Linear Equations

Linear Equations

An equation 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 equation and then "moving the coefficient $a$ to the denominator of $b

quot;.)

  • 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).

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*} $

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: $\frac{2 \cdot 5}{8}-\frac{5-2}{2} \equiv \frac{5}{8}-(4-5).$

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*} $