word problems of linear equation Related: Short Notes - Linear Equati...
Read the problem carefully and note what is given and what is required and what is given.
● Denote the unknown by the variables as x, y, …….
● Translate the problem to the language of mathematics or mathematical statements.
● Form the linear equation in one variable using the conditions given in the problems.
● Solve the equation for the unknown.
● Verify to be sure whether the answer satisfies the conditions of the problem.
word problems on linear equations with solutions explained step-by-step in different types of examples.
There are several problems which involve relations among known and unknown numbers and can be put in the form of equations
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word problems of linear equation Related: Short Notes - Linear Equati...
Introduction to Linear Equations in One Variable
Linear equations in one variable are mathematical equations that involve only one variable and have a degree of 1. These equations can be represented in the form of ax + b = 0, where 'a' and 'b' are constants.
Word Problems of Linear Equations
Problem 1:
The sum of two consecutive integers is 35. Find the integers.
Solution:
Let's assume the first integer as 'x'. The next consecutive integer would be 'x + 1'.
According to the problem, the sum of the two consecutive integers is 35. So, we can write the equation as:
x + (x + 1) = 35
Simplifying the equation, we get:
2x + 1 = 35
Subtracting 1 from both sides, we get:
2x = 34
Dividing both sides by 2, we get:
x = 17
Therefore, the first integer is 17 and the next consecutive integer is 17 + 1 = 18.
Problem 2:
The perimeter of a rectangle is 60 cm. The length of the rectangle is 5 cm more than twice its width. Find the length and width of the rectangle.
Solution:
Let's assume the width of the rectangle as 'x'. According to the problem, the length is 5 cm more than twice its width, which can be represented as '2x + 5'.
The perimeter of a rectangle is given by the formula: 2(length + width)
Using this formula, we can write the equation as:
2(2x + 5 + x) = 60
Simplifying the equation, we get:
2(3x + 5) = 60
Expanding and simplifying further, we get:
6x + 10 = 60
Subtracting 10 from both sides, we get:
6x = 50
Dividing both sides by 6, we get:
x = 8.33
Since the width cannot be in decimal form, we round it to the nearest whole number. Therefore, the width of the rectangle is 8 cm.
Substituting the value of the width in the equation '2x + 5', we get:
2(8) + 5 = 21
Therefore, the length of the rectangle is 21 cm.
Conclusion
Linear equations in one variable can be solved using algebraic methods to find the values of the variables involved. These equations can be applied to solve real-life problems, such as finding unknown numbers or dimensions of shapes. By understanding the concepts and applying the appropriate mathematical techniques, we can solve word problems related to linear equations efficiently.
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