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.