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If \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\], then the pair of linear equation has ………... solutions.
Answer
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Hint: These \[a,b,c\] are the coefficients of \[x,y\] and constant terms in the equations \[1\] and \[2\]. As it is given that the ratios of these coefficients are equal that means these two equations are the same just the equivalent fraction where we multiply the numerator and denominator with the same number after this the overall fraction is the same. Similarly, the overall equation will be the same just the written form is different. When we reduce the equation in simplest form that means the equation is in simplified form but that equation is the same. These two same equations when we plot on the graph, they will both coincide hence there are a lot of or infinity points where these two have the same value. Hence it has infinite solutions.
Complete Step by Step Solution:
Let consider two linear equations of standard form
\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0~{ }--{ }\left( 1 \right)\]
\[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0{ }--(2)\]
As it is given that \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
Since the ratio of coefficients are same, the other one can be written in terms of other
Let this ratio be equal to \[k\]
\[\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k\]
\[\Rightarrow {{a}_{1}}={{a}_{2}}k{ },{ }{{b}_{1}}={{b}_{2}}k{ },{ }{{c}_{1}}={{c}_{2}}k\]
Now replacing the coefficients of first equation in terms of a single variable \[k\] as per above
\[\Rightarrow k{{a}_{2}}x+k{{b}_{2}}y+k{{c}_{2}}=0\]
Now taking common \[k\] from above equation
\[\Rightarrow k({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0){ }--(3)\]
\[\Rightarrow ({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0){ }--(4)\]
Since, \[Eq(4)\] and \[Eq(1)\] are same, as we have just modified
Here, we clearly see that \[Eq(1)\] and \[Eq(2)\] are the same means when plotting these two on graphs they will coincide and thus an infinite number of common points.
Hence, an infinite number of solutions.
Note:
As in this question the ratio of their coefficients is the same that means one equation can be written in terms of another and furthermore it leads to the same equation. When the ratio of coefficients is equal but the ratio of coefficients of constant term is not equal means half part of equation is same only constant term differs hence this will lead to no solution.
Complete Step by Step Solution:
Let consider two linear equations of standard form
\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0~{ }--{ }\left( 1 \right)\]
\[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0{ }--(2)\]
As it is given that \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
Since the ratio of coefficients are same, the other one can be written in terms of other
Let this ratio be equal to \[k\]
\[\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k\]
\[\Rightarrow {{a}_{1}}={{a}_{2}}k{ },{ }{{b}_{1}}={{b}_{2}}k{ },{ }{{c}_{1}}={{c}_{2}}k\]
Now replacing the coefficients of first equation in terms of a single variable \[k\] as per above
\[\Rightarrow k{{a}_{2}}x+k{{b}_{2}}y+k{{c}_{2}}=0\]
Now taking common \[k\] from above equation
\[\Rightarrow k({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0){ }--(3)\]
\[\Rightarrow ({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0){ }--(4)\]
Since, \[Eq(4)\] and \[Eq(1)\] are same, as we have just modified
Here, we clearly see that \[Eq(1)\] and \[Eq(2)\] are the same means when plotting these two on graphs they will coincide and thus an infinite number of common points.
Hence, an infinite number of solutions.
Note:
As in this question the ratio of their coefficients is the same that means one equation can be written in terms of another and furthermore it leads to the same equation. When the ratio of coefficients is equal but the ratio of coefficients of constant term is not equal means half part of equation is same only constant term differs hence this will lead to no solution.
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