Question

Solve the following equation and check your results: \[x = \dfrac{4}{5}\left( {x + 10} \right)\]

Answer 1

Hint: Here, we will first add or subtract the above equation to get all the variable terms on one side and the constants on the other. Then we will simplify the obtained equation to find the required value. Now we will substitute the obtained value of \[x\] in the left hand side and right hand side of the equation separately to check our result if both the sides are equal.

Complete step-by-step answer:
We are given that the equation is
\[x = \dfrac{4}{5}\left( {x + 10} \right){\text{ ......eq(1)}}\]
We know that an equation tells us that two of sides are equal with some variables and constants.
Multiplying the above equation by 5 on both sides, we get
\[
   \Rightarrow 5x = 5\left[ {\dfrac{4}{5}\left( {x + 10} \right)} \right] \\
   \Rightarrow 5x = 4\left( {x + 10} \right) \\
   \Rightarrow 5x = 4 \times x + 4 \times 10 \\
   \Rightarrow 5x = 4x + 40 \\
 \]
Subtracting the above equation by \[4x\] on both sides, we get
\[
   \Rightarrow 5x - 4x = 4x + 40 - 4x \\
   \Rightarrow x = 40 \\
 \]
We will now check the above value of \[x\] in the equation (1) is the solution of this equation.
We know that a solution is a value we can put in place of a variable that makes the equation true, that is, the left hand side is equal to the right hand side in the equation.
Substituting the value of \[x\] in the left hand side of the equation (1), we get
\[ \Rightarrow 40\]
Replacing 40 for \[x\] in the right hand side of the equation (1), we get
\[
   \Rightarrow \dfrac{4}{5}\left( {40 + 10} \right) \\
   \Rightarrow \dfrac{4}{5}\left( {50} \right) \\
   \Rightarrow 4 \times 10 \\
   \Rightarrow 40 \\
 \]
Therefore, LHS is equal to RHS.
Hence, proved.

Note: Students forget to check the answer, which is an incomplete solution for this problem. One should know that if there is only one variable in the equation with the general form \[ax + b = c\], where \[a\], \[b\] and \[c\] are real numbers and \[a \ne 0\]. We have used here the balancing method to solve it, as we have added or subtracted with the same number on both sides without disturbing the balance to find the solution. Avoid calculation mistakes.