Question

solve the following
I) 3 (n-2) = 5 (n-4)
II) -2 -7 = 5n +3​

Answer 1

Given:

$i) \ 3 (n-2) = 5 (n-4)\\ii) \ -2 -7 = 5n +3$

We have to solve the above equations.

• By using the Bodmas rule, we are solving the above equation.
• As we know that the Bodmas rule is used to remember the order of operations to be followed while solving expressions in mathematics.

Where,

$\begin{array}{l}\mathrm{B}=\text{brackets}\\\mathrm{O}=\text { order of powers or rules } \\\mathrm{D}=\text { division } \\\mathrm{M}=\text { multiplication } \\\mathrm{A}=\text { addition } \\\mathrm{S}=\text { subtraction }\end{array}$

We are solving in the following way:

We have,

$i) \ 3 (n-2) = 5 (n-4)\\ii) \ -2 -7 = 5n +3$

now,

$i) \ 3 (n-2) = 5 (n-4)\\\Rightarrow 3n-6=5n-20\\\Rightarrow 3n-5n=-20+6\\\Rightarrow-2n=-14\\\Rightarrow 2n=14\\\\\Rightarrow n=\frac{14}{2} \\\\\Rightarrow n=7$

$ii) \ -2 -7 = 5n +3\\\Rightarrow -9=5n+3\\\Rightarrow -9-3=5n\\\Rightarrow -12=5n\\\\\Rightarrow n=\frac{-12}{5} \\\\\Rightarrow n=-2.4$

Hence, we get, the solution of the equation is:

$i) \ n=7\\ii)\ n=-2.4$

Answer 2

1]

$3n - 6 = 5(n - 4) \\ 3n - 6 = 5n - 2 \\ add \: 6 \: to \: both \: the \: sides \: \\ 3n - 6 + 6 = 5n - 20 + 6 \\ 3n = 5n - 14 \\ subract \: 5n \: from \: both \: the \: sides \\ 3n - 5n = 5n - 14 - 5n \: \\ - 2n = - 14 \\ divide \: - 2n \: from \: both \: sides \\ \frac{ - 2n}{ - 2} = \frac{ - 14}{ - 2} r \\ n = 7$

therefore n=7