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

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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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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