Linear Equations

Understanding Linear Equations is crucial at the secondary 1 because they serve as a cornerstone for more advanced mathematical concepts. Linear Equations introduce students to fundamental algebraic concepts such as variables, coefficients, and constants. Mastering these basics is essential for progressing to more complex algebraic topics.

Linear Equation

A linear equation is a math statement where the variables (like \(x\) or \(y\)) are not raised to any power other than \(1\). In simpler terms, no squared or cubed variables.

Each linear equation has an equals sign, showing that both sides are exactly equal.

When graphed, linear equations make straight lines. They're often written in the form \(ax + by + c = 0\), or more commonly, \(y = ax + b\).

For example:

• \(3x - 2y + 6 = 0\)
• \(y = 4x + 3\)
• \(5y - 9x = 6\)
• \(5x = 15\)

Solving Linear Equation

In this chapter, we will be discussing the below-mentioned topics in detail:

• Solving a Linear Equation by Balancing the Equation.
• Solving a Linear Equation with Fractional Coefficients.
• Solving Simple Fractional Equations that can be Reduced to Linear Equations. 

1. Solving a Linear Equation by balancing the equation.

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 1: } &x + a = c, \\[2ex] &\text{where }a \text{ and } c \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 1:

Solve:  \(\begin{align} x + 5 = 8 \end{align}\).

Solution:

\(\begin{align} x + 5 &= 8\\[2ex] x + 5 – 5 &= 8 – 5 \\[2ex] x &=3 \end{align}\)

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 2: } &ax + b = c,\\[2ex] &\text{where }a, b \text{ and } c \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 2:

Solve:  \(3x − 1 = 5\).

Solution:

\(\begin{align} 3x − 1 &= 5\\[2ex] 3x − 1 + 1 &= 5 + 1 \\[2ex] 3x &=6 \end{align}\)

Dividing both the sides by \(3\)

     \(\begin{align} 3x \div 3 &= 6 \div 3 \\[2ex] x&=2 \end{align}\)

Hence, \(\begin{align} x = 2. \end{align}\)

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 3: } &ax + c = bx + d,\\[2ex] &\text{where }a, b, c \text{ and } d \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 3:

Solve:  \(4x + 1 = 2x − 7\).

Solution:

\(\begin{align*} 4x + 1 &= 2x − 7\\[2ex] 4x + 1 \;– 1 &= 2x \;– 7 – 1\\[2ex] 4x &= 2x \;– 8\\[2ex] 4x \;– 2x &= − 8\\[2ex] 2x &= − 8\\[2ex] x &= − 4 \end{align*}\)

Hence, \(x = − 4\).

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 4: } &a(bx + c) = px + q,\\[2ex] &\text{where }a, b, c, p \text{ and } q \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 4:

Solve:  \(3 (x + 2) = x + 14\).

Solution: 

\(\begin{align} 3 (x + 2) &= x + 14 \end{align}\)

Expanding the equation,

\(\begin{align*} 3x + 6 &= x + 14\\[2ex] 3x \;– x &= 14 \;– 6\\[2ex] 2x &= 8\\[2ex] x &= 4 \end{align*}\)

Hence, \(x = 4\).

2.  Solving a linear equation with fractional coefficients.

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 5: } &\frac{x}{a} = b, \\[2ex] &\text{where }a \text{ and } b \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 5:

Solve:  \(\begin{align} \frac{x}{3} &= \;– 5 \end{align}\).

Solution: 

\(\begin{align} \frac{x}{3} &= \;– 5 \end{align}\)

Multiplying both sides by 3

\(\begin{align*} \frac{x}{3} (3) &= (–5) (3)\\[2ex] x &= \;–15 \end{align*}\)

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 6: } &\frac{x}{a} +b = c, \\[2ex] &\text{where }a, b \text{ and } c \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 6:

Solve:  \( \begin{align*} \frac{x}{4} \;– \;2 = 3 \end{align*}\).

Solution: 

\(\begin{align*} \frac{x}{4} \;– \;2 &= 3\\[2ex] \frac{x}{4}  \;–\; 2 + 2 &= 3 + 2\\[2ex] \frac{x}{4}  &= 5\\[2ex] x &= 5\times4\\[2ex] &= 20 \end{align*} \)

Hence, \(x = 20\).

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 7: } &\frac{a}{b}x +c = d, \\[2ex] &\text{where }a, b, c \text{ and } d \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 7:

Solve:  \(\begin{align*} \frac{2}{3} x + 1 &= 7 \end{align*}\).

Solution: 

\(\begin{align*} \frac{2}{3} x + 1 &= 7\\[2ex] \frac{2}{3} x + 1 \;– 1 &= 7 \;– 1\\[2ex] \frac{2}{3} x &= 6\\[2ex] 2x &= 18\\[2ex] x &= 9 \end{align*}\)

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 8: } &\frac{a}{b}x +c = \frac{p}{q}x +r, \\[2ex] &\text{where }a, b, c, p, q \text{ and } r \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 8:

Solve:  \(\begin{align*} \frac{3}{4}x \;– \;3 = \frac{x}{5} + 8 \end{align*}\)

Solution:

\(\begin{align*} \frac{3}{4}x \;– 3 &= \frac{x}{5} + 8\\[2ex] \frac{3}{4} x \;– 3 + 3 &= \frac{x}5 + 8 + 3\\[2ex] \frac{3}{4}x &= \frac{x}5 + 11\\[2ex] \frac{3}{4} x \;– 15 x &= 11\\[2ex] \frac{11}{20}x &= 11\\[2ex] 11x &= 11 \times 20\\[2ex] 11x &= 220\\[2ex] x &= 20 \end{align*} \)        

3. Solving Simple Fractional Equations that can be reduced to linear equations.

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 9: } &\frac{ax + b}{c} = d, \\[2ex] &\text{where }a, b, c, \text{ and } d \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 9:

Solve:  \(\begin{align*} \frac{2x + 3}{5}  = 7 \end{align*}\).

Solution:

 \(\begin{align*} \frac{2x+3}5 &= 7\\[2ex] 2x + 3 &= 7 (5)\\[2ex]      2x + 3 &= 35\\[2ex] 2x + 3 \;– 3 &= 35 \;– 3\\[2ex]     2x &= 32\\[2ex]               x &= 16 \end{align*} \)

Hence, \(x = 16\).

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 10: } &\frac{ax + b}{c} = px + q, \\[2ex] &\text{where }a, b, c, p \text{ and } q \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 10:

Solve:  \(\begin{align*} \frac{4x+1}{3} = 2x– 3 \end{align*}\).

Solution: 

  \(\begin{align*} \frac{4x+1}3 &= 2x \;– 3\\[2ex]  4x + 1 &= 3 (2x \;– 3)\\[2ex]  4x + 1 &= 6x \;– 9\\[2ex] 4x – 6x &= \;– 9 \;– 1\\[2ex]   – 2x &= \;–10\\[2ex]          x &= 5 \end{align*} \)

Hence, \(x = 5\).

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 11: } &\frac{ax + b}{c} = \frac{px + q}{r}, \\[2ex] &\text{where }a, b, c, p, q \text{ and } r \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 11: 

Solve:  \(\begin{align*} \frac{2x–3}{5}  = \frac{x+1}{4 } \end{align*}\).

Solution:   

  \(\begin{align*} \frac{2x – 3}5  &= \frac{x + 1}4\\[2ex] 4(2x – 3) &= 5 (x + 1)\\[2ex]    8x – 12 &= 5x + 5\\[2ex]            8x &= 5x + 17\\[2ex]             3x &= 17\\[2ex]              x &= \frac{17}3\\[2ex]       &= 5\frac{2}3 \end{align*} \)

Hence, \(\begin{align} x &= 5\frac{2}{3} \end{align} ​\).

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 12: } &\frac{ax + b}{c} = \frac{p}{q}, \\[2ex] &\text{where }a, b, c, p \text{ and } q \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 12:

Solve:  \(\begin{align} \frac{x + 1}{2x - 3}  &= -\frac{1}{5} \end{align}\).

Solution:  

 \(\begin{align*} \frac{x + 1}{2x - 3}  &= -\frac15\\[2ex] 5(x + 1) &= (–1)(2x \;– 3)\\[2ex]    5x + 5 &= \;–2x + 3\\[2ex]         5x &= \;–2x \;– 2\\[2ex]          7x &= \;–2\\[2ex]         x &=\; -{{2}\over{7}}\\ \end{align*} \)

\(\bbox[5px,border:2px solid #262262]{ \small\begin{align} \textbf{Case 13: } &\frac{ax+b}{c}+\frac{px+q}{r} = d,\\[2ex] &\text{where }a, b, c, d, p, q \text{ and } r \text{ are constants.} \end{align}}\)

Let’s understand this with the help of some examples:

Question 13:

Solve:  \(\begin{align} \frac{x+1}{2}-\frac{x-1}{3}  = 1 \end{align}\).

Solution:  

   \(\begin{align*} \frac{x \;+ \;1}2-\frac{x\; - \;1}3  &= 1\\[2ex] \frac{3(x + 1) - 2 (x - 1)}6  &= 1\\[2ex]     \frac{3x + 9 - 2x + 2}6  &= 1\\[2ex]                 \frac{x + 11}6  &= 1\\[2ex]         x + 11 &= 6\\[2ex]                          x &= \;– 5 \end{align*} \)