Question

Solve the following equations and find the value of x (i) \[\dfrac{3x}{5}=15\] (ii) \[8x-4(x-3)=2x\] (iii) \[2-3(3x+1)=2(7-6x)\].

Answer 1

Hint: In question (i), just multiply by 5 in LHS and RHS of the equation. In question (ii), first of all, just expand the given equation. Take all the variable terms of x in LHS and constants to the RHS of the equation and then solve. Similarly, in question (iii), first of all, expand the given equation. Take all the variable terms of x in LHS and constants to the RHS of the equation and then solve it further.

Complete step-by-step solution -
In question (i), we have the equation \[\dfrac{3x}{5}=15\] ………….(1)
Multiplying by 5 in LHS and RHS of equation (1), we get
\[\begin{align}
  & \dfrac{3x}{5}=15 \\
 & \Rightarrow 3x=75 \\
 & \Rightarrow x=25 \\
\end{align}\]
So, the value of x is 25.

In question (ii), we have the equation \[8x-4(x-3)=2x\] ……………….(2)
We have a linear equation. In that linear equation, we have an unknown variable x. We have to find the value of x using the given equation \[8x-4(x-3)=2x\] .
Now, expanding equation (2), we get
\[8x-4(x-3)=2x\]
\[\Rightarrow 8x-4x+12=2x\] ………….…….(3)
Now, taking the term 2x to the LHS and constant term 12 to the RHS of the equation (3), we get
\[\begin{align}
  & 8x-4x-2x=-12 \\
 & \Rightarrow 2x=-12 \\
 & \Rightarrow x=-6 \\
\end{align}\]
So, the value of x is -6.

In question (iii), we have the equation \[2-3(3x+1)=2(7-6x)\] ……………….(4)
We have a linear equation. In that linear equation, we have an unknown variable x. We have to find the value of x using the given equation \[2-3(3x+1)=2(7-6x)\].
Now, expanding equation (4), we get
\[2-3(3x+1)=2(7-6x)\]
\[\Rightarrow 2-9x-3=14-12x\] …………….(5)
Now, taking the term -12x to the LHS and constant term 2,-3 to the RHS of the equation (5), we get
\[\begin{align}
  & 2-9x-3=14-12x \\
 & \Rightarrow 12x-9x=14-2+3 \\
 & \Rightarrow 3x=15 \\
 & \Rightarrow x=5 \\
\end{align}\]
So, the value of x is 5.

Note: In question (ii), we can make a mistake in taking +2x of RHS to the LHS and write +2x in LHS too, which is wrong. If we are moving some terms of RHS to LHS then its sign is changed and vice-versa. Here we are moving +2x to the LHS. So, we have to write -2x in the LHS of the equation. A similar mistake can be done in question (iii). We must take care of signs while opening the brackets also.