Turn your child’s weaknesses into strengths
Claim your free demo today!
chevron icon chevron icon chevron icon

Linear Equations

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.

 

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

 

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

 

Question 1:

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

 

Solution:

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

 

 

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

 

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

 

Question 2:

Solve:  \(3x − 1 = 5\).

 

Solution:

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

Dividing both the sides by \(3\)

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

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

 

 

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

 

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

 

Question 3:

Solve:  \(4x + 1 = 2x − 7\).

 

Solution:

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

Hence, \(x = − 4\).

 

 

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

 

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,

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

Hence, \(x = 4\).

 

 

2.  Solving a linear equation with fractional coefficients.

 

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

 

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

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

 

 

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

 

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

 

Question 6:

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

 

Solution: 

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

Hence, \(x = 20\).

 

 

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

 

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

 

Question 7:

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

 

Solution: 

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

 

 

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

 

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:

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

 

 

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

 

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

 

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

 

Question 9:

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

 

Solution:

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

Hence, \(x = 16\).

 

 

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

 

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

 

Question 10:

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

 

Solution: 

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

Hence, \(x = 5\).

 

 

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

 

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:   

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

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

 

 

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

 

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:  

 \(\begin{align*} \frac{x + 1}{2x - 3}  &= -\frac15\\ 5(x + 1) &= (–1)(2x \;– 3)\\    5x + 5 &= \;–2x + 3\\          5x &= \;–2x \;– 2\\          7x &= \;–2\\         x &=\; –27\\ \end{align*} \)

 

 

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

 

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:  

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

 

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
Resources - Academic Topics
icon expand icon collapse Primary
icon expand icon collapse Secondary
icon expand icon collapse
+ More
Sign up for a free demo
(P1 to S4 levels)
select dropdown icon
Our Education Consultants will get in touch with you to offer your child a complimentary Strength Analysis.
Sign up for a free demo
(P1 to S4 levels)
Geniebook CTA Illustration Geniebook CTA Illustration
Turn your child's weaknesses into strengths
Geniebook CTA Illustration Geniebook CTA Illustration
Geniebook CTA Illustration Geniebook CTA Illustration
Turn your child's weaknesses into strengths
Let our Education Consultants show you how.
 
Arrow Down Arrow Down
 
Arrow Down Arrow Down
 
Error
Oops! Something went wrong.
Let’s refresh the page!
Error
Oops! Something went wrong.
Let’s refresh the page!
We got your request!
A consultant will be contacting you in the next few days to schedule a demo!
*By submitting your phone number, we have your permission to contact you regarding Geniebook. See our Privacy Policy.
Gain access to 300,000 questions aligned to MOE syllabus
Designed by our team of MOE/NIE-trained teachers.
Trusted by over 220,000 students.
Arrow Down Arrow Down
 
Error
Oops! Something went wrong.
Let’s refresh the page!
Error
Oops! Something went wrong.
Let’s refresh the page!
We got your request!
A consultant will be contacting you in the next few days to schedule a demo!
*By submitting your phone number, we have your permission to contact you regarding Geniebook. See our Privacy Policy.
media logo
Geniebook CTA Illustration Geniebook CTA Illustration
icon close
Default Wrong Input
Get instant access to
our educational content
Start practising and learning.
No Error
arrow down arrow down
No Error
*By submitting your phone number, we have
your permission to contact you regarding
Geniebook. See our Privacy Policy.
Success
Let’s get learning!
Download our educational
resources now.
icon close
Error
Error
Oops! Something went wrong.
Let’s refresh the page!