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:
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$
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}$
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$
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$
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$
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