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Sagot :
Step-by-step explanation:
To find - How can you determine the number of solutions of a system
of linear equations by inspecting its equations?
Proof -
There exists three types of solution for a system of linear equations.
1. No solution.
2. Unique solution.
3. Infinite many solutions.
Now,
Let the 2 linear equations are as follows :
y = ax + b
y = cx + d
1. For no solution -
a = c and b ≠d
For example-
y = -2x + 5
y = -2x - 1
Subtract first equation from second , we get
y - y = -2x - 1 - ( -2x + 5)
⇒0 = -2x - 1 + 2x - 5
⇒0 = -6
Not possible
So, There exist no solution.
2. For unique solution -
a ≠ c and b , d can be any value
For example -
If b and d are same ,
y = 2x - 4
y = -3x - 4
Subtract equation 1st from 2nd , we get
y - y = -3x - 4 - ( 2x - 4)
⇒0 = -3x - 4 - 2x + 4
⇒0 = -5x
⇒x = 0
⇒y = 2(0) - 4 = -4
∴ we get
x = 0, y = -4
If b and d are different
y = 2x - 4
y = -3x + 5
Subtract equation 1st from second
y - y = -3x + 5 - 2x + 4
⇒0 = -5x + 9
⇒-5x = -9
⇒x = [tex]\frac{9}{5}[/tex]
⇒y =2([tex]\frac{9}{5}[/tex]) - 4 = [tex]\frac{18}{5} - 4 = \frac{18 - 20}{5} = -\frac{2}{5}[/tex]
∴ we get
x = [tex]\frac{9}{5}[/tex] , y = [tex]-\frac{2}{5}[/tex]
3. Infinitely many solution -
a = c and b = d
For example -
y = -4x + 1
y = -4x + 1
Subtract equation first from equation second
y - y = -4x + 1 - ( -4x + 1)
⇒0 = -4x + 1 + 4x - 1
⇒0 = 0
Satisfied
So, Infinite many solution possible.
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