Algebra can seem intimidating at first, but it is built on a few simple, logical rules. At its core, solving a linear equation is like balancing a scale. Whatever you do to one side, you must do to the other to find the value of the unknown variable (usually $x$).
To move a number to the other side of the equals sign ($=$), you simply perform the opposite operation.
Addition becomes Subtraction: If $x + a = b$, then $x = b – a$
Subtraction becomes Addition: If $x – a = b$, then $x = b + a$
Multiplication becomes Division: If $x(a) = b$, then $x = \frac{b}{a}$
Division becomes Multiplication: If $\frac{x}{a} = b$, then $x = b \times a$
When equations involve fractions, we use the same “opposite” logic, often involving cross-multiplication or reciprocal multiplication.
Handling Coefficients: If $\frac{ax}{b} = \frac{c}{d}$, then $x = \frac{c}{d} \times \frac{b}{a}$ (which simplifies to $\frac{c \times b}{d \times a}$)
Moving a Multiplier to the Denominator: If $ax = \frac{c}{d}$, then $x = \frac{c}{d \times a}$
Moving a Divisor to the Numerator: If $\frac{x}{a} = \frac{c}{d}$, then $x = \frac{c \times a}{d}$
Sometimes, you need to isolate a specific variable in a long string. The rule remains the same: move terms across the “=” sign by changing their sign (+ to – or vice versa).
Given the equation: $x + a – b = p + q – r$
To solve for $x$: $x = p + q – r – a + b$
To solve for $b$: $b = x + a – p – q + r$
To solve for $r$: $r = p + q – x – a + b$
Let’s put these rules into practice with a common algebraic problem.
Step 1: Isolate the $x$ term. Move $+\frac{2}{5}$ to the right side by subtracting it.
Step 2: Solve the subtraction.
To subtract $\frac{2}{5}$ from $1$, convert $1$ to $\frac{5}{5}$.
Step 3: Isolate $x$ completely.
To move the fraction $\frac{5}{3}$ to the other side, multiply by its reciprocal ($\frac{3}{5}$).
Step 4: Final calculation.
Multiply the numerators and the denominators.
Problem: Solve $\frac{2}{3}(x-5) – \frac{1}{4}(x-2) = \frac{9}{2}$
Step 1: Find a Common Denominator. The denominators on the left side are $3$ and $4$, so use $12$ as the common denominator.
$\frac{8(x-5) – 3(x-2)}{12} = \frac{9}{2}$
Step 2: Simplify and Clear Denominators. Multiply by $12$ to clear the fraction.
$8x – 40 – 3x + 6 = 54$
Step 3: Combine Like Terms.
$5x – 34 = 54$
Step 4: Isolate $x$.
$5x = 54 + 34 \Rightarrow 5x = 88$
Result: $x = \frac{88}{5}$
Problem: Solve $\frac{17-3x}{5} – \frac{4x+2}{3} = 5 – 6x + \frac{7x+14}{3}$
Step 1: Simplify both sides. Use common denominators ($15$ for the left side and $3$ for the right side).
$\frac{3(17-3x) – 5(4x+2)}{15} = \frac{15 – 18x + (7x+14)}{3}$
Step 2: Expand and combine.
$\frac{51 – 9x – 20x – 10}{5} = 15 – 18x + 7x + 14$ (after reducing the denominators $15$ and $3$).
$\frac{-29x + 41}{5} = -11x + 29$
Step 3: Eliminate the fraction. Multiply both sides by $5$.
$-29x + 41 = 5(-11x + 29) \Rightarrow -29x + 41 = -55x + 145$
Step 4: Group like terms. Move $x$ terms to one side and constants to the other.
$-29x + 55x = 145 – 41 \Rightarrow 26x = 104$
Step 5: Solve for $x$.
$x = \frac{104}{26}$
Result: $x = 4$
Problem: Solve $(2x+3)^2 + (2x-3)^2 = (8x+6)(x-1) + 22$
This problem utilizes the identity $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$ to simplify the left side quickly.
Result: $x = -1$
In this example, we combine fractions on one side to create a single ratio, then cross-multiply.
The Equation:
The Process:
$x = -\frac{26}{17}$
Sometimes, an equation has many terms. A great strategy is to move terms with similar denominators (or constants) to one side to simplify the math.
The Equation:
The Process:
Final Steps: From here, you would simplify $\frac{50}{36}$ to $\frac{25}{18}$ and cross-multiply to solve for $x$.
The Challenge: After $12$ years, I shall be $3$ times as old as I was $4$ years ago. Find my present age.
Define the Variable: Let the present age be $x$ years.
Translate to an Equation: * Age in $12$ years: $x + 12$
Age $4$ years ago: $x – 4$
The relationship: $x + 12 = 3(x – 4)$
Simplify the Equation:
Distribute the $3$: $x + 12 = 3x – 12$
Rearrange terms: $x – 3x = -12 – 12$
Combine like terms: $-2x = -24$
Final Result: Divide by $-2$ to find $2x = 24$, which means $x = 12$.
The Challenge: The perimeter of a rectangle is $13\text{ cm}$ and its width is $2\frac{3}{4}\text{ cm}$. Find the length ($l$).
Use the Perimeter Formula: The perimeter ($P$) of a rectangle is $2l + 2w = P$.
Substitute the Known Values: * $2l + 2(2\frac{3}{4}) = 13$
Convert to Improper Fractions: * $2\frac{3}{4}$ becomes $\frac{11}{4}$.
The equation becomes: $2l + 2(\frac{11}{4}) = 13$.
Simplify and Isolate $l$:
Multiply $2$ by $\frac{11}{4}$: $2l + \frac{11}{2} = 13$
Subtract $\frac{11}{2}$ from both sides: $2l = 13 – \frac{11}{2}$
Solve the subtraction: $2l = \frac{15}{2}$
Final Result: Divide by $2$ to get $l = \frac{15}{4}$, which simplifies to $3\frac{3}{4}\text{ cm}$.
Problem: Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has a total sum of Rs. 77, how many coins of each denomination does he have?
The Solution Strategy:
Define the unknown: Let the number of five-rupee coins be $x$.
Express other values: Therefore, the number of two-rupee coins is $3x$.
Calculate the total value:
Value from 5-rupee coins: $5 \times x = 5x$
Value from 2-rupee coins: $2 \times 3x = 6x$
Result: Bansi has 7 five-rupee coins and 21 two-rupee coins.
Problem: The sum of three consecutive multiples of 11 is 363. Find these multiples.
The Solution Strategy:
Let the first multiple be $x$.
Since they are consecutive multiples of 11, the next two are $(x + 11)$ and $(x + 22)$.
Result: The three multiples are 110, 121, and 132.
Problem: The difference between two whole numbers is 66. The ratio of the two numbers is 2:5. What are the two numbers?
The Solution Strategy:
When given a ratio like 2:5, we can represent the numbers as $2x$ and $5x$.
Calculate the numbers:
$2 \times 22 = 44$
$5 \times 22 = 110$
Result: The numbers are 44 and 110.
Problem: Deveshi has a total of Rs. 590 in denominations of Rs. 50, Rs. 20, and Rs. 10. The ratio of the number of Rs. 50 notes to Rs. 20 notes is 3:5. If she has a total of 25 notes, how many notes of each denomination does she have?
The Solution Strategy:
Identify the counts:
Rs. 50 notes = $3x$
Rs. 20 notes = $5x$
Rs. 10 notes = $25 – (3x + 5x) = 25 – 8x$
Calculate the values:
From Rs. 50: $3x \times 50 = 150x$
From Rs. 20: $5x \times 20 = 100x$
From Rs. 10: $(25 – 8x) \times 10 = 250 – 80x$
Result:
Rs. 50 notes: $3 \times 2 =$ 6
Rs. 20 notes: $5 \times 2 =$ 10
Rs. 10 notes: $25 – (6 + 10) =$ 9
Problem: Arjun is twice as old as Shriya. Five years ago, his age was three times Shriya’s age. Find their present ages.
To solve this, we establish their ages today and then look at what they were five years ago.
1. Define the Present Ages
Let the present age of Shriya be $x$.
Since Arjun is twice as old as Shriya, Arjun\’s present age is $2x$.
2. Look at the Ages Five Years Ago
Five years ago, everyone was 5 years younger.
Shriya\’s age was $x – 5$.
Arjun\’s age was $2x – 5$.
3. Set Up the Equation
According to the problem, five years ago Arjun was three times as old as Shriya:
4. Solve for $x$
Expand the right side: $2x – 5 = 3x – 15$
Rearrange to solve for $x$: $2x – 3x = -15 + 5$
Simplify: $-x = -10$
Therefore: $x = 10$
Present age of Shriya: 10 years
Present age of Arjun: 20 years