# 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*}

Continue Learning
Basic Geometry Linear Equations
Number Patterns Percentage
Prime Numbers Ratio, Rate And Speed
Functions & Linear Graphs 1 Integers, Rational Numbers And Real Numbers
Basic Algebra And Algebraic Manipulation 1 Approximation And Estimation

Primary
Secondary
Book a free product demo
Suitable for primary & secondary
Our Education Consultants will get in touch with you to offer your child a complimentary Strength Analysis.
Book a free product demo
Suitable for primary & secondary
our educational content
Start practising and learning.
No Error
No Error
*By submitting your phone number, we have
your permission to contact you regarding
Let’s get learning!
resources now.
Error
Oops! Something went wrong.
Let’s refresh the page!
Turn your child's weaknesses into strengths
Turn your child's weaknesses into strengths
Trusted by over 220,000 students.

Error
Oops! Something went wrong.
Let’s refresh the page!
Error
Oops! Something went wrong.
Let’s refresh the page!