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Question

Answers

x + y = 1 and 2x + 3y = 3

A. Consistent

B. Inconsistent

C. Dependent

D. All of the above

Answer

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Hint: To check the consistency of the equation we had to only compare that with standard linear equations \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\]. And after comparing the ratio \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\] we will get the consistency of the equations.

Complete step-by-step answer:

As we know that the standard form of pair of linear equation of two variable x and y is written as,

\[{a_1}x + {b_1}y + {c_1} = 0\] -----(1)

\[{a_2}x + {b_2}y + {c_2} = 0\] -----(2)

And to check whether the system of equations is consistent or not. We compare the ratio,

\[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\]

And if, \[\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}\] then we can say the equation has exactly one solution.

And as we know that if the system of equations has exactly one solution then the system of equations is consistent.

And if the system of equations has infinite solutions (both equations are of the parallel) then the system of equations will be inconsistent.

Now as we know that the given pair of equations are,

x + y = 1 ------ (3)

2x + 3y = 3 ------(4)

So, to find the value of \[{a_1},{\text{ }}{a_2},{\text{ }}{b_1},{\text{ }}{b_2},{\text{ }}{c_1}\] and \[{c_2}\].

Let us compare equation 1 with equation 3 and equation 2 with equation 4.

So, on comparing. We get,

\[{a_1} = 1,{\text{ }}{a_2} = 2,{\text{ }}{b_1} = 1,{\text{ }}{b_2} = 3,{\text{ }}{c_1} = 1\] and \[{c_2} = 3\].

So, now comparing \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\]. We get,

\[\dfrac{1}{2} = \dfrac{1}{3} = \dfrac{1}{3}\] As we can see that for the given system of equations \[\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}\] i.e. \[\dfrac{1}{2} \ne \dfrac{1}{3}\]

So, the system of equations will have exactly one solution. So, these equations are consistent.

Now, we had to find the solution of given equations.

So, multiplying equation 3 by 2 and then subtracting from equation 4. We get,

2x – 2x + 3y – 2y = 3 – 2

y = 1

Now putting the value of y in equation 3. We get,

x + 1 = 1

x = 0

So, the solution of the system of equations will be x = 0 and y = 1.

Hence, the correct option will be A.

Note: Whenever we come up with this type of problem we can also find the solution of the pair of equations by substitution or elimination method and if they have a unique solution then the system of equations will be consistent otherwise if they are parallel or have infinite solutions then the system of equations will be inconsistent.

Complete step-by-step answer:

As we know that the standard form of pair of linear equation of two variable x and y is written as,

\[{a_1}x + {b_1}y + {c_1} = 0\] -----(1)

\[{a_2}x + {b_2}y + {c_2} = 0\] -----(2)

And to check whether the system of equations is consistent or not. We compare the ratio,

\[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\]

And if, \[\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}\] then we can say the equation has exactly one solution.

And as we know that if the system of equations has exactly one solution then the system of equations is consistent.

And if the system of equations has infinite solutions (both equations are of the parallel) then the system of equations will be inconsistent.

Now as we know that the given pair of equations are,

x + y = 1 ------ (3)

2x + 3y = 3 ------(4)

So, to find the value of \[{a_1},{\text{ }}{a_2},{\text{ }}{b_1},{\text{ }}{b_2},{\text{ }}{c_1}\] and \[{c_2}\].

Let us compare equation 1 with equation 3 and equation 2 with equation 4.

So, on comparing. We get,

\[{a_1} = 1,{\text{ }}{a_2} = 2,{\text{ }}{b_1} = 1,{\text{ }}{b_2} = 3,{\text{ }}{c_1} = 1\] and \[{c_2} = 3\].

So, now comparing \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\]. We get,

\[\dfrac{1}{2} = \dfrac{1}{3} = \dfrac{1}{3}\] As we can see that for the given system of equations \[\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}\] i.e. \[\dfrac{1}{2} \ne \dfrac{1}{3}\]

So, the system of equations will have exactly one solution. So, these equations are consistent.

Now, we had to find the solution of given equations.

So, multiplying equation 3 by 2 and then subtracting from equation 4. We get,

2x – 2x + 3y – 2y = 3 – 2

y = 1

Now putting the value of y in equation 3. We get,

x + 1 = 1

x = 0

So, the solution of the system of equations will be x = 0 and y = 1.

Hence, the correct option will be A.

Note: Whenever we come up with this type of problem we can also find the solution of the pair of equations by substitution or elimination method and if they have a unique solution then the system of equations will be consistent otherwise if they are parallel or have infinite solutions then the system of equations will be inconsistent.

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