Respuesta :
we have
[tex]\begin{gathered} \frac{1}{5}+\frac{2}{3}\cdot\mleft|2-x\mright|=\frac{4}{15} \\ \text{solve for x} \\ \text{Multiply by 15 both sides} \\ 3+10\cdot|2-x|=4 \\ 10\cdot|2-x|=4-3 \\ 10\cdot|2-x|=1 \\ |2-x|=\frac{1}{10} \\ \text{First solution (positive value)} \\ +(2-x)=\frac{1}{10} \\ x=2-\frac{1}{10} \\ x=\frac{19}{10} \\ \text{Second solution (negative case)} \\ -(2-x)=\frac{1}{10} \\ \text{Multiply by -1 both sides} \\ (2-x)=-\frac{1}{10} \\ x=2+\frac{1}{10} \\ x=\frac{21}{10} \end{gathered}[/tex]Verify each equation
Remember that
If the equations are equivalent, then the solutions are the same
N 1
we have
[tex]\begin{gathered} \frac{2}{3}\cdot\mleft|2-x\mright|=\frac{1}{15} \\ \text{Multiply by 15 both sides} \\ 10\cdot|2-x|=1 \\ |2-x|=\frac{1}{10} \end{gathered}[/tex]The equation N 1 is equivalent to the given expression
N 2
we have
[tex]\begin{gathered} \frac{13}{15}\cdot|2-x|=\frac{4}{15} \\ \text{Multiply by 15 both sides} \\ 13\cdot|2-x|=4 \\ |2-x|=\frac{4}{13} \end{gathered}[/tex]The equation N 2 is not equivalent to the given expression
N 3
we have
[tex]|2-x|=\frac{1}{10}[/tex]The equation N 3 is equivalent to the given expression
N 4
we have
[tex]\begin{gathered} \frac{1}{5}+\mleft|\frac{4}{3}-\frac{2}{3}x\mright|=\frac{4}{15} \\ \frac{1}{5}+|\frac{4-2x}{3}|=\frac{4}{15} \\ \\ \text{Multiply by 15 both sides} \\ 3+15\cdot|\frac{4-2x}{3}|=4 \\ 15\cdot|\frac{4-2x}{3}|=1 \\ \frac{15}{3}\mleft|4-2x\mright|=1 \\ \\ 5\cdot|4-2x|=1 \\ \text{Factor 2} \\ 5\cdot(2)\cdot|2-x|=1 \\ |2-x|=\frac{1}{10} \end{gathered}[/tex]The equation N 4 is equivalent to the given expression
N 5
we have
[tex]|2-x|=\frac{4}{13}[/tex]The equation N 5 is not equivalent to the given expression
