+\dfrac{1-x}{3}\) .
Solution
\[\begin{align} \dfrac{3x+7}{4} & =1+\dfrac{1-x}{3}\tag{given equation}\\ 12\cdot\frac{3x+7}{4} & =12\cdot\left(1+\frac{1-x}{3}\right)\tag{multiply both sides by $LCD = 12$}\\ 3\cdot(3x+7) & =12+4\cdot(1-x)\tag{simplify}\\ 9x+21 & =12+4-4x\tag{expand}\\ 9x+4x & =12+4-21\tag{add $4x$ to and subtract $21$ from both sides}\\ 13x & =-5\tag{simplify}\\ x & =-\frac{5}{13}\tag{divide both sides by $13$} \end{align}\]