Simple Equations Class 7 CBSE Mathematics

Introduction to Simple Equations

Simple Equations – An equation is a mathematical statement that shows the equality of two expressions. It consists of two sides: the Left Hand Side (LHS) and the Right Hand Side (RHS), connected by an equal sign (=). For example, in the equation $2x+3=7$, $2x+3$ is the LHS, and $7$ is the RHS. The goal of solving an equation is to find the value of the unknown variable (like $x$) that makes the equation true.

A simple equation is a linear equation with one variable, and it does not involve exponents higher than 1. Simple equations are foundational in algebra and help us solve real-life problems, such as finding an unknown quantity.

Key Terms

1. Variable: A symbol (usually a letter like $x,y,z$) that represents an unknown value.
2. Constant: A fixed number (e.g., 5, -3, 10) that does not change.
3. Coefficient: The number multiplied by the variable (e.g., in $4x$, 4 is the coefficient of $x$).
4. Expression: A combination of variables, constants, and operations (e.g., $3x+2$).
5. Equation: A statement showing two equal expressions (e.g., $5x-2=8$).
6. Solution: The value of the variable that satisfies the equation.

Setting Up Equations

In real-life situations, we often translate word problems into equations. To do this:

• Identify the unknown quantity and assign it a variable (e.g., let the unknown number be $x$).
• Use the given information to form an equation.

Example:
Ravi is 5 years older than his brother. If Ravi’s age is 12, what is his brother’s age?

• Let the brother’s age be $x$.
• Ravi’s age = $x+5$.
• Given Ravi’s age = 12, so the equation is $x+5=12$.
• Solving: $x=12-5=7$.
• Brother’s age = 7 years.

Solving Simple Equations

To solve an equation, we need to isolate the variable on one side of the equation. This is done by performing the same operation on both sides to keep the equation balanced. The basic operations are:

• Addition or Subtraction
• Multiplication or Division

Rules for Solving Equations

1. Add or subtract the same number on both sides to move constants.
2. Multiply or divide both sides by the same non-zero number to remove coefficients.
3. Perform operations step-by-step and simplify.

Example 1: Solve $3x+4=13$

• Step 1: Subtract 4 from both sides to isolate the term with $x$.
  $3x+4-4=13-4$
  $3x=9$
• Step 2: Divide both sides by 3 to solve for $x$.
  3×3=93\frac{3x}{3} = \frac{9}{3} 33x​=39​
  $x=3$
• Verification: Substitute $x=3$ in the original equation:
  LHS = $3(3)+4=9+4=13$, RHS = 13.
  LHS = RHS, so the solution is correct.

Example 2: Solve $5x-7=8$

• Step 1: Add 7 to both sides.
  $5x-7+7=8+7$
  $5x=15$
• Step 2: Divide both sides by 5.
  5×5=155\frac{5x}{5} = \frac{15}{5} 55x​=515​
  $x=3$
• Verification: $5(3)-7=15-7=8$, which matches the RHS.

Equations with Variables on Both Sides

Sometimes, variables appear on both sides of the equation. To solve:

• Bring all terms with the variable to one side and constants to the other.
• Simplify and solve.

Example: Solve $4x+5=2x+13$

• Step 1: Subtract $2x$ from both sides.
  $4x-2x+5=2x-2x+13$
  $2x+5=13$
• Step 2: Subtract 5 from both sides.
  $2x+5-5=13-5$
  $2x=8$
• Step 3: Divide both sides by 2.
  2×2=82\frac{2x}{2} = \frac{8}{2} 22x​=28​
  $x=4$
• Verification: LHS = $4(4)+5=16+5=21$, RHS = $2(4)+13=8+13=21$.
  LHS = RHS, so $x=4$ is correct.

Equations Involving Fractions

When equations have fractions, multiply both sides by the Least Common Multiple (LCM) of the denominators to eliminate them.

Example: Solve x2+3=5 \frac{x}{2} + 3 = 5 2x​+3=5

• Step 1: Subtract 3 from both sides.
  x2+3−3=5−3\frac{x}{2} + 3 – 3 = 5 – 3 2x​+3−3=5−3
  x2=2\frac{x}{2} = 2 2x​=2
• Step 2: Multiply both sides by 2.
  2×x2=2×22 \times \frac{x}{2} = 2 \times 2 2×2x​=2×2
  $x=4$
• Verification: 42+3=2+3=5 \frac{4}{2} + 3 = 2 + 3 = 5 24​+3=2+3=5, which matches the RHS.

Example: Solve 2×3−1=5 \frac{2x}{3} – 1 = 5 32x​−1=5

• Step 1: Add 1 to both sides.
  2×3−1+1=5+1\frac{2x}{3} – 1 + 1 = 5 + 1 32x​−1+1=5+1
  2×3=6\frac{2x}{3} = 6 32x​=6
• Step 2: Multiply both sides by 3.
  3×2×3=3×63 \times \frac{2x}{3} = 3 \times 6 3×32x​=3×6
  $2x=18$
• Step 3: Divide both sides by 2.
  2×2=182\frac{2x}{2} = \frac{18}{2} 22x​=218​
  $x=9$
• Verification: 2(9)3−1=183−1=6−1=5 \frac{2(9)}{3} – 1 = \frac{18}{3} – 1 = 6 – 1 = 5 32(9)​−1=318​−1=6−1=5, which is correct.

Applications of Simple Equations

Simple equations are used to solve real-world problems like calculating ages, distances, costs, or quantities.

Example Problem:

The sum of two numbers is 25, and one number is 7 more than the other. Find the numbers.

• Let the smaller number be $x$.
• The larger number = $x+7$.
• Equation: $x+(x+7)=25$.
• Simplify: $2x+7=25$.
• Subtract 7: $2x=18$.
• Divide by 2: $x=9$.
• Smaller number = 9, larger number = $9+7=16$.
• Check: $9+16=25$, which is true.

Tips for Solving Equations

1. Always perform the same operation on both sides to maintain equality.
2. Simplify step-by-step and avoid errors in arithmetic.
3. Verify the solution by substituting it back into the original equation.
4. For word problems, clearly define the variable and double-check the translation into an equation.

Practice Questions

1. Solve: $7x-3=18$.
2. Solve: $2x+9=5x-6$.
3. Solve: x4+2=6 \frac{x}{4} + 2 = 6 4x​+2=6.
4. The perimeter of a rectangle is 26 cm. If the length is 3 cm more than the breadth, find the dimensions.

Conclusion

Chapter 4 introduces the concept of simple equations, a key skill in algebra. By mastering how to form and solve equations, students can tackle a variety of mathematical and real-life problems. The chapter emphasizes logical steps, verification, and practical applications, laying a strong foundation for future topics like linear equations in higher classes.

Download pdf notes and Sulutions of the Chapter:

Chapter 5: Lines and Angles