
Equation x+y=2, 2x+2y=3 will have
A) No solution
B) Only one solution
C) Many finite solution
D) Trivial solution
Answer
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Hint: In this question it is given that we have to find these two equations x+y=2, 2x+2y=3 has any solution or not. So to find whether the given equations have any solution or not we have to use some method, which says that if $$a_{1}x+b_{1}y=c_{1}$$ and $$a_{2}x+b_{2}y=c_{2}$$ be any two linear equation then-
1) The equation has one solution, if $$\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$$.
2) Equation has many finite solution, if $$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} =\dfrac{c_{1}}{c_{2}}$$.
3) Equation has zero solution, if $$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$$
4) Equation has trivial solution, if $$c_{1}=c_{2}=0$$
Complete step-by-step solution:
Comparing the given equation x+y=2 and 2x+2y=3 with $$a_{1}x+b_{1}y=c_{1}$$ and $$a_{2}x+b_{2}y=c_{2}$$, we can write,
$$a_{1}=1,b_{1}=1,c_{1}=2$$
$$a_{2}=2,b_{2}=2,c_{2}=3$$
Now,
$$\dfrac{a_{1}}{a_{2}} =\dfrac{1}{2} $$..............(1)
$$\dfrac{b_{1}}{b_{2}} =\dfrac{1}{2} $$..............(2)
$$\dfrac{c_{1}}{c_{2}} =\dfrac{2}{3}$$...................(3)
So from (1),(2) and (3) we can write,
$$\dfrac{a_{1}}{a_{1}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$$
Thus the equation has zero solution.
Hence the correct option is option A.
Note: If any pair of linear equations gives zero solution then it defines that the lines do not have any common point or we can say they do not have any intersecting point, that implies that the given two lines must be parallel to each other.
1) The equation has one solution, if $$\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$$.
2) Equation has many finite solution, if $$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} =\dfrac{c_{1}}{c_{2}}$$.
3) Equation has zero solution, if $$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$$
4) Equation has trivial solution, if $$c_{1}=c_{2}=0$$
Complete step-by-step solution:
Comparing the given equation x+y=2 and 2x+2y=3 with $$a_{1}x+b_{1}y=c_{1}$$ and $$a_{2}x+b_{2}y=c_{2}$$, we can write,
$$a_{1}=1,b_{1}=1,c_{1}=2$$
$$a_{2}=2,b_{2}=2,c_{2}=3$$
Now,
$$\dfrac{a_{1}}{a_{2}} =\dfrac{1}{2} $$..............(1)
$$\dfrac{b_{1}}{b_{2}} =\dfrac{1}{2} $$..............(2)
$$\dfrac{c_{1}}{c_{2}} =\dfrac{2}{3}$$...................(3)
So from (1),(2) and (3) we can write,
$$\dfrac{a_{1}}{a_{1}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$$
Thus the equation has zero solution.
Hence the correct option is option A.
Note: If any pair of linear equations gives zero solution then it defines that the lines do not have any common point or we can say they do not have any intersecting point, that implies that the given two lines must be parallel to each other.
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