Check whether the following equations are consistent or inconsistent solve them graphically x-y=8

Check whether the following pair of linear equations are consistent or inconsistent.
3x + 2y = 5, 2x - 3y = 7.

Answer

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.

(i) x + y = 5 ⇒ x + y – 5 = 0 

2x + 2y = 10 ⇒ 2x + 2y – 10 = 0 

a1 = 1, b1 = 1, c1 = -5 

a2 = 2, b2 = 2, c2 = -10 

Here, 

∴ Pair of equations are consistent 

(i) x + y = 5 

y = 5 – x

(ii) 2x + 2y = 10
x + y = 5
y = 5 – x

∴ We can give any value for ‘x’, 

i.e., solutions are infinite.
∴ P (5, 0) x = 5, y = 0

(ii) x – y = 8 ⇒ x – y – 8 = 0 

3x – 3y = 16 

⇒ 3x – 3y – 16 = 0

∴ Linear equations are in consistent tent. 

∴ Algebraically it has no solution. 

Graphical representation → Parallel Lines. 

(i) x – y = 8 

-y = 8 – x 

y = -8 + x

(ii) 3x – 3y = 16
-3y = 16 – 3x
3y = -16 + 3x

No solution because it is inconsistent

(iii) 2x + y – 6 = 0 

4x – 2y – 4 = 0 

Here a1 = 2, b1 = 1, c1 = -6 

a2 = 4, b2 = -2, c2 = -4

Pair of equations are consistent. 

Algebraically both lines intersect. 

Graphical Representation : 

(i) 2x + y = 6 

y = 6 – 2x

(ii) 4x – 2y – 4 = 0
4x – 2y = 4
-2y = 4 – 4x
2y = -4 + 4x

Solution: intersecting point, P (2, 2) 

i.e., x = 2, y = 2

(iv) 2x – 2y – 2 = 0 

4x – 3y – 5 = 0 

a1 = 2, b1 = -2, c1 = -2 

a2 = 4, b2 = -3, c2 = -5

Pair of equations are consistent. 

∴ Algebraically both lines intersect. 

Graphical Representation : 

(i) 2x – 2y – 2 =0 

2x – 2y = 2 

-2y = 2 – 2x 

2y = -2 + 2x

∴ y =\( \frac{-2 + 2x}{2}\)

∴ y = – 1 + x

(ii) 4x – 3y – 5 = 0 

4x – 3y = 5 

-3y = 5 – 4x 

3y = -5 + 4x

∴ y = \(\frac{-5 + 4x}{3}\)

Solution: P(2, 1) 

i.e., x = 2, y = 1