Mathematics
Grade10
Easy
Question
Solve the system:
y = x + 4
5y = 5x + 20
- No solution
- (11, 8)
- (5, 15)
- Infinitely many solutions.
Hint:
we have given two equation, we have to solve the system. We have two equation which is y = x + 4 and 5y = 5x + 20 .Make two equation , if (a1/a2) = (b1/b2) = (c1/c2) then it is infinitely many solutions, if (a1/a2) = (b1/b2) ≠ (c1/c2) and if (a1/a2) ≠(b1/b2) then it is unique solution.
The correct answer is: Infinitely many solutions.
Here we have to find the system of equation.
Firstly, we have given equation y = x + 4 and 5y = 5x + 20.
y = x + 4 --(1)
5y = 5x + 20 --(2)
We have a1 = 1 , b1 = 1 and c1 = 4
And a2 = 5 , b2 = 5 and c2 = 20,
Now , a1/a2 = 1/5 ,
b1/b2 = 1/5
and c1/ c2 = 4/20 = 1/5
therefore, a1/a2 = b1/b2 = c1/c2
Therefore , it solution having infinitely many solution.
The correct answer is Infinitely many solution.
Or,
y = x + 4 …(i)
5y = 5x + 20 …(ii)
Substituting y from (i) in (ii), we get
5(x + 4) = 5x + 20
5x + 20 = 5x + 20
20 = 20
The statement 20 = 20 is an identity, so the system of equations has infinitely many solutions.
In this question, we have solve this question by system of equation we have , if (a1/a2) = (b1/b2) = (c1/c2) then it is infinitely many solutions, if (a1/a2) = (b1/b2) ≠ (c1/c2) and if (a1/a2) ≠(b1/b2) then it is unique solution.
