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Exercise 2.1 · Q1

Q.If (x3+1, y−23)=(53, 13)\left(\dfrac{x}{3} + 1,\ y - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right), find the values of xx and yy.

Uttar Pradesh UpmspTextbookSubjective· 2mImportance★★★★★est
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Ordered pairs are equal only when their corresponding components are equal. Equating the xx-coordinates gives x=2x = 2, and equating the yy-coordinates gives y=1y = 1.

The idea is simple: an ordered pair is a pair of numbers where order matters.

So (a,b)=(c,d)(a, b) = (c, d) means a=ca = c and b=db = d — both must hold at the same time.

This is called the equality of ordered pairs, and it’s the only rule we need here.

We have:

(x3+1, y−23)=(53, 13)\left(\frac{x}{3} + 1,\ y - \frac{2}{3}\right) = \left(\frac{5}{3},\ \frac{1}{3}\right)

Let’s break it down.

  1. Equate the first components (the xx-coordinates):

x3+1=53\frac{x}{3} + 1 = \frac{5}{3}

Subtract 11 from both sides. Write 11 as 33\frac{3}{3} to keep fractions consistent:

x3=53−33=23\frac{x}{3} = \frac{5}{3} - \frac{3}{3} = \frac{2}{3}

Multiply both sides by 33:

x=2x = 2

  1. Equate the second components (the yy-coordinates):

y−23=13y - \frac{2}{3} = \frac{1}{3}

Add 23\frac{2}{3} to both sides:

y=13+23=33=1y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1

Watch out

A common mistake is to forget that both equalities must hold. Some students solve only one equation and assume the other automatically works — but here each coordinate gives a separate condition. Always check both.

Tip

When fractions look messy, rewrite whole numbers as fractions with the same denominator. Here 1=331 = \frac{3}{3} made the subtraction clean.

✓Final answer

The values are x=2x = 2 and y=1y = 1.

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