
Check whether the following pair of linear equations are consistent or inconsistent.
3x + 2y = 5, 2x - 3y = 7.
Answer
531.1k+ views
Hint: For checking whether the pair of linear equations are consistent or inconsistent, we try to obtain values of x and y. If both x and y have a unique value then the system is consistent. The system becomes inconsistent when there exist no values of x and y that satisfy both the equations.
Complete step-by-step answer:
According to the given system of equations, we assign equations corresponding to the expression.
Let the first expression be: $3x+2y=5\ldots (1)$
The second expression will be: $2x-3y=7\ldots (2)$
Now, we try to eliminate one of the variables x or y by using both the equations.
To do so, we multiply the equation (1) with 3 and multiply the equation (2) with 2.
$\begin{align}
& \left( 3x+2y=5 \right)\times 3 \\
& 9x+6y=15\ldots (3) \\
& \left( 2x-3y=7 \right)\times 2 \\
& 4x-6y=14\ldots (4) \\
\end{align}$
Since both the equations have the same value of y, it can be eliminated. Now, adding equation (3) and (4), we get
$\begin{align}
& 9x+6y-15+\left( 4x-6y-14 \right)=0 \\
& 9x+4x+6y+6y-15-14=0 \\
& 13x-29=0 \\
& x=\dfrac{29}{13} \\
\end{align}$
So, the obtained value of x is $\dfrac{29}{13}$.
Putting the value of x in equation 1, we get
$\begin{align}
& 3\times \dfrac{29}{13}+2y=5 \\
& 2y=5-\dfrac{87}{13} \\
& 2y=\dfrac{65-87}{13} \\
& 2y=-\dfrac{22}{13} \\
& y=-\dfrac{11}{13} \\
\end{align}$
Hence, the value of y is $-\dfrac{11}{13}$.
Since there exists a unique value of x and y, therefore the system is consistent.
Note: This problem can alternatively be solved by using the coefficient analysis method for determination of consistent system. In this method the coefficients of x and y i.e. a and b, are compare and if the condition \[\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\] is satisfied, then the system is consistent.
Complete step-by-step answer:
According to the given system of equations, we assign equations corresponding to the expression.
Let the first expression be: $3x+2y=5\ldots (1)$
The second expression will be: $2x-3y=7\ldots (2)$
Now, we try to eliminate one of the variables x or y by using both the equations.
To do so, we multiply the equation (1) with 3 and multiply the equation (2) with 2.
$\begin{align}
& \left( 3x+2y=5 \right)\times 3 \\
& 9x+6y=15\ldots (3) \\
& \left( 2x-3y=7 \right)\times 2 \\
& 4x-6y=14\ldots (4) \\
\end{align}$
Since both the equations have the same value of y, it can be eliminated. Now, adding equation (3) and (4), we get
$\begin{align}
& 9x+6y-15+\left( 4x-6y-14 \right)=0 \\
& 9x+4x+6y+6y-15-14=0 \\
& 13x-29=0 \\
& x=\dfrac{29}{13} \\
\end{align}$
So, the obtained value of x is $\dfrac{29}{13}$.
Putting the value of x in equation 1, we get
$\begin{align}
& 3\times \dfrac{29}{13}+2y=5 \\
& 2y=5-\dfrac{87}{13} \\
& 2y=\dfrac{65-87}{13} \\
& 2y=-\dfrac{22}{13} \\
& y=-\dfrac{11}{13} \\
\end{align}$
Hence, the value of y is $-\dfrac{11}{13}$.
Since there exists a unique value of x and y, therefore the system is consistent.
Note: This problem can alternatively be solved by using the coefficient analysis method for determination of consistent system. In this method the coefficients of x and y i.e. a and b, are compare and if the condition \[\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\] is satisfied, then the system is consistent.
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