Answer
Verified
456.9k+ views
Hint: We can take the cost of ball pen as one variable and cost of fountain pen as another variable. By reading the given statement, we can understand that half of the cost of a fountain pen minus 5 gives the cost of a ball pen. We get the required equation by writing the relation in terms to two variables and simplifying.
Complete step by step answer:
We are given two objects, a ball pen and fountain pen. We can assign any two variables to the cost of ball pen and cost of fountain pen
Let \[{\text{x = }}\] cost of fountain pen
And let \[{\text{y = }}\] cost of ball pen
According to the question, half of the cost of fountain pen minus 5 gives the cost of a ball pen. Writing this in terms of variables, we get,
${\text{y = }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - 5}}$
Rearranging the equation, we get
$
{\text{y + 5 = }}\dfrac{{\text{x}}}{{\text{2}}} \\
{\text{2y + 10 = x}} \\
$
Therefore, the required linear equation in 2 variables is ${\text{x = 2y + 10}}$.
Note: As the variable is not mentioned, we can use any variable. We must make sure that it is clearly written what each of the variables used refers to, and we should not mix them up. In this problem we convert a statement into a mathematical equation. This method of solving is known as mathematical modeling. Using mathematical modeling we can form equations from a statement and we can solve for the variable if the number of equations formed is greater than the number of variables, we can solve for the variables. The number of variables used must be minimum to avoid errors. We must read and understand the given statement completely to form the equations.
Complete step by step answer:
We are given two objects, a ball pen and fountain pen. We can assign any two variables to the cost of ball pen and cost of fountain pen
Let \[{\text{x = }}\] cost of fountain pen
And let \[{\text{y = }}\] cost of ball pen
According to the question, half of the cost of fountain pen minus 5 gives the cost of a ball pen. Writing this in terms of variables, we get,
${\text{y = }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - 5}}$
Rearranging the equation, we get
$
{\text{y + 5 = }}\dfrac{{\text{x}}}{{\text{2}}} \\
{\text{2y + 10 = x}} \\
$
Therefore, the required linear equation in 2 variables is ${\text{x = 2y + 10}}$.
Note: As the variable is not mentioned, we can use any variable. We must make sure that it is clearly written what each of the variables used refers to, and we should not mix them up. In this problem we convert a statement into a mathematical equation. This method of solving is known as mathematical modeling. Using mathematical modeling we can form equations from a statement and we can solve for the variable if the number of equations formed is greater than the number of variables, we can solve for the variables. The number of variables used must be minimum to avoid errors. We must read and understand the given statement completely to form the equations.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Derive an expression for drift velocity of free electrons class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
The energy of a charged conductor is given by the expression class 12 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Derive an expression for electric field intensity due class 12 physics CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Derive an expression for electric potential at point class 12 physics CBSE