Linear Equations in One Variable

Introduction

In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are:

5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x ² + 1, y + y ²

Some examples of equations are:

5x = 25, 2x – 3 = 9, \(2y+\frac{5}{2}=\frac{37}{2}\), 6z+10=-2

You would remember that equations use the equality (=) sign; it is missing in expressions. Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover, the expressions we use to form equations are linear. This means that the highest power of the variable appearing in the expression is 1. These are linear expressions:

               2x, 2x + 1, 3y – 7, 12 – 5z, \(\frac{5}{4}(x-4)+10\)

These are not linear expressions: x ² + 1, y + y ², 1 + z + z ² + z ³ (since highest power of variable > 1) Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type.

Let us briefly revise what we know:

(a) An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS).

(b) In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation.

(c) How to find the solution of an equation? We assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. A few such steps give the solution.

Linear Equations in One Variable

Solving Equations having the Variable on both Sides

An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7, the two expressions are 2x – 3 and 7. In most examples that we have come across so far, the RHS is just a number. But this need not always be so; both sides could have expressions with variables. For example, the equation 2x – 3 = x + 2 has expressions with a variable on both sides; the expression on the LHS is (2x – 3) and the expression on the RHS is (x + 2).

• We now discuss how to solve such equations which have expressions with the variable on both sides.

Example 1: Solve 2x – 3 = x + 2

Solution: We have 2x = x + 2 + 3

or                             2x = x + 5

or                             2x – x = x + 5 – x (subtracting x from both sides)

or                             x = 5 (solution)

Here we subtracted from both sides of the equation, not a number (constant), but a term involving the variable. We can do this as variables are also numbers. Also, note that subtracting x from both sides amounts to transposing x to LHS.

Example 2: Solve \(5x+\frac{7}{2}=\frac{3}{2}x-14\)

Solution: Multiply both sides of the equation by 2.

We get  \(\begin{align} & 2\times \left( 5x+\frac{7}{2} \right)=2\times \left( \frac{3}{2}x-14 \right) \\ & (2\times 5x)+\left( 2\times \frac{7}{2} \right)=\left( 2\times \frac{3}{2}x \right)-\left( 2\times 14 \right) \\ \end{align}\)

or                 10x + 7 = 3x – 28

or                 10x – 3x + 7 = – 28 (transposing 3x to LHS)

or                 7x + 7 = – 28

or                 7x = – 28 – 7

or                 7x = – 35

or x = \(\frac{-35}{7}\)                             or             x = – 5 (solution)

Linear Equations in One Variable

Reducing Equations to Simpler Form

Example:

Solve \(\frac{6x+1}{3}+1=\frac{x-3}{6}\)

Solution: Multiplying both sides of the equation by 6,

                  \(\frac{6(6x+1)}{3}+6\times 1=\frac{6(x-3)}{6}\)              

or               2 (6x + 1) + 6 = x – 3

or              12x + 2 + 6 = x – 3 (opening the brackets )

or               12x + 8 = x – 3

or               12x – x + 8 = – 3

or               11x + 8 = – 3

or               11x = –3 – 8

or               11x = –11

or                  x = – 1 (required solution)

Check: LHS = \(\frac{6(-1)+1}{3}+1=\frac{-6+1}{3}+1=\frac{-5}{3}+\frac{3}{3}=\frac{-5+3}{3}=\frac{-2}{3}\)

RHS \(\frac{(-1)-3}{6}=\frac{-4}{6}=\frac{-2}{3}\)

LHS = RHS.            (as required)

Example :

Solve 5x – 2 (2x – 7) = 2 (3x – 1) + \(\frac{7}{2}\)

Solution: Let us open the brackets, LHS = 5x – 4x + 14 = x + 14

RHS = \(6x-2+\frac{7}{2}=6x-\frac{4}{2}+\frac{7}{2}=6x+\frac{3}{2}\)

The equation is x + 14 = 6x + \(\frac{3}{2}\)

or 14 = 6x – x + \(\frac{3}{2}\)

or 14 = 5x + \(\frac{3}{2}\)

or 14 – \(\frac{3}{2}\) = 5x (transposing \(\frac{3}{2}\) )

or \(\frac{28-3}{2}\)= 5x

or \(\frac{25}{2}\) = 5x

or x = \(\frac{25}{2}\times \frac{1}{5}=\frac{5\times 5}{2\times 5}=\frac{5}{2}\)

Therefore, required solution is x = \(\frac{5}{2}\).

LHS = \(\begin{align}   & 5\times \frac{5}{2}-2\left( \frac{5}{2}\times 2-7 \right) \\  & =\frac{25}{2}-2\left( 5-7 \right)=\frac{25}{2}-2\left( -2 \right)=\frac{25}{2}+4=\frac{25+8}{2}=\frac{33}{2} \\ \end{align}\)

RHS =\(\begin{align}   & 2\left( \frac{5}{2}\times 3-1 \right)+\frac{7}{2}=2\left( \frac{15}{2}-\frac{2}{2} \right)+\frac{7}{2}=\frac{2\times 13}{2}+\frac{7}{2} \\  & =\frac{26+7}{2}=\frac{33}{2} \\ \end{align}\)

           =LHS     (as required)