Question

Solve the following:
\[\dfrac{{2x - 13}}{5} - \dfrac{{x - 3}}{{11}} = \dfrac{{x - 9}}{5} + 1\]
a). 8
b). 10
c). 12
d). 14

Answer 1

Hint: Here we have a linear equation with a variable ‘x’. Here we need to solve for ‘x’. We can solve this using the transposition method. That is, we group the ‘x’ terms on one side and constants on the other side of the equation. We take LCM on both sides of the equation and after that we cross multiply it.

Complete step-by-step solution:
Given, \[\dfrac{{2x - 13}}{5} - \dfrac{{x - 3}}{{11}} = \dfrac{{x - 9}}{5} + 1\].
Taking LCM, we now LCM of 5, 11 is 55
\[\dfrac{{11\left( {2x - 13} \right) - 5\left( {x - 3} \right)}}{{55}} = \dfrac{{\left( {x - 9} \right) + 5}}{5}\]
\[\Rightarrow \dfrac{{22x - 143 - 5x + 15}}{{55}} = \dfrac{{x - 9 + 5}}{5}\]
\[\Rightarrow \dfrac{{17x - 128}}{{55}} = \dfrac{{x - 4}}{5}\]
Cross multiplying, we have,
\[5\left( {17x - 128} \right) = 55\left( {x - 4} \right)\]
Expanding the brackets we have,
\[85x - 640 = 55x - 220\]
We transpose \[ - 640\] which is present in the left-hand side of the equation to the right-hand side of the equation by adding \[640\]on the right-hand side of the equation.
\[85x = 55x - 220 + 640\]
We transpose ‘55x’ to the LHS of the equation by subtracting ‘55x’ on LHS
\[85x - 55x = - 220 + 640\]
\[30x = 420\]
Divide the whole equation by 30,
\[x = \dfrac{{420}}{{30}}\]
\[ \Rightarrow x = 14\].
Hence the correct answer is option is (d).

Note: By simplifying we have obtained the answer for ‘x’. We can check whether the obtained value of ‘x’ is correct or not. To check we simply substitute the obtained value of ‘p’ in the given problem. If L.H.S is equal to R.H.S. then our answer is correct. Here we have obtained the answer which is present in the given option, so no need to check again.
In the above, we did the transpose of addition and subtraction. Similarly, if we have multiplication, we use division to transpose. If we have division, we use multiplication to transpose. Follow the same procedure for these kinds of problems.