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Question: How do you tell whether the system \(y=-2x+1\) and \(y=-\dfrac{1}{3}x-3\) has no solution or infinit...

How do you tell whether the system y=2x+1y=-2x+1 and y=13x3y=-\dfrac{1}{3}x-3 has no solution or infinitely many solutions?

Explanation

Solution

These questions can be solved by comparing the coefficients of the given system of equations. One can also find whether the given system of equations has no solution or infinitely many solutions by plotting the graph for the equations.

Complete step-by-step solution:
The given equations are:
y=2x+1y=-2x+1 and y=13x3y=-\dfrac{1}{3}x-3
The above equations can also be written as:
2x+y1=0\Rightarrow 2x+y-1=0 And 13x+y+3=0\dfrac{1}{3}x+y+3=0 .
Comparing the above equations with the pair of general equations 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 we get
a1=2,b1=1,c1=1\Rightarrow {{a}_{1}}=2,{{b}_{1}}=1,{{c}_{1}}=-1 And a2=13,b2=1,c2=3{{a}_{2}}=\dfrac{1}{3},{{b}_{2}}=1,{{c}_{2}}=3
From the above values, we can see that ,
a1a2=213=6\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{2}{\dfrac{1}{3}}=6 and b1b2=11\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{1}{1} .
a1a2b1b2\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}, therefore, the given system of equations neither has infinitely many solutions nor no solution, but has a unique solution, since the graphs of the given equations will intersect at a distinct point.

Note: These questions can be solved easily by comparing the coefficients of the given equations and then by checking what cases are satisfied by the coefficients. The cases that usually occur are:
Case 1: If a1a2=b1b2=c1c2\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}} then there will be infinitely many solutions of the given system of equations. If we plot the graph of these equations, then the lines will coincide, hence giving infinite solutions. These equations are also called as a consistent system of equations and are dependent.
Case 2: If a1a2=b1b2c1c2\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}} then there will be no solution and the system of equations is said to be inconsistent. The graphs of the equations do not intersect; hence the lines are parallel and there is no solution.
Case 3: If a1a2b1b2\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}} then there is exactly one solution and the system of equations is considered to be a consistent and independent system. If we plot the graphs of the equations, then they will intersect at a single point, giving a unique solution.