# Write five pairs of integers (a, b) such that $\dfrac{a}{b}=-3$. One such pair is (6, -2) because $\dfrac{6}{-2}=-3$.

Answer

Verified

363.9k+ views

Hint: We can find pairs by keeping the value of one variable in a given equation and another variable can easily be calculated as it’s just a two-variable equation. Generally, sequential integers are to be used for our ease of understanding.

Complete step-by-step answer:

Here, we have to form such pairs using integers, be it positive or negative, such that $\dfrac{a}{b}=-3$. We can form these pairs by keeping integers on one variable sequentially or randomly, i.e.,

We can start substituting integral values in one variable, let’s say for b

$\dfrac{a}{b}=-3$

Substituting b = 1 in above equation, we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{1}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-3\]

One such pair formed by the values calculated above will be i.e., pair 1: (-3, 1).

Again, substituting sequential or random values for b in the given equation i.e., b = -3, 5, 6, 10, we get

For $b=-3$, we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{-3}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[\begin{align}

&a=\left( -3 \right)\times \left( -3 \right) \\

&a=9 \\

\end{align}\]

Thus, pairs formed from similar operations in a given equation, we get pair 2: (9, -3).

Similarly, for$b=5$, we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{5}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-15\]

Thus, pairs formed from similar operations in a given equation, we get pair 3: (-15, 5).

Now for \[b=6\], we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{6}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-18\]

Thus, pairs formed from similar operations in a given equation, we get pair 4: (-18, 6).

Now for \[b=10\], we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{10}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-30\]

Thus, pairs formed from similar operations in a given equation, we get pair 5: (-30, 10).

Hence, pairs formed using given conditionality, we get (-3 ,1); (9, -3); (-15, 5); (-18, 6); (-30, 10).

Note: Student might perform a mistake in the sign conventions while substituting values of either a or b in given equation like, \[\dfrac{6}{-2}=\dfrac{-6}{2}=-3\], hence pairs so formed could change from (6, -2) to (-6, 2), unknowingly.

Complete step-by-step answer:

Here, we have to form such pairs using integers, be it positive or negative, such that $\dfrac{a}{b}=-3$. We can form these pairs by keeping integers on one variable sequentially or randomly, i.e.,

We can start substituting integral values in one variable, let’s say for b

$\dfrac{a}{b}=-3$

Substituting b = 1 in above equation, we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{1}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-3\]

One such pair formed by the values calculated above will be i.e., pair 1: (-3, 1).

Again, substituting sequential or random values for b in the given equation i.e., b = -3, 5, 6, 10, we get

For $b=-3$, we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{-3}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[\begin{align}

&a=\left( -3 \right)\times \left( -3 \right) \\

&a=9 \\

\end{align}\]

Thus, pairs formed from similar operations in a given equation, we get pair 2: (9, -3).

Similarly, for$b=5$, we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{5}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-15\]

Thus, pairs formed from similar operations in a given equation, we get pair 3: (-15, 5).

Now for \[b=6\], we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{6}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-18\]

Thus, pairs formed from similar operations in a given equation, we get pair 4: (-18, 6).

Now for \[b=10\], we get

\[\begin{align}

&\dfrac{a}{b}=-3 \\

&\dfrac{a}{10}=-3 \\

\end{align}\]

On cross-multiplying above equations, we get

\[a=-30\]

Thus, pairs formed from similar operations in a given equation, we get pair 5: (-30, 10).

Hence, pairs formed using given conditionality, we get (-3 ,1); (9, -3); (-15, 5); (-18, 6); (-30, 10).

Note: Student might perform a mistake in the sign conventions while substituting values of either a or b in given equation like, \[\dfrac{6}{-2}=\dfrac{-6}{2}=-3\], hence pairs so formed could change from (6, -2) to (-6, 2), unknowingly.

Last updated date: 04th Oct 2023

•

Total views: 363.9k

•

Views today: 5.63k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Who had given the title of Mahatma to Gandhi Ji A Bal class 10 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE