
How do you solve the system of equations $2y+x-4=0$ and $2y=-x+4$?
Answer
464.4k+ views
Hint: We will write both the linear equations in standard form. Then we will check the ratios of the corresponding coefficients in both the equations. We will check which condition for the solution of a system of linear equations is being satisfied by the ratios of the coefficients. After verifying the condition, we can arrive at a conclusion about the solution of the system of linear equations.
Complete step-by-step solution:
The system of linear equations consists of
$2y+x-4=0....(i)$
and the second equation,
$2y=-x+4$
This equation can be rewritten in the standard form as
$-x-2y+4=0....(ii)$
Now, we will compare both these equations to the standard linear equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$. So, we get the coefficients as ${{a}_{1}}=1$, ${{b}_{1}}=2$, ${{c}_{1}}=-4$, ${{a}_{2}}=-1$, ${{b}_{2}}=-2$ and ${{c}_{2}}=4$. We will look at the ratios of the corresponding coefficients of both the linear equations. So, we have the following ratios,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{1}{-1}=-1$
$\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{2}{-2}=-1$
$\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-4}{4}=-1$
Therefore, we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=-1$. Now, we know that if we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k$ then the system of linear equations have infinite number of solutions. Hence, we can conclude that the given system of equations has an infinite number of solutions.
Note: The conditions for the solution of a system of linear equations is the key aspect in this type of questions. There are three conditions. If we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k$ then the system of linear equations have infinite number of solutions. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ then the system of linear equations has no solutions, that is, the system of linear equations is inconsistent. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then the system of linear equations has a unique solution.
Complete step-by-step solution:
The system of linear equations consists of
$2y+x-4=0....(i)$
and the second equation,
$2y=-x+4$
This equation can be rewritten in the standard form as
$-x-2y+4=0....(ii)$
Now, we will compare both these equations to the standard linear equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$. So, we get the coefficients as ${{a}_{1}}=1$, ${{b}_{1}}=2$, ${{c}_{1}}=-4$, ${{a}_{2}}=-1$, ${{b}_{2}}=-2$ and ${{c}_{2}}=4$. We will look at the ratios of the corresponding coefficients of both the linear equations. So, we have the following ratios,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{1}{-1}=-1$
$\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{2}{-2}=-1$
$\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-4}{4}=-1$
Therefore, we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=-1$. Now, we know that if we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k$ then the system of linear equations have infinite number of solutions. Hence, we can conclude that the given system of equations has an infinite number of solutions.
Note: The conditions for the solution of a system of linear equations is the key aspect in this type of questions. There are three conditions. If we have $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k$ then the system of linear equations have infinite number of solutions. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ then the system of linear equations has no solutions, that is, the system of linear equations is inconsistent. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then the system of linear equations has a unique solution.
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