Q.If $\left(\dfrac{x}{3} + 1,\ y - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$, find the values of $x$ and $y$.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\left(\dfrac{x}{3}+1, y-\dfrac{2}{3}\right) = \left(\dfrac{5}{3}, \dfrac{1}{3}\right)$ then the value of $x$ is

   (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:

   $\dfrac{x}{3} + 1 = \dfrac{5}{3}$

   $\dfrac{x}{3} = \dfrac{5}{3} - 1 = \dfrac{2}{3}$

   $x = 2$

   (Equating second coordinates similarly gives $y - \tfrac{2}{3} = \tfrac{1}{3} \implies y = 1$, though only $x$ is asked.)

   ✓Final answer

   (b) 2.

2. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: If $(x+1, y-2) = (3, 1)$, then the value of $y$ is ______.

   ›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 $(x+1, y-2) = (3, 1)$.

   Equating the first components: $x+1 = 3 \Rightarrow x = 2$.

   Equating the second components: $y-2 = 1 \Rightarrow y = 3$.

   ✓Final answer

   $y = 3$.

3. CBSE 2025Set ANNUAL1 markMCQ
   Q.$(x - 2, 11) = (5, 11) \Rightarrow x =$

   (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 $b = d$.

   Given $(x - 2, 11) = (5, 11)$: the second components already match ($11 = 11$). Equating the first components: $x - 2 = 5 \Rightarrow x = 7$.

   ✓Final answer

   The correct option is (b) 7.

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Find $a$ and $b$ when $(a - 2b, 13) = (7, 2a - 3b)$.

   (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):

   $13 = 2(7+2b) - 3b = 14 + 4b - 3b = 14 + b$

   $b = 13 - 14 = -1$

   Then $a = 7 + 2(-1) = 7 - 2 = 5$.

   Check: $a-2b = 5-2(-1) = 5+2=7$ correct; $2a-3b = 10-(-3)=10+3=13$ correct.

   ✓Final answer

   (c) $a = 5, b = -1$

5. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $\left(\dfrac{x}{3}+1,\ y-\dfrac{2}{3}\right) = \left(\dfrac{5}{3}, \dfrac{1}{3}\right)$, then the values of $x$ and $y$ are

   (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: $\dfrac{x}{3} + 1 = \dfrac{5}{3}$

   $\dfrac{x}{3} = \dfrac{5}{3} - 1 = \dfrac{5-3}{3} = \dfrac{2}{3}$

   $x = 2$

   Second components: $y - \dfrac{2}{3} = \dfrac{1}{3}$

   $y = \dfrac{1}{3} + \dfrac{2}{3} = \dfrac{3}{3} = 1$

   So $x = 2$ and $y = 1$.

   ✓Final answer

   (a) $x=2,\ y=1$.

6. 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 × 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:

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

   The reason statement lists only 6 pairs and omits $(3,3),(4,3),(4,4)$ — so as stated, $A\times A$ is described incompletely, making Reason (R) false.

   So (A) is true, (R) is false.

   ✓Final answer

   (A) is true, (R) is false — option (c).

7. CBSE 2023Set ANNUAL1 mark
   Q.If $\left(\dfrac{x}{3}+1,\ y-\dfrac{2}{3}\right)=\left(\dfrac{5}{3},\dfrac{1}{3}\right)$, find the values of $x$ and $y$.

   ›Reveal solutionSolution

   $x=2,\ y=1$.

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

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

   From the first: $\dfrac x3=\dfrac53-1=\dfrac23 \Rightarrow x=2$.

   From the second: $y=\dfrac13+\dfrac23=1$.

   ✓Final answer

   $x=2,\ y=1$.

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.If $\left(\frac{x}{3} + 1,\ y - \frac{2}{3}\right) = \left(\frac{5}{3},\ \frac{1}{3}\right)$ then value of $x$ and $y$ is

   (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):

   $\dfrac{x}{3}+1=\dfrac53\Rightarrow \dfrac{x}{3}=\dfrac23\Rightarrow x=2,$

   $y-\dfrac23=\dfrac13\Rightarrow y=1.$

   ✓Final answer

   (a) $x=2,\ y=1$.

9. CBSE 2022Set TERM11 markMCQ
   Q.If $(x+y, x-y) = (3, 5)$ then $x$ and $y$ will be

   (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.

   From $(x+y,\ x-y)=(3,5)$: $x+y=3$ ...(i), and $x-y=5$ ...(ii). Adding (i) and (ii): $2x=8\Rightarrow x=4$. Substituting into (i): $4+y=3\Rightarrow y=-1$.

   ✓Final answer

   (c) $x=4,\ y=-1$.

10. 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.

    Real parts: $x - 2 = 3 \Rightarrow x = 5$

    Imaginary parts: $y + 1 = 4 \Rightarrow y = 3$

    ✓Final answer

    $x = 5$, $y = 3$.

11. 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: $4x = 3 \Rightarrow x = \dfrac{3}{4}$

    Imaginary parts: $3x - y = -6 \Rightarrow y = 3x + 6 = 3\left(\dfrac{3}{4}\right) + 6 = \dfrac{9}{4} + \dfrac{24}{4} = \dfrac{33}{4}$

    ✓Final answer

    $x = \dfrac{3}{4}$, $y = \dfrac{33}{4}$.

12. CBSE 2020Set ANNUAL1 markMCQ
    Q.If $(x + y, x - y) = (3, 5)$ then $x$ and $y$ will be:

    (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): $2x = 8 \Rightarrow x = 4$.

    Substitute back into (i): $4 + y = 3 \Rightarrow y = -1$.

    Check: $x-y = 4-(-1)=5$ ✓.

    ✓Final answer

    (c) $x=4, y=-1$