In each of the following cases, find $a$ and $b$. (i) $(2a + b,\ a - b) = (8, 3)$ (ii) $\left(\dfrac{a}{4},\ a - 2b\right) = (0,\ 6 + b)$

> [!TLDR] > Two ordered pairs are equal if and only if their corresponding components are equal. Equate first components, equate second components, then solve the resulting system. **(i)** $a = \frac{11}{3}, b = \frac{2}{3}$; **(ii)** $a = 0, b = -2$. The fundamental property of ordered pairs is that $(x_1, y_1) = (x_2, y_2)$ if and only if $x_1 = x_2$ **and** $y_1 = y_2$. This gives us two independent equations from a single ordered-pair equality, turning what looks like one condition into a system we can solve. --- ## (i) $(2a + b,\ a - b) = (8, 3)$ The ordered pairs are equal, so their first components must match and their second components must match. 1. **Equate the first components:** The first entry on the left is $2a + b$, and on the right is $8$. Therefore $$2a + b = 8. \tag{1}$$ 2. **Equate the second components:** The second entry on the left is $a - b$, and on the right is $3$. Therefore $$a - b = 3. \tag{2}$$ 3. **Solve the system by elimination:** Add equations (1) and (2) to eliminate $b$: $$(2a + b) + (a - b) = 8 + 3$$ $$3a = 11$$ $$a = \frac{11}{3}.$$ 4. **Substitute back to find $b$:** Use equation (2): $$\frac{11}{3} - b = 3$$ $$b = \frac{11}{3} - 3 = \frac{11}{3} - \frac{9}{3} = \frac{2}{3}.$$ 5. **Verify (optional but good practice):** Check in equation (1): $2 \cdot \frac{11}{3} + \frac{2}{3} = \frac{22}{3} + \frac{2}{3} = \frac{24}{3} = 8$. ✓ Check in equation (2): $\frac{11}{3} - \frac{2}{3} = \frac{9}{3} = 3$. ✓ --- ## (ii) $\left(\dfrac{a}{4},\ a - 2b\right) = (0,\ 6 + b)$ …

In each of the following cases, find a and b. (i) – Class 11 Mathematics

NCERT Exemplar · Q4

a

(i)

(2a + b,\ a - b) = (8, 3)

(ii)

\left(\dfrac{a}{4},\ a - 2b\right) = (0,\ 6 + b)

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Concept understanding — Ordered Pair Equality

Ordered Pair Equality: From Intuition to Precision

Think about a simple list of two things — say, your name and your age. If I write (Ravi, 15), that's an ordered pair. The word "ordered" is the key: the first position and the second position mean different things. (Ravi, 15) is not the same as (15, Ravi), because the first one tells you a name first, the second tells you an age first.

Now, when are two such pairs equal? Intuitively, they are equal only when both the first things match and both the second things match, in that exact order.

So (Ravi, 15) equals (Ravi, 15), but it does not equal (15, Ravi) — even though both contain the same two items. The order matters.

The Precise Statement

(a, b) = (c, d) \quad \iff \quad a = c \ \text{and} \ b = d

Read this as: "The ordered pair (a, b) equals the ordered pair (c, d) if and only if a equals c and b equals d."

Two conditions must hold simultaneously:

The first components are equal: $a = c$
The second components are equal: $b = d$

If either condition fails, the pairs are different.

Why This Matters

This definition is the foundation for everything that uses ordered pairs — coordinates in the plane, relations, functions, and even complex numbers. When you plot the point $(3, 5)$ on a graph, you are implicitly using this rule: $(3, 5)$ is a different point from $(5, 3)$ because the first coordinates differ.

Watch out

A common mistake is to think $(a, b) = (b, a)$ just because the same two objects appear. That is false unless $a = b$ . For example, $(2, 3) \neq (3, 2)$ .

Quick Check

Which of these are true?

$(4, 7) = (4, 7)$ → True (both components match)
$(4, 7) = (7, 4)$ → False (first components differ: $4 \neq 7$ )
$(x, 5) = (3, 5)$ → True only if $x = 3$
$(p, q) = (q, p)$ → True only if $p = q$

The last one surprises many students. If $p = q$ , then the pair becomes $(p, p)$ and swapping gives the same thing. But if $p \neq q$ , they are different.

One More Layer: Why "Ordered"? …

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Two ordered pairs are equal if and only if their corresponding components are equal. Equate first components, equate second components, then solve the resulting system. (i) $a = \frac{11}{3}, b = \frac{2}{3}$ ; (ii) $a = 0, b = -2$ .

The fundamental property of ordered pairs is that $(x_1, y_1) = (x_2, y_2)$ if and only if $x_1 = x_2$ and $y_1 = y_2$ . This gives us two independent equations from a single ordered-pair equality, turning what looks like one condition into a system we can solve.

(i) $(2a + b,\ a - b) = (8, 3)$

The ordered pairs are equal, so their first components must match and their second components must match.

Equate the first components: The first entry on the left is $2a + b$ , and on the right is $8$ . Therefore

2a + b = 8. \qquad(1)

Equate the second components: The second entry on the left is $a - b$ , and on the right is $3$ . Therefore

a - b = 3. \qquad(2)

Solve the system by elimination: Add equations (1) and (2) to eliminate $b$ :

(2a + b) + (a - b) = 8 + 3

3a = 11

a = \frac{11}{3}.

Substitute back to find $b$ : Use equation (2):

\frac{11}{3} - b = 3

b = \frac{11}{3} - 3 = \frac{11}{3} - \frac{9}{3} = \frac{2}{3}.

Verify (optional but good practice): Check in equation (1): $2 \cdot \frac{11}{3} + \frac{2}{3} = \frac{22}{3} + \frac{2}{3} = \frac{24}{3} = 8$ . ✓ Check in equation (2): $\frac{11}{3} - \frac{2}{3} = \frac{9}{3} = 3$ . ✓

(ii) $\left(\dfrac{a}{4},\ a - 2b\right) = (0,\ 6 + b)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\left(\dfrac{x}{3}+1, y-\dfrac{2}{3}\right) = \left(\dfrac{5}{3}, \dfrac{1}{3}\right)$
(a) 1
(b) 2
(c) 3
(d) None of these
›Reveal solutionSolution

Two ordered pairs are equal only if their corresponding coordinates are equal; equate the first coordinates and solve for $x$ .

Two ordered pairs $(a,b) = (c,d)$ are equal if and only if $a = c$ and $b = d$ . Here $\left(\dfrac{x}{3}+1,\ y-\dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$ , so equating first coordinates:

$\dfrac{x}{3} + 1 = \dfrac{5}{3}$ …

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CBSE 2026Set ANNUAL1 mark
$(x+1, y-2) = (3, 1)$
›Reveal solutionSolution

Ordered pairs $(x_1,y_1)=(x_2,y_2)$ iff $x_1=x_2$ and $y_1=y_2$ ; matching components gives the value of $y$ .

Given $(x+1, y-2) = (3, 1)$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$(x - 2, 11) = (5, 11) \Rightarrow x =$
(a) 5
(b) 7
(c) 9
(d) 11
›Reveal solutionSolution

$x = 7$ , found by equating the first components of the two equal ordered pairs.

Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if $a = c$ and $b = d$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$a$
(a) $a = 4, b = 1$
(b) $a = 5, b = 2$
(c) $a = 5, b = -1$
(d) $a = 6, b = 7$
›Reveal solutionSolution

Two ordered pairs are equal only when their corresponding entries are equal. Setting up and solving the resulting linear system gives $a=5$ , $b=-1$ .

Since $(a-2b, 13) = (7, 2a-3b)$ , equate the corresponding components:

$a - 2b = 7 \quad \text{...(1)}$

$13 = 2a - 3b \quad \text{...(2)}$

From (1): $a = 7 + 2b$ .

Substitute into (2): …

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CBSE 2024Set ANNUAL1 markMCQ
$\left(\dfrac{x}{3}+1,\ y-\dfrac{2}{3}\right) = \left(\dfrac{5}{3}, \dfrac{1}{3}\right)$
(a) $x = 2,\ y = 1$
(b) $x = 1,\ y = 2$
(c) $x = -1,\ y = -3$
(d) None of these
›Reveal solutionSolution

Equate the first components and the second components of the two ordered pairs separately, then solve each simple equation.

Two ordered pairs are equal iff their first components are equal AND their second components are equal. We are given:

$\left(\frac{x}{3}+1,\ y-\frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$

First components: $\dfrac{x}{3} + 1 = \dfrac{5}{3}$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Assertion (A): If (4x + 3, y) = (3x + 5, -2), then x = 2 and y = -2 Reason (R): If A = {-1, 3, 4}, then A \times A is {(-1, -1), (-1, 3), (-1, 4), (3, -1), (4, -1), (3, 4)}$
(a) (A) is true, (R) is true; (R) is correct explanation of (A)
(b) (A) is true, (R) is true; (R) is not a correct explanation of (A)
(c) (A) is true, (R) is false
(d) (A) is false, (R) is true
›Reveal solutionSolution

(A) is verified true by solving the equal-ordered-pair equations; (R) is false because it lists only 6 of the 9 elements of $A\times A$ .

Checking Assertion (A): Two ordered pairs are equal iff corresponding components are equal.

$(4x+3,\,y) = (3x+5,\,-2)$ gives:

$4x+3 = 3x+5 \Rightarrow x=2$ , and $y=-2$ .

So $x=2,y=-2$ — Assertion (A) is true.

Checking Reason (R): For $A=\{-1,3,4\}$ , $A\times A$ should contain all $3\times3=9$ ordered pairs:

$(-1,-1),(-1,3),(-1,4),(3,-1),(3,3),(3,4),(4,-1),(4,3),(4,4)$ . …

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CBSE 2023Set ANNUAL1 mark
$\left(\dfrac{x}{3}+1,\ y-\dfrac{2}{3}\right)=\left(\dfrac{5}{3},\dfrac{1}{3}\right)$
›Reveal solutionSolution

$x=2,\ y=1$ .

By equality of ordered pairs, $\left(\dfrac x3+1,\ y-\dfrac23\right)=\left(\dfrac53,\dfrac13\right)$ means:

$\dfrac x3+1=\dfrac53 \quad\text{and}\quad y-\dfrac23=\dfrac13.$

…

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CBSE 2023Set ANNUAL1 markMCQ
$\left(\frac{x}{3} + 1,\ y - \frac{2}{3}\right) = \left(\frac{5}{3},\ \frac{1}{3}\right)$
(a) $x = 2,\ y = 1$
(b) $x = 1,\ y = 1$
(c) $x = 1,\ y = 1$
(d) $x = 2,\ y = 2$
›Reveal solutionSolution

Equating components gives $x=2,\ y=1$ ; option (a).

By equality of ordered pairs (NCERT Class 11 Relations and Functions):

…

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CBSE 2022Set TERM11 markMCQ
$(x+y, x-y) = (3, 5)$
(a) $x=3, y=1$
(b) $x=1, y=3$
(c) $x=4, y=-1$
(d) $x=5, y=2$
›Reveal solutionSolution

Two ordered pairs are equal exactly when their corresponding components are equal, giving a linear system to solve.

…

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CBSE 2022Set ANNUAL1 mark
$Q. If (x - 2) + (y + 1)i = 3 + 4i, then the value of x and y are ............ and ............ .$
›Reveal solutionSolution

Comparing real and imaginary parts directly gives $x = 5$ , $y = 3$ .

$(x - 2) + (y+1)i = 3 + 4i$

Two complex numbers are equal only when their real parts are equal AND their imaginary parts are equal.

…

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CBSE 2021Set ANNUAL1 mark
$Q. If 4x + i(3x - y) = 3 - 6i, then the value of x and y are ............., ............. respectively.$
›Reveal solutionSolution

Matching real and imaginary parts gives $x = 3/4$ and $y = 33/4$ .

Two complex numbers are equal only if their real parts match and their imaginary parts match.

$4x + i(3x-y) = 3 - 6i$

Real parts: $4x = 3 \Rightarrow x = \dfrac{3}{4}$

…

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CBSE 2020Set ANNUAL1 markMCQ
$(x + y, x - y) = (3, 5)$
(a) $x = 3, y = 1$
(b) $x = 1, y = 3$
(c) $x = 4, y = -1$
(d) $x = 5, y = 2$
›Reveal solutionSolution

Two ordered pairs are equal exactly when their corresponding components match, giving two linear equations to solve together.

$(x+y, x-y) = (3,5)$ means:

$x + y = 3 \quad \text{...(i)}$

$x - y = 5 \quad \text{...(ii)}$

Add (i) and (ii): $2x = 8 \Rightarrow x = 4$ . …

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Q.In each of the following cases, find aaa and bbb.

(i) (2a+b, a−b)=(8,3)(2a + b,\ a - b) = (8, 3)(2a+b, a−b)=(8,3)

(ii) (a4, a−2b)=(0, 6+b)\left(\dfrac{a}{4},\ a - 2b\right) = (0,\ 6 + b)(4a​, a−2b)=(0, 6+b) …

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Q.In each of the following cases, find $a$ and $b$ .

(i) $(2a + b,\ a - b) = (8, 3)$

(ii) $\left(\dfrac{a}{4},\ a - 2b\right) = (0,\ 6 + b)$ …

Unlock everything free for 14 days