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Maths > Linear Equations in One Variable > Solving Linear Equation in One Variable
Linear Equations in One Variable

Solving Linear Equation in One Variable

Solving equations means having variable terms on one side and number on the other side. Following examples will illustrate the method of solving linear equations in one variable having variable terms on one side and numbers on the other side. Let us study the topic solving linear equation in one variable in detail.

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Let’s Begin with an Example

Example 1: Solve  \( \frac { x }{ 2 } -\frac { x }{ 3 } =8 \)

Solution: we have, \( \frac { x }{ 2 } -\frac { x }{ 3 } =8 \)
LCM of denominators 2 and 3 on L.H.S. is 6. So, \(\frac { 3x-2x }{ 6 } =8 \)

Multiplying both sides by 6, we get
\(3x-2x=6\times 8\\ x=48 \)

Check   Substituting x=48 in the given equation, we get
\(LHS=\frac { 48 }{ 2 } -\frac { 48 }{ 3 } \\ LHS=24-16\\ LHS=8 \)
Therefore L.H.S. =R.H.S. Hence, x =48 is the solution of the given equation.

Transposition Method for Solving Linear Equations in One Variable

Sometimes the two sides of an equation contain both variable and constants. In such cases, we first simplify two sides in their simplest forms and then transpose terms containing variable on R.H.S. to L.H.S. and constant terms on L.H.S. to R.H.S.

By transposing a term from one side to another side, we mean changing its sign and carrying it to the other side. In transposition, the plus sign of the term changes into minus sign on the other side and vice –versa. The transposition method involves the following steps:

  1. Obtain the linear equation.
  2. Identify the variable and constants.
  3. Simplify the L.H.S. and R.H.S. to their simplest forms by removing brackets.
  4. Transpose all terms containing variable on L.H.S. and constant term on R.H.S. Note that the sign of the terms will change in shifting them from L.H.S. to R.H.S. and vice- versa.
  5. Simplify L.H.S and R.H.S. in the simplest form so that each side contains just one term.
  6. Solve the equation obtained in Step V by dividing both sides by the coefficient of the variable on L.H.S.

Following examples will illustrate the above procedure.

Example 2: Solve 0.16(5x-2) =0.4x +7

Sol: we have, \( 0.16\left( 5x-2 \right) =0.4x+7 \)

on expanding the bracket on L.H.S. –
\( 0.16\times 5x-0.16\times 2=0.4x+7\quad \qquad \\ 0.8x-0.32=0.4x+7 \)

transposing 0.4x to L.H.S. and-0.32 To RHS –
\( 0.8x-0.32=0.4x+7\\ 0.8x-0.4x=7+0.32\\ 0.4x=7.32\\ x=\frac { 7.32 }{ 0.4 } \\ x=18.3 \)

Cross-Multiplication Method for Solving Linear Equations of the Form

Consider the equation $$\frac { 2x+5 }{ 3x+7 } =\frac { 3 }{ 5 } $$

Evidently, it is an equation in one variable x but it is not a linear equation, because the LHS is not a linear polynomial. However, it can be converted into a linear equation by applying the rules for solving an equation as discussed below.


As x represents a number, so 3x+7 also represents a number. Multiplying both sides by (3x+7) x5 i.e. the product of numbers in the denominators on LHS and RHS, we get

$$5\times \left( 3x+7 \right) \times \frac { 2x+5 }{ 3x+7 } =5\times \left( 3x+7 \right) \times \frac { 3 }{ 5 } \\ 5\times \left( 2x+5 \right) =\left( 3x+7 \right) \times 3$$

After expanding the brackets –

$$10x+25=9x+21$$

transposing 9x to L.H.S. and 25 To RHS –

$$10x-9x=21-25\\ x=-4$$

This is the required solution of the equation. Note that in solving this equation, we have first converted it into a linear equation by applying the rules of solving equations. We can also obtain direct equation by equating the product of L.H.S and denominator of R.H.S. to the product of the denominator of L.H.S and numerator of R.H.S. this can exhibit as follows:

Linear Equations

This process of multiplying the numerator on L.H.S. with the denominator on R.H.S. and equating it to the product of the denominator on L.H.S. with the numerator on R.H.S. is called cross multiplication.

Solved Example for You on Linear Equations

Question 1. Solve $$\frac { 3y+20 }{ 3y+2 } =2$$

Answer : We have –

$$\frac { 3y+20 }{ 3y+2 } =\frac { 2 }{ 1 } \\ 1\times \left( 3y+20 \right) =\left( 3y+2 \right) \times 2\\ 3y+20=6y+4\\ 3y-6y=4-20\\ -3y=-16\\ y=\frac { 16 }{ 3 } $$

Question 2: Explain the formula for Linear Equations?

Answer: Solving a linear equation means finding out the value of y for a particular value of x. In case the equation happens to be in the form y = mx + b, then one can solve the equation in algebraic terms.

Question 3: In maths, explain the concept of linear equations with example?

Answer: Linear equation refers to an algebraic equation in which each term consist of an exponent of one. Furthermore, the graphing of the equation provides us with a straight line. A good example of linear equation is y=mx + b.

Question 4: How can one determine if the equation is linear or non-linear?

Answer: First of all, one must simplify the equation as near as possible to the form of y = mx + b. Afterward, one must check if the particular equation has exponents. If it happens to have exponents, it is nonlinear. If the equation happens to have no exponents, it is linear.

Question 5: What is meant by simultaneous linear equations?

Answer: Simultaneous linear equations refer to two linear equations in two variables that are taken together.

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