Answer
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Hint: In this question, we need to determine the solution for the equations $ 2x + 5y = 10,2x + 3y = 6 $ . For this, we will use the arithmetic operations on the given equations and evaluate the solution of the equations.
Complete step-by-step answer:
A solution to the equation are values such that the equation is satisfied.
Take the equation $ 2x + 5y = 10 $ and substituting the values of $ x = a,y = 0 $ , we will get
$
\Rightarrow 2a + 5(0) = 10 \\
\Rightarrow a = 5 \\
$
So, \[\left( {5,0} \right)\] is one solution of $ 2x + 5y = 10 $ .
Now, put $ x = 0,y = b $ in $ 2x + 5y = 10 $ , we will get
$
\Rightarrow 2(0) + 5b = 10 \\
\Rightarrow b = 2 \\
$
So, \[\left( {0,2} \right)\] is another solution to $ 2x + 5y = 10 $
Now, we will find the solution to $ 2x + 3y = 6 $
This time we are putting $ x = a,y = 0 $ , and $ x = 0,y = b $ in one step, we get
$ 2a + 3(0) = 6{\text{ and }}2(0) + 3b = 6 $
On solving the above two equations, we get
$
\Rightarrow 2a + 3(0) = 6{\text{ and }}2(0) + 3b = 6 \\
\Rightarrow a = 3{\text{ and }}b = 2 \\
$
So \[\left( {3,0} \right)\] and \[\left( {0,2} \right)\] are solutions of $ 2x + 3y = 6 $
For the common solution, we will check whether the condition $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $ is satisfied or not.
From equations $ 2x + 5y = 10,2x + 3y = 6 $ , the values of the constants are
$ {a_1} = 2,{b_1} = 5,{c_1} = 10,{a_2} = 2,{b_2} = 3{\text{ and }}{c_2} = 6 $
Now, the ratio of $ {a_1} $ and $ {a_2} $ is $ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{2}{2} = 1 $
Similarly, the ratio of $ {b_1} $ and $ {b_2} $ is $ \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{5}{3} $
As, $ 1 \ne \dfrac{5}{3} $
So, $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $
So, we can say that the given equations have one common solution.
So, the correct answer is “Option C”.
Note: There are many ways to find the solution of simultaneous linear equations like substitution method, elimination method etc.
We can also find a common solution by plotting equations graphically. If two lines intersect at one point, then they have a unique solution. The condition of the common solution of simultaneous linear equations $ {a_1}x + {b_1}y + {c_1} = 0,{a_2}x + {b_2}y + {c_2} = 0 $ is $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $ .
Complete step-by-step answer:
A solution to the equation are values such that the equation is satisfied.
Take the equation $ 2x + 5y = 10 $ and substituting the values of $ x = a,y = 0 $ , we will get
$
\Rightarrow 2a + 5(0) = 10 \\
\Rightarrow a = 5 \\
$
So, \[\left( {5,0} \right)\] is one solution of $ 2x + 5y = 10 $ .
Now, put $ x = 0,y = b $ in $ 2x + 5y = 10 $ , we will get
$
\Rightarrow 2(0) + 5b = 10 \\
\Rightarrow b = 2 \\
$
So, \[\left( {0,2} \right)\] is another solution to $ 2x + 5y = 10 $
Now, we will find the solution to $ 2x + 3y = 6 $
This time we are putting $ x = a,y = 0 $ , and $ x = 0,y = b $ in one step, we get
$ 2a + 3(0) = 6{\text{ and }}2(0) + 3b = 6 $
On solving the above two equations, we get
$
\Rightarrow 2a + 3(0) = 6{\text{ and }}2(0) + 3b = 6 \\
\Rightarrow a = 3{\text{ and }}b = 2 \\
$
So \[\left( {3,0} \right)\] and \[\left( {0,2} \right)\] are solutions of $ 2x + 3y = 6 $
For the common solution, we will check whether the condition $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $ is satisfied or not.
From equations $ 2x + 5y = 10,2x + 3y = 6 $ , the values of the constants are
$ {a_1} = 2,{b_1} = 5,{c_1} = 10,{a_2} = 2,{b_2} = 3{\text{ and }}{c_2} = 6 $
Now, the ratio of $ {a_1} $ and $ {a_2} $ is $ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{2}{2} = 1 $
Similarly, the ratio of $ {b_1} $ and $ {b_2} $ is $ \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{5}{3} $
As, $ 1 \ne \dfrac{5}{3} $
So, $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $
So, we can say that the given equations have one common solution.
So, the correct answer is “Option C”.
Note: There are many ways to find the solution of simultaneous linear equations like substitution method, elimination method etc.
We can also find a common solution by plotting equations graphically. If two lines intersect at one point, then they have a unique solution. The condition of the common solution of simultaneous linear equations $ {a_1}x + {b_1}y + {c_1} = 0,{a_2}x + {b_2}y + {c_2} = 0 $ is $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $ .
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