Question

Show that the following equation has a unique solution: \[3x + 5y = 12\] , \[5x + 3y = 4\] .

Answer 1

Hint: First, compare the given pair of equations with ${a_1}x + {b_1}y = {c_1}$ and ${a_2}x + {b_2}y = {c_2}$ .
Now, find the values of ${a_1},{a_2},{b_1},{b_2},{c_1},{c_2}$ .
If the pair of equations has a unique solution, then $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ .
See whether it is the case in the given question.

Complete step-by-step answer:
The given pair of equations is:
 \[3x + 5y = 12\] and
 \[5x + 3y = 4\]
The above equations are of the type ${a_1}x + {b_1}y = {c_1}$ and ${a_2}x + {b_2}y = {c_2}$ .
 $\therefore {a_1} = 3,{b_1} = 5,{c_1} = 12$ and ${a_2} = 5,{b_2} = 3,{c_2} = 4$ .
Now, for having a unique solution to the pair of equations, it must be $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ .
Here ${a_1} = 3,{b_1} = 5,{c_1} = 12$ and ${a_2} = 5,{b_2} = 3,{c_2} = 4$ .
 \[\therefore \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{3}{5}\] and $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{5}{3}$ .
Thus, $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ .

Note: Alternate method:
The given pair of equations is:
 \[3x + 5y = 12\] ... (i)
 \[5x + 3y = 4\] ... (ii)
Now, we will solve the given pair of equations by the method of elimination.
Thus, multiplying equation (i) by 3, we get
 \[9x + 15y = 36\] ... (iii)
Also, multiplying equation (ii) by 5, we get
 \[25x + 15y = 20\] ... (iv)
On subtracting (iii) from (iv), we get
  \[\left( {25x + 15y} \right)-\left( {9x + 15y} \right) = 20-36\]
 \[\therefore 25x + 15y-9x-15y = - 16\]
 $\therefore 16x = - 16$
 $\therefore x = - 1$
On substituting the value of x in equation (ii), we get
\[5\left( { - 1} \right) + 3y = 4\]
 \[\therefore - 5 + 3y = 4\]
 $\therefore 3y = 4 + 5$
 $\therefore 3y = 9$
 $\therefore y = 3$
Thus, we get \[x = - 1\] and \[y = 3\] , i.e. (x,y) \[ = \left( { - 1,3} \right)\] , which is a unique solution.