Question
By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them:
$x + y = 3, 3x + 3y = 9.$
$x + y = 3, 3x + 3y = 9.$
✓
Answer
Given pair of equations is x + y = 3 .....(i) 3x + 3y = 9 .....(ii)
On comparing with $ax + by + c = 0$,
we get $a_1 = 1, b_1 = 1$ and $c_1 = -1$
[from Eq. (i)] $a_2 = 3, b_2 = 3$ and $c_2 = -9$
[from Eq. (ii)]Here, $\frac{\text{a}_1}{\text{a}_1}=\frac{1}{3},$
$\frac{\text{b}_1}{\text{b}_2}=\frac{1}{3}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-9}=\frac{1}{3}$
$\Rightarrow\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of lines is coincident.
Therefore, these lines have infinitely many solutions.
Hence, the given pair of linear equations is consistent.
Now, $x + y = 3$
$\Rightarrow y = 3 - x$
If x = 0, then y = 3, if x = 3, then y = 0
and $3x + 3y = 9$
$\Rightarrow 3y = 9 - 3x$
$\Rightarrow\ \text{y}=\frac{9-3\text{x}}{3}$
If x = 0, then y = 3,
if x = 1, then y = 2 and
if x = 3, then y = 0
Potting the points A(0, 3) and B(3, 0), we get the line AB. Again, plotting the points C(0, 3) D(1, 2) and E(3, 0), we get the line CDE. We observe that the lines represented by Eqs. (i) and (ii) are coincident.
On comparing with $ax + by + c = 0$,
we get $a_1 = 1, b_1 = 1$ and $c_1 = -1$
[from Eq. (i)] $a_2 = 3, b_2 = 3$ and $c_2 = -9$
[from Eq. (ii)]Here, $\frac{\text{a}_1}{\text{a}_1}=\frac{1}{3},$
$\frac{\text{b}_1}{\text{b}_2}=\frac{1}{3}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-9}=\frac{1}{3}$
$\Rightarrow\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of lines is coincident.
Therefore, these lines have infinitely many solutions.
Hence, the given pair of linear equations is consistent.
Now, $x + y = 3$
$\Rightarrow y = 3 - x$
If x = 0, then y = 3, if x = 3, then y = 0
| x | 0 | 3 |
| y | 3 | 0 |
| Points | A | B |
$\Rightarrow 3y = 9 - 3x$
$\Rightarrow\ \text{y}=\frac{9-3\text{x}}{3}$
If x = 0, then y = 3,
if x = 1, then y = 2 and
if x = 3, then y = 0
| x | 0 | 1 | 3 |
| y | 3 | 2 | 0 |
| Points | C | D | E |
Potting the points A(0, 3) and B(3, 0), we get the line AB. Again, plotting the points C(0, 3) D(1, 2) and E(3, 0), we get the line CDE. We observe that the lines represented by Eqs. (i) and (ii) are coincident.
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