1. If <math> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>y</mi> <mo>−</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> , find the values of x and y.

1. Since the ordered pairs are equal, the corresponding elements are equal. ∴ x 3 + 1 = 5 3 . and y−23=13 x 3 = 5 3 − 1 y = 1 3 + 2 3 ⇒ x 3 = 5 − 3 3 ⇒ y = 1 + 2 3 ⇒ x 3 = 2 3 ⇒ y = 3 3 ⇒ x = 2 y = 1.

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

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Payal Gupta | Contributor-Level 10

4 months ago

1. Since the ordered pairs are equal, the corresponding elements are equal.

$∴ \frac{x}{3} + 1 = \frac{5}{3} .$ and $y - \frac{2}{3} = \frac{1}{3}$

$\frac{x}{3} = \frac{5}{3} - 1$ $y = \frac{1}{3} + \frac{2}{3}$

$\Rightarrow \frac{x}{3} = \frac{5 - 3}{3}$ $\Rightarrow y = \frac{1 + 2}{3}$

$\Rightarrow \frac{x}{3} = \frac{2}{3}$ $\Rightarrow y = \frac{3}{3}$

$\Rightarrow x = 2$ y = 1.

1. Since the ordered pairs are equal, the corresponding elements are equal. <math> <mrow> <mo>∴</mo> <mtext> </mtext> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mn>1</mn> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>.</mo> </mrow> </math> and <math> <mrow> <mi>y</mi><mo>−</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow> </math> <math> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>−</mo> <mn>1</mn> </mrow> </math> <math> <mrow> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> <mo>−</mo> <mn>3</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>x</mi> <mo>=</mo> <mn>2</mn> </mrow> </math> y = 1.

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