##### Worked Examples – First degree equations

Based on what we have explained to the article First degree Equations, we are going to solve some equations below:

Example 1

Solve the equation:

a) $\displaystyle 5x=-60$
$\displaystyle x=\frac{{-60}}{5}$
$\displaystyle x=-12$

b) $\displaystyle -3x=-9$
$\displaystyle x=\frac{{-9}}{{-3}}$
$\displaystyle x=3$

c) $\displaystyle \frac{1}{3}x=12$
$\displaystyle x=\frac{{12}}{{\frac{1}{3}}}=12\cdot 3$
$\displaystyle x=36$

d) $\displaystyle x=-x$
$\displaystyle x+x=0$
$\displaystyle 2x=0$
$\displaystyle x=\frac{0}{2}$
$\displaystyle x=0$

e) $\displaystyle x-4x=6$
$\displaystyle -3x=6$
$\displaystyle x=\frac{6}{{-3}}$
$\displaystyle x=-2$

f) $\displaystyle \frac{1}{7}x=0$
$\displaystyle x=\frac{0}{{\frac{1}{7}}}$
$\displaystyle x=0\cdot 7$
$\displaystyle x=0$

Example 2

Solve the equation (Using the Cross Multiplication Method):

a) $\displaystyle \frac{{3x-2}}{2}=x+1$

$\displaystyle 3x-2=2\left( {x+1} \right)$

$\displaystyle 3x-2=2x+2$

$\displaystyle 3x-2x=2+2$

$\displaystyle x=4$

b) $\displaystyle \frac{{3x}}{4}=\frac{{x-1}}{3}$

$\displaystyle 3x\cdot 3=4\left( {x-1} \right)$

$\displaystyle 9x=4x-4$

$\displaystyle 9x-4x=-4$\

$\displaystyle 5x=-4$

$\displaystyle x=\frac{{-4}}{5}$

c) $\displaystyle \frac{1}{2}=\frac{{3\left( {x-1} \right)}}{5}$

$\displaystyle 1\cdot 5=2\cdot 3\left( {x-1} \right)$

$\displaystyle 5=6x-6$

$\displaystyle 5+6=6x$

$\displaystyle 6x=11$

$\displaystyle x=\frac{{11}}{6}$

Example 3

Solve the equation(Expanding brackets and formulas):

a) $\displaystyle {{\left( {x+2} \right)}^{2}}-{{\left( {x-2} \right)}^{2}}=3\left( {x+1} \right)$

$\displaystyle {{x}^{2}}+4x+4-\left( {{{x}^{2}}-4x+4} \right)=3x+3$

$\displaystyle {{{\cancel{x}}}^{2}}+4x+\cancel{4}-{{{\cancel{x}}}^{2}}+4x-\cancel{4}=3x+3$

$\displaystyle 8x=3x+3$

$\displaystyle 5x=3$

$\displaystyle x=\frac{3}{5}$

b) $\displaystyle \left( {3x-1} \right)\left( {x+3} \right)=0$

$\displaystyle 3x-1=0$ or $\displaystyle x+3=0$

$\displaystyle x=\frac{1}{3}$ or $\displaystyle x=-3$

c) $\displaystyle \left( {x-3} \right)\left( {x+3} \right)=\left( {x-2} \right)\left( {x-6} \right)$

$\displaystyle {{{\cancel{x}}}^{2}}-9={{{\cancel{x}}}^{2}}-6x-2x+12$

$\displaystyle 8x=12+9$

$\displaystyle x=\frac{{21}}{{8}}$

d) $\displaystyle \left( {x-5} \right)\left( {x+6} \right)=0$

$\displaystyle x-5=0$ or $\displaystyle x+6=0$

$\displaystyle x=5$ or $\displaystyle x=-6$

Example 4

Solve the equation (Factoring)

a) $\displaystyle {{x}^{2}}-x=0$

$\displaystyle x\left( {x-1} \right)=0$

$\displaystyle x=0$ or $\displaystyle x=1$

b) $\displaystyle {{x}^{2}}=2x$

$\displaystyle {{x}^{2}}-2x=0$

$\displaystyle x\left( {x-2} \right)=0$

$\displaystyle x=0$ or $\displaystyle x=2$

c) $\displaystyle \left( {x-1} \right)\left( {x-2} \right)\left( {x-3} \right)=0$

$\displaystyle \left( {x-1} \right)=0$ or $\displaystyle \left( {x-2} \right)=0$ or $\displaystyle \left( {x-3} \right)=0$

$\displaystyle x=1$ or $\displaystyle x=2$ or $\displaystyle x=3$

Example 5

Solve the equation (Firstly determine the
domain
$\displaystyle D=\left\{ {x\in R/\left. {x\ne 0} \right\}} \right.$ ):

a) $\displaystyle \frac{{x-3}}{x}=\frac{{2x+3}}{x}$

$\displaystyle x\left( {x-3} \right)=x\left( {2x+3} \right)$

$\displaystyle {{x}^{2}}-3x=2{{x}^{2}}+3x$

$\displaystyle {{x}^{2}}-2{{x}^{2}}-3x-3x=0$

$\displaystyle -{{x}^{2}}-6x=0$

$\displaystyle -x\left( {x-6} \right)=0$

$\displaystyle x=6$

b) $\displaystyle \frac{{x-3}}{{2x}}=\frac{{2x+3}}{x}$

$\displaystyle x\left( {x-3} \right)=2x\left( {2x+3} \right)$

$\displaystyle {{x}^{2}}-3x=4{{x}^{2}}+6x$

$\displaystyle {{x}^{2}}-3x-4{{x}^{2}}-6x=0$

$\displaystyle -3{{x}^{2}}-9x=0$

$\displaystyle 3{{x}^{2}}+9x=0$

$\displaystyle 3x\left( {x+3} \right)=0$

$\displaystyle x=-3$

c) $\displaystyle \frac{{x-1}}{x}=\frac{2}{3}$

$\displaystyle 3\left( {x-1} \right)=2x$

$\displaystyle 3x-3=2x$

$\displaystyle 3x-2x=3$

$\displaystyle x=3$

Example 6  Solve the equation.

a) $\displaystyle \sqrt{{x+7}}=-1$

Square the both sides of the equation

$\displaystyle {{(\sqrt{{x+7}})}^{2}}={{(-1)}^{2}}$

$\displaystyle x+7=1$

Isolate the variable x

$\displaystyle x=1-7$

$\displaystyle x=-6$

b) $\displaystyle \sqrt{{5+\sqrt{x}}}=\sqrt{8}$

Square the both sides of the equation

$\displaystyle {{\left( {\sqrt{{5+\sqrt{x}}}} \right)}^{2}}={{\left( {\sqrt{8}} \right)}^{2}}$

$\displaystyle 5+\sqrt{x}=8$

Isolate the square root

$\displaystyle \sqrt{x}=8-5=3$

Square the both sides again to find x

$\displaystyle {{(\sqrt{x})}^{2}}={{(3)}^{2}}$

$\displaystyle x=9$

Reminder: To learn equations read First degree Equations