Question

What is the value of \[x\] in the equation \[\dfrac{1}{5}x-\dfrac{2}{3}y=30\] , when \[y=15\]?

Answer 1

Hint: In this problem we have found the value of the variable \[x\] from the given linear equation. Where the value of another variable \[y\] is 15. Now we have to solve this equation and find the value of the variable \[x\]. The coefficients of the variable \[x\] and \[y\] are given in fractional form. We have to substitute the value of y in the given equation, to solve and find the value of x.

Complete step by step solution:
We know that the given linear equation is,
\[\Rightarrow \dfrac{1}{5}x-\dfrac{2}{3}y=30\]
Where the value of the variable \[y\] is given, \[y=15\].
We can now substitute the value of \[y\] in the given linear equation we get,
\[\Rightarrow \dfrac{1}{5}x-\dfrac{2}{3}15=30\]
Now simplifying this we get,
\[\Rightarrow \dfrac{1}{5}x-2\times 5=30\]
\[\Rightarrow \dfrac{1}{5}x-10=30\]
Now we can add 10 on both sides to eliminate the -10 on the left hand side of the term. By adding 10 on both sides we get,
\[\Rightarrow \dfrac{1}{5}x-10+10=30+10\]
\[\Rightarrow \dfrac{1}{5}x=40\]
Now we can multiply both sides by 5 to eliminate the coefficient of \[x\] in the left hand side of the linear equation. By multiplying we get,
\[\Rightarrow x=200\]
Therefore, the value of x is 200.

Note: In these types of problems we have to read the question properly, whether the given value is given for the variable \[x\] or \[y\]. Students will make mistakes in substituting the value of the given variable. They may interchange and substitute. To check either the solution is correct or wrong. We can substitute the solution and the given variable value in the left-hand side of the given linear equation. If the solution we get from this matches the right-hand side of the linear equation, then the problem we have is correct.