To find the equation that does not belong with the other three we have to solve each equation.
Given the equation:
[tex]4\cdot x\cdot7=56[/tex]
Dividing by 4 at both sides:
[tex]\begin{gathered} \frac{4\cdot x\cdot7}{4}=\frac{56}{4} \\ x\cdot7=14 \end{gathered}[/tex]
7:
[tex]\begin{gathered} \frac{x\cdot7}{7}=\frac{14}{7} \\ x=2 \end{gathered}[/tex]
[tex]5\cdot2\cdot x=40[/tex]
Simplifying on the left:
[tex]10\cdot x=40[/tex]
10:
[tex]\begin{gathered} \frac{10\cdot x}{10}=\frac{40}{10} \\ x=4 \end{gathered}[/tex]
:
[tex]x\cdot3\cdot9=54[/tex]
[tex]x\cdot27=54[/tex]
27:
[tex]\begin{gathered} \frac{x\cdot27}{27}=\frac{54}{27} \\ x=2 \end{gathered}[/tex]
[tex]4\cdot x\cdot5=40[/tex]
4 :
[tex]\begin{gathered} \frac{4\cdot x\cdot5}{4}=\frac{40}{4} \\ x\cdot5=10 \end{gathered}[/tex]
5
[tex]\begin{gathered} \frac{x\cdot5}{5}=\frac{10}{5} \\ x=2 \end{gathered}[/tex]
In consequence, the equation that does not belong with the other three is:
[tex]5\cdot2\cdot x=40[/tex]
because its solution, 4, is different from the solution to the other equations (2).
If we change the 5 in the equation by a 10 and solve the equation we get:
[tex]\begin{gathered} 10\cdot2\cdot x=40 \\ \frac{10\cdot2\cdot x}{10}=\frac{40}{10} \\ 2\cdot x=4 \\ \frac{2\cdot x}{2}=\frac{4}{2} \\ x=2 \end{gathered}[/tex]
And now the equation (with a 10) belongs with the other three
