Question

Solve the following equation and check your results: \[\dfrac{{2x}}{3} + 1 = \dfrac{{7x}}{{15}} + 3\].

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
\[\dfrac{{2x}}{3} + 1 = \dfrac{{7x}}{{15}} + 3{\text{ ......eq(1)}}\]
We know that an equation tells us that two of sides are equal with some variables and constants.
Multiplying the above equation by 15 on both sides, we get
\[
   \Rightarrow 15\left( {\dfrac{{2x}}{3} + 1} \right) = 15\left( {\dfrac{{7x}}{{15}} + 3} \right) \\
   \Rightarrow 15 \times \dfrac{{2x}}{3} + 15 \times 1 = 15 \times \dfrac{{7x}}{{15}} + 15 \times 3 \\
   \Rightarrow 5 \times 2x + 15 = 7x + 45 \\
   \Rightarrow 10x + 15 = 7x + 45 \\
 \]
Subtracting the above equation by \[7x\] on both sides, we get
\[
   \Rightarrow 10x + 15 - 7x = 7x + 45 - 7x \\
   \Rightarrow 3x + 15 = 45 \\
 \]
Subtracting the above equation by 15 on both sides, we get
\[
   \Rightarrow 3x + 15 - 15 = 45 - 15 \\
   \Rightarrow 3x = 30 \\
 \]
Dividing the above equation by 3 on both sides, we get
\[
   \Rightarrow \dfrac{{3x}}{3} = \dfrac{{30}}{3} \\
   \Rightarrow x = 10 \\
 \]
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
\[
   \Rightarrow \dfrac{{2\left( {10} \right)}}{3} + 1 \\
   \Rightarrow \dfrac{{20}}{3} + 1 \\
   \Rightarrow \dfrac{{20 + 3}}{3} \\
   \Rightarrow \dfrac{{23}}{3} \\
 \]
Replacing 10 for \[x\] in the right hand side of the equation (1), we get
\[
   \Rightarrow \dfrac{{7\left( {10} \right)}}{{15}} + 3 \\
   \Rightarrow \dfrac{{70}}{5} + 3 \\
   \Rightarrow \dfrac{{14}}{3} + 3 \\
   \Rightarrow \dfrac{{14 + 9}}{3} \\
   \Rightarrow \dfrac{{23}}{3} \\
 \]
Therefore, LHS is equal to RHS.
Hence, proved.

Note: One should know that a linear expression is an expression whose highest power of the variable is one only. Students forget to check the answer, which is an incomplete solution for this problem. Since there is an equality sign in the linear equation, the expression on the left of the equal sign is always equals to the right of the equal, but this happens only for some values only and these values are the solution of these linear equations.