Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77, but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A?
Rs. 13. Rs. 17
This problem involves finding the cost of bus tickets between different cities based on given purchase scenarios. We can solve this using a system of linear equations.
Let's define the unknown fares:
We are given two conditions:
$2x + 3y = 77$ (Equation 1)
$3x + 2y = 73$ (Equation 2)
We need to solve these two equations simultaneously to find the values of $x$ and $y$. We can use the elimination method.
To eliminate one variable, we can multiply the equations by suitable numbers. Let's multiply Equation 1 by 3 and Equation 2 by 2, so the coefficients of '$x$' become the same.
Multiplying Equation 1 by 3:
$3 \times (2x + 3y) = 3 \times 77$
$6x + 9y = 231$ (Equation 3)
Multiplying Equation 2 by 2:
$2 \times (3x + 2y) = 2 \times 73$
$6x + 4y = 146$ (Equation 4)
Now, subtract Equation 4 from Equation 3 to eliminate '$x$':
$(6x + 9y) - (6x + 4y) = 231 - 146$
$6x + 9y - 6x - 4y = 85$
$5y = 85$
Now, solve for '$y$':
$y = \frac{85}{5}$
$y = 17$
Substitute the value of '$y$' (which is 17) back into either Equation 1 or Equation 2 to find '$x$'. Let's use Equation 1:
$2x + 3y = 77$
$2x + 3(17) = 77$
$2x + 51 = 77$
Now, solve for '$x$':
$2x = 77 - 51$
$2x = 26$
$x = \frac{26}{2}$
$x = 13$
So, the fare for a bus ticket from city A to B is Rs. 13, and the fare from city A to C is Rs. 17.
The fares for cities B and C from A are Rs. 13 and Rs. 17, respectively.
If 2 x + 3 y = 17;
2 x+2 - 3 y+1 = 5
then the values of x and y are:
The solution of pair of linear equations \(\dfrac{1}{2}x+\dfrac{2}{3}y=-1,x-\dfrac{1}{3}y=3\) by the elimination method, is:
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