Question

A taxi's charges consist of fixed charges and the remaining depending upon the distance traveled in kilometers. If a person travels 10km, he pays Rs.68 and for traveling 15 km, he pays Rs.98 Express the above statements with the help of simultaneous equations and hence find the fixed charges and the rate per km.

Answer 1

Hint: A linear system of two equations with two variables is any system that can be written in the form.
\[
  ax + by = p \\
  cx + dy = q \\
 \]
where any of the constants can be zero with the exception that each equation must have at least one variable in it. Use the Linear equation with two variables to solve the question.

Complete step by step solution:
A way to solve a linear system algebraically is to use the substitution method. The substitution method functions by substituting the one y-value with the other.
Let taxi charges which are consist of fixed charge be $x$ and running charge be $y$.
So, according to the question, If a person travels 10km, he pays Rs.68 and for traveling 15 km, he pays Rs.98
\[x + 10y = 68 \ldots (1)\]
\[x + 15y = 98 \ldots (2)\]
Solve one of the two equations for one of the variables in terms of the other.
Finding the value of $x$ from equation (1)
\[ \Rightarrow x = 68 - 10y \ldots (3)\]
 Substitute the expression for this variable into the second equation, then solve for the remaining variable. Putting the value of $x$ to equation (2)
\[ \Rightarrow x = 68 - 10y\]
\[ \Rightarrow 68 - 10y + 15y = 98\]
\[ \Rightarrow 5y = 98 - 68\]
\[ \Rightarrow 5y = 30\]
\[ \Rightarrow y = \dfrac{{30}}{5} = 6\]
Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair. Putting the value of $x$ in equation 3
\[ \Rightarrow x = 68 - 10y\]
\[ \Rightarrow x = 68 - 10\left( 6 \right)\]
\[ \Rightarrow x = 68 - 60 = 8\]

Therefore, the fixed charge is Rs 8, and the rate per km is Rs 6.

Note:
You can use the substitution method even if both equations of the linear system are in standard form. Just begin by solving one of the equations for one of its variables.
You must check your solutions, inn every linear equation. To do this "checking", you need only plug your answer into the original equation, and make sure that you end up with a true statement.
Substituting the value of $x=8$ and $y=6$ in equation (1)
\[x + 10y = 68 \ldots (1)\]
$ \Rightarrow 8 + 10(6) = 8 + 60 = 68$
So, therefore my L.H.S. = R.H.S.. Hence our solution is absolutely correct.