Mettl Equation Quiz 1

Let the larger number be x and the smaller be y.
Then, 2x – 20 = y and 2y – 5 = x
∴ 2x – y = 20 & x – 2y = –5.
by solving these equation, we get:x = 15 & y = 10.
Hence, the larger number = 15.

Let the number be x and y. then,
x + y = 80 ...... (i)
and x - 4y = 5. ....... (ii)
Solving these equation, we get:
y = 15.

Let there be x hens and y cows. Then,
x + y = 48 and 2x + 4y = 140
Solving x + y = 48 and x + 2y = 70, we get: x = 26

Let ten's digit = x & unit digit = y. Then,
x + y = 10 & 10x + y - 18 = 10y + x
∴ x + y = 10 & x - y = 2. So, x = 6 and y = 4.
So, the number is 64.

The given equations are 3x + y = 1 & 2(3x + y) = 2.
or 3x + y = 1 & 3x + y = 1
Thus, there is one equation in two variables.

Therefore this system of equations does not have any solutions

Video solution:

I. 2x² – 12x + 16 = 0
⇒ 2x² – 4x – 8x + 16 = 0
⇒ 2x (x – 2) – 8 (x – 2) = 0
⇒ (2x – 8) (x – 2) = 0
x = 4, 2
II. y² – 7y + 12 = 0
⇒ y² – 4y – 3y + 12 = 0
⇒ y (y – 4) – 3 (y – 4) = 0
⇒ (y – 3) (y – 4) = 0
y = 4, 3
While comparing the root values of x and y, we find that one root value of y lies between the value of x's root values. Hence, the relationship between x and y can't be established.

I. x² + 7x + 12 = 0
⇒ x² + 3x + 4x + 12 = 0
⇒ x (x + 3) + 4 (x + 3) = 0
⇒ (x + 3) (x + 4) = 0
x = - 3, - 4
II. y² - 2y - 15 = 0
⇒ y² - 5y + 3y - 15 = 0
⇒ y (y - 5) + 3 (y - 5) = 0
⇒ (y + 3) (y - 5) = 0
y = – 3, 5
While comparing the root values of x and y, we find that one root value of y is equal to x and the other one is less than y's root values. Hence, x ≤ y.

I. 2x² – 10x + 12 = 0
⇒ 2x² – 6x – 4x + 12 = 0
⇒ 2x (x – 3) – 4 (x – 3) = 0
⇒ (2x – 4) (x – 3) = 0
x = 2, 3

II. 2y² – 19y + 35 = 0
⇒ 2y² – 14y – 5y + 35 = 0
⇒ 2y (y – 7) – 5 (y – 7) = 0
⇒ (2y – 5) (y – 7) = 0
y = 2.5, 7
While comparing the root values of x and y, we find that one root value of x lies between the values of y's. Hence, the relationship between x and y can't be established.

I. x² - 6x - 16 = 0
⇒ x² + 2x - 8x - 16 = 0
⇒ x (x + 2) - 8 (x + 2) = 0
⇒ (x - 8) (x + 2) = 0
x = 8, - 2
II. y² + 5y + 6 = 0
⇒ y² + 3y + 2y + 6 = 0
⇒ y (y + 3) + 2 (y + 3) = 0
⇒ (y + 3) (y + 2) = 0
y = - 3, - 2

While comparing the root values of x and y, we find that one root value of x is equal to the value of y's and another one is greater than y's root values. Hence, x.≥ y.

I. x² - x - 20 = 0
⇒ x² - 5x + 4x - 20 = 0
⇒ x (x - 5) + 4 (x - 5) = 0
⇒ (x + 4) (x - 5) = 0
x = – 4, 5
II. y² - 3y + 2 = 0
⇒ y² - 2y - y + 2 = 0
⇒ y (y - 2) - 1 (y - 2) = 0
⇒ (y - 1) (y - 2) = 0
y = 1, 2
While comparing the root values of x and y, we find that both the root values of y lies between the values of x's. Hence, the relationship between x and y can't be established.