Linear equations in one variable are the basic building blocks of algebra. They are simple equations that have only one variable (like x or y), and the highest power of the variable is 1. These equations usually look like this: ax + b = 0, where a and b are numbers.

Understanding linear equations is very important, especially for students preparing for competitive exams or school tests. In many government exams, questions from Algebra appear in the Quantitative Aptitude section. That’s why learning this topic well is a must.

What are Linear Equations in One Variable?

A linear equation in one variable is a type of equation that contains only one unknown number, usually written as x, and the highest power of this variable is 1. These equations show that two expressions are equal, and we try to find the value of the variable that makes this equation true.

Every linear equation in one variable has only one solution or root. This means there is only one number that, when placed in place of the variable, makes both sides of the equation equal.

Example:
Take the equation 4x + 6 = 14.
We want to find the number that we can put in place of x to make both sides equal.

Try x = 2:
4 × 2 + 6 = 8 + 6 = 14.
This is a true statement, so x = 2 is the solution.

If you try any other value, like x = 1 or x = 3, the result will not be 14. So, only x = 2 makes the equation correct.
This value is called the solution or root of the equation.

Standard Form of Linear Equations in One Variable

Standard form of linear equations in one variable is written as ax+b=0, (a is not equal to zero) where x is the one variable while a and b are any real numbers.

Therefore, the formula of the linear equations in one variable is formulated as ax + b = 0, such that ‘a’ and ‘b’ are not equivalent to zero. When a graph is drawn for such an equation it seems to be a straight line either horizontally or vertically.

Linear Equations in One Variable Formula

A linear equation is a straightforward approach to represent a mathematical statement. Moreover, solving a linear equation comprises a collection of simple methods. For this, we separate the variables on one side of the equation and the constants on another side of the given equation and conclude the final value of the unknown quantity/elements.

The important formulas of linear equations in one variable are:

How to Solve Linear Equations in One Variable?

We can solve linear equations in one variable by following the below steps:

• Step 1 – Clear the equation of fractions (or decimals) by multiplying both sides by the LCM of the denominators (or by a power of 10 in the case of decimals).
• Step 2 – Remove group symbols from both sides, if any, and simplify the equation.
• Step 3 – Bring all variable terms on one side and all the constants on the other. Combine the like terms.
• Step 4 – Divide the terms into both sides by the numerical coefficient of the variable.

Let us understand how to solve linear equations in one variable with the help of the following example:

Example: Consider the following equation and solve for ‘x’ \( \frac{x}{4}=28-\frac{x}{3}\)

Step 1. Clear the fraction by multiplying both sides by the LCM of denominators (4 and 3), which is 12.

\(12\frac{x}{4}=12*28-12\frac{x}{3}\)

Step 2. Simplify the equation

\(3x=336-4x\)

Step 3. Isolate all variable terms on one side and all the constants on the other and combine the like terms.

\(3x+4x=336\)

\(7x=336\)

Step 4. Divide the terms into both sides by the numerical coefficient of the variable, which is 7.

\(x=48\)

When you’ve finished with linear equations in one variable, you can learn about Linear Equations in Two variables

Rules for Solving Linear Equations in One Variable

In these types of linear equations in one variable questions, a problem (or situation) will be given in verbal language involving certain numbers. You will have to find the solution to the problem by translating words into a linear equation.

There are some specific rules for solving linear equations in one variable which are mentioned as follows:

• Rule 1: Adding the same number (positive or negative) to both sides of an equation, then the resulting equation will be equivalent to the original equation.
• Rule 2: Multiplying or dividing both sides of an equation by the same non-zero number, then the resulting equation will be equivalent to the original equation.
• Rule 3: Transposing a term from one side to the other side of the equation, then the resulting equation will be equivalent to the original equation, but remember that the sign of the transposed term changes.

Applications of Linear Equations in One Variable

1. Age Problems
   Linear equations help solve problems where the ages of people are compared or related.
   Example: If a person's age is 5 years more than twice their sister's age, we can use a linear equation to find the exact ages.
2. Money and Cost Problems
   They are used to calculate expenses, income, savings, or to split costs.
   Example: If you buy pens for ₹10 each and you spend ₹100, how many pens did you buy? The equation 10x = 100 solves it.
3. Distance, Speed, and Time
   Linear equations help solve travel-related problems using the formula: Distance = Speed × Time.
   Example: If a car travels at 60 km/h for x hours and covers 180 km, the equation 60x = 180 helps find the time.
4. Solving Number Puzzles
   They are used in finding unknown numbers based on given conditions.
   Example: “A number added to 7 gives 15” becomes x + 7 = 15.
5. Geometry and Measurement
   They help in finding unknown angles, side lengths, or other measurements when some values are known.

Tips on Linear Equations:

1. Solution or Root: The number that makes a linear equation true when you put it in place of the variable is called the solution or root of the equation.
2. Balance Rule: If you add, subtract, multiply, or divide the same number on both sides of a linear equation, the answer doesn’t change. This helps in solving equations easily.
3. Graph Shape: When you draw a linear equation on a graph, it always forms a straight line—whether it has one variable or two.

So far we have seen the linear equations in one variable definition and how to solve them. Now let us also learn about non-linear equations. As even non-linear equations have multiple applications in geometry, trigonometry, as well in calculus.

The following table shows the comparison of linear equations in one variable vs non-linear equations:

Solved Examples of Linear Equations in One Variable

The following are a few examples of questions that can come in examinations, related to Linear Equations in one variable.

Example 1. The sum of the two numbers is 45 and their ratio is 2 : 3. Find the numbers.

Solution: Let x be one of the numbers. Then, the other number is 45-x.

Since the two numbers are in the ratio 2 : 3, we have

\(\frac{x}{45-x}=\frac{2}{3}\)

\(3x=2(45-x)\)

\(3x=90-2x\)

\(5x=90\)

\(x=\frac{90}{5}\)

\(x=18\)

Thus, one number is 18 and the other number is 45-18=27.

Example 2. Solve the following equations for \(x : \) \(4(x+5)=14x+50\)

Solution: \(4x+20=14x+50\) [Removing the brackets]

\(4x-14x=50-20\) [Transposing like terms to same side]

\(-10x=30\)[Dividing both sides by -10]

\(x= -3\)

Example 3. Solve the following equation for x:
2(x - 1) - 3(x - 2) = 4(x - 5)

Solution: Expand both sides:
2x - 2 - 3x + 6 = 4x - 20

Now combine like terms:
(2x - 3x) + (-2 + 6) = 4x - 20
-1x + 4 = 4x - 20

Bring like terms to one side:
-1x - 4x = -20 - 4
-5x = -24

Divide both sides by -5:
x = 24 / 5
x = 4.8

Final Answer:
x = 4.8

Example 4. Solve the following equations for \(x : \) \( \frac{3x}{8}+\frac{1}{4}=\frac{x-3}{3}\)

Solution: \(24(\frac{3x}{8}+\frac{1}{4})=24\frac{x-3}{3}\)

\(9x+6=8x-24\)

\(9x-8x= -6-24\)

\(x = -30\)

Example 5. Solve the following equations for \(x : \) \(\frac{5}{x}=\frac{9}{x-8}\)

Solution: Multiplying throughout by \(x(x – 8)\), LCM of x and (x – 4), we get

\(5(x-8)=9x\)

\(5x-40=9x\)

\(5x-9x=40\)

\(-4x=40\)

\(x= -10\)

Example 6. Solve the following equations for \(x : \) \(\frac{x+b}{a+b}=\frac{x-b}{a-b}\)

Solution: Multiplying throughout by \((a + b)(a – b)\), LCM of the denominators

\(x(a-b)+b(a-b)=a(x-b)+b(x-b)\)

\(xa-xb+ba-b^2=ax-ab+bx-b^2\)

\(ax-bx-ax-bx= -ab-b^2-ab+b^2\)

\(- 2bx= -2ab\)

\(x=a\)

Example 7. A woman is five times as old as her daughter. In two years’ time, she will be four times as old as her daughter. Find their present ages.

Solution: Let the present age of the daughter be \(x\) years.

Then, the present age of woman = \(5x\) years

After two years,

Daughter’s age = \((x+2)\) years

Woman’s age = \((5x+ 2)\)years

According to the question,

\(5x+2=4(x+2)\)

\(5x+2=4x+8\)

\(5x-4x=8-2\)

\(x=6\)

Thus, the daughter’s present age is 6 years and the woman’s present age is (56) years, i.e., 30 years

Example 8. A number consists of two digits whose sum is 9. If 9 is subtracted from the number its digits are interchanged. Find the number.

Solution: Let the unit digit of the number be x.

Then, the tens digit =9-x

Therefore, Number =\(10(9-x)+x\)

\(=90-10x+x\)

\(=90-9x\)

Number with interchanged digits =10 x+9-x

\(=9x+9\)

According to the question,

\((90-9x)-9=9x+9\)

\(81-9x=9x+9\)

\(9x+9x=81-9\)

\(18x=72\)

\(x=4\)

Therefore, the units digit \(=4\) and the tens digit \(=9-4=5\)

Hence, the number is \(54\).

Example 9. Rahul has 3 boxes of different vegetables. Box A weighs 2.50 kg more than box B and box C weighs 10.25 kg more than box B. The total weight of the three boxes is 48.75 kg. How much does box A weighs?

Solution:  Let the weight of box B be x kg.

Then,

• Weight of box A = \((x+2.50)\)kg
• Weight of box C = \((x+10.25)\)kg

According to the question,

\((x+2.50)+x+(x+10.25)=48.75\)

\(x+x+x=48.75-2.50-10.25\)

\(3x=36\)

\(x=12\)

Therefore, Weight of box A =(x+2.50) kg=14.50 kg.

Example 10. The distance between the two stations is 340 km. two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/h. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.

Solution: Let the speed of the train starting from A be \(x\) km/h.

Therefore, The speed of the second train starting from B will be \((x + 5)\) km/h.

The first train reaches C and the second train reaches D in 2 hours.

According to the question,

\(x2+30+(x+5)2=340\)

\(2x+30+2x+10=340\)

\(2x+2x=340-40\)

\(4x=300\)

\(x=75\)

Therefore, The speeds of the trains are 75 km/h and 80 km/h respectively.

We hope you found this article useful, informative and helpful in clearing all your doubts regarding linear equations in one variable and how to solve questions regarding the same.

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