
How do you solve for $k$ in $\dfrac{9}{k} - 7 = \dfrac{6}{k}$ ?
Answer
453.3k+ views
Hint: In the given question, we have been asked to find the value of ‘k’ and it is given that $\dfrac{9}{k} - 7 = \dfrac{6}{k}$ . To solve this question, we need to get ‘k’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘k’, we have to undo the mathematical operations such as addition, subtraction, multiplication, and division that have been done to the variables.
Complete step-by-step solution:
We have given that,
$\dfrac{9}{k} - 7 = \dfrac{6}{k}$
We can also write this as ,
$ \Rightarrow \dfrac{9}{k} - \dfrac{6}{k} = 7$
Now , take $\dfrac{1}{k}$ common from every term present on left side ,
$ \Rightarrow \dfrac{1}{k}(9 - 6) = 7$
Now , multiply by $k$ to both the side of the equation ,
$
\Rightarrow k \times \dfrac{1}{k}(9 - 6) = k \times 7 \\
\Rightarrow 9 - 6 = 7k \\
$
Or $7k = 9 - 6$
$ \Rightarrow 7k = 3$
Divide by $7$ on both the sides of the equation
$ \Rightarrow k = \dfrac{3}{7}$
Therefore, the value of $k$ is equal to $\dfrac{3}{7}$ .
It is the required answer.
Additional information: In the given question, no mathematical formula is being used only the mathematical operations such as addition, subtraction, multiplication and division are used. Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to one.
Note: The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.
Complete step-by-step solution:
We have given that,
$\dfrac{9}{k} - 7 = \dfrac{6}{k}$
We can also write this as ,
$ \Rightarrow \dfrac{9}{k} - \dfrac{6}{k} = 7$
Now , take $\dfrac{1}{k}$ common from every term present on left side ,
$ \Rightarrow \dfrac{1}{k}(9 - 6) = 7$
Now , multiply by $k$ to both the side of the equation ,
$
\Rightarrow k \times \dfrac{1}{k}(9 - 6) = k \times 7 \\
\Rightarrow 9 - 6 = 7k \\
$
Or $7k = 9 - 6$
$ \Rightarrow 7k = 3$
Divide by $7$ on both the sides of the equation
$ \Rightarrow k = \dfrac{3}{7}$
Therefore, the value of $k$ is equal to $\dfrac{3}{7}$ .
It is the required answer.
Additional information: In the given question, no mathematical formula is being used only the mathematical operations such as addition, subtraction, multiplication and division are used. Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to one.
Note: The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Trending doubts
A number is chosen from 1 to 20 Find the probabili-class-10-maths-CBSE

Find the area of the minor segment of a circle of radius class 10 maths CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

A gulab jamun contains sugar syrup up to about 30 of class 10 maths CBSE

Leap year has days A 365 B 366 C 367 D 368 class 10 maths CBSE
