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Question: Find the value of \(M\) for which the given system of equation has only one solution (i.e. unique so...

Find the value of MM for which the given system of equation has only one solution (i.e. unique solution)
Mx2y=9Mx - 2y = 9 and 4xy=74x - y = 7
A.MM can have all real values except 5
B.MM can have all real values except 2
C.MM can have all real values except 8
D.MM can have all real values except 10

Explanation

Solution

Here we will first compare the given equations with the standard equation and then we will find the value of the coefficient. The given system will have unique solution when the ratio of coefficients of the two equations is not equal.

Complete step-by-step answer:
The given equations are;
Mx2y=9Mx - 2y = 9
4xy=74x - y = 7
The given system can be written as:
Mx2y9=0\Rightarrow Mx - 2y - 9 = 0 ………. (1)\left( 1 \right)
4xy7=0\Rightarrow 4x - y - 7 = 0 ……….. (2)\left( 2 \right)
Now, we will write standard form of linear equations;
a1x+b1y+c1=0\Rightarrow {a_1}x + {b_1}y + {c_1} = 0 ……….. (3)\left( 3 \right)
a2x+b2y+c2=0\Rightarrow {a_2}x + {b_2}y + {c_2} = 0 ………… (4)\left( 4 \right)
Now, we will compare equation (1)\left( 1 \right) with equation (3)\left( 3 \right).
a1=M{a_1} = M , b1=2{b_1} = - 2 and c1=9{c_1} = - 9
Now, we will compare equation (2)\left( 2 \right) with equation (4)\left( 4 \right).
a2=4{a_2} = 4 , b2=1{b_2} = - 1 and c2=7{c_2} = - 7
We know the condition for a system of equation to have a unique solution is a1a2b1b2\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}.
Now, we will substitute a1=M{a_1} = M , b1=2{b_1} = - 2, a2=4{a_2} = 4 and b2=1{b_2} = - 1 in a1a2b1b2\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}. Therefore, we get
M421\Rightarrow \dfrac{M}{4} \ne \dfrac{{ - 2}}{{ - 1}}
Multiplying 4 on both sides, we get
M4×421×4 M8 \Rightarrow \dfrac{M}{4} \times 4 \ne \dfrac{{ - 2}}{{ - 1}} \times 4 \\\ \Rightarrow M \ne 8 \\\
Therefore, MM can have all real values other than 8.
Hence, the correct option is C.

Note: Pair of linear equations which are given by a1x+b1y+c1=0{a_1}x + {b_1}y + {c_1} = 0 and a2x+b2y+c2=0{a_2}x + {b_2}y + {c_2} = 0 can also have infinitely many solutions, the condition a1a2=b1b2=c1c2\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} should also be satisfied. In this problem, the condition doesn’t depend on the constant terms for the pair of linear equations to have a unique solution.
Two linear equations have a unique solution when these two lines represented by these equations intersect at only one point i.e. when these two lines are not parallel and not coincident. Also, the slopes of these two lines are different. We need to remember all these conditions to solve such types of problems.