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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:  

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

 

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

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:  

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

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