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Solving Equations Using Method of Substitution

Assertion: If...

Question

Assertion: If $x \neq y$ , then $(x, y) \neq (y, x)$
Reason: If x=y, then $(x, y) \neq (y, x)$
Which of the following is correct?

A

Both A and R are true and R is the correct explanation of A.

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B

Both A and R are true and R is not the correct explanation of A.

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C

A is true and R is false.

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D

A is false and R is true.

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Solution

The correct option is C
A is true and R is false.

$A : x \neq y$ , it is clear $(x, y) \neq (y, x)$
So, A is true.
R : R is false.

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Similar questions

Q. Assertion (A)
The system of equations

x + y - 8 = 0 and x - y - 2 = 0

has a unique solutions.
Reason (R)
The system of equations

a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0

has a unique solution when

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} .

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Q. Assertion (A)
The point P(−1, 6) divides the join of A(−3, 10) and B(6, −8) in the ratio 2 : 7.

Reason (R)
If the point C(x, y) divides the join of A(x₁, y₁) and B(x₂, y₂) in the ratio m : n, then

x = \frac{m x_{2} + n x_{1}}{m + n} and y = \frac{m y_{2} + n y_{1}}{m + n} .

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Q. Assertion (A)
If on dividing the polynomial p(x) = x² − 3ax + 3a − 7 by (x + 1), we get 6 as remainder, then a= 3.

Reason (R)
When a polynomial p(x) is divided by (x − α), then the remainder is p(α).

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Q. Assertion (A)
The equation x² + x + 1 = 0 has both real roots.

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The equation ax2 + bx + c = 0, (a ≠ 0) has both real roots, if (b2 − 4ac) > 0.

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