Linear Equations in One Variable Chapter Notes - Class 6 ICSE PDF Download
Introduction
Imagine you are on a treasure hunt and the map gives you clues in the form of equations. A linear equation in one variable is a simple puzzle where you must find the value of a single unknown, usually denoted by a letter such as x or y. These equations are called linear because, when drawn on a graph, they form a straight line. They involve the variable only to the first power (no squares or higher powers).
To Write a Given Statement in Algebraic Form
- Choose a letter (variable) to represent the unknown quantity.
- Identify keywords in the sentence to determine operations: added to means +, subtracted from means -, multiplied by means ×, divided by means ÷, gives or equals means =.
- Write the algebraic expression or equation that matches the sentence exactly.
- Check the sentence and the algebraic form by reading the algebraic expression back in words.
Solution:
The phrase "seven added to y" corresponds to the expression:
\[7 + y\]
The phrase "gives 20" means equality to 20, so the algebraic form is:
\[7 + y = 20\]
To Write a Statement for a Given Algebraic Form
- Read the symbols and replace the variable by a phrase such as "a number" or "the number x."
- Translate mathematical signs into words: + as "added to", - as "subtracted from", × as "multiplied by", ÷ as "divided by", = as "gives" or "is equal to".
- Construct a clear sentence that matches the equation.
Solution:
Read as: "x subtracted from 7 gives 5."
Linear Equations in One Variable
- An equation is a statement that two expressions are equal, shown by the equals sign =.
- A linear equation has the variable raised only to the first power (no \(x^2\), \(x^3\), etc.).
- A linear equation in one variable contains exactly one unknown, for example \(x\), \(n\), or \(y\).
- Examples: \(2n + 3 = 7\), \(x + 5 = 4x\).
Properties of Equations
- Addition Property: If two expressions are equal, adding the same number to both sides keeps them equal. If \(x = y\), then \(x + z = y + z\).
- Subtraction Property: If two expressions are equal, subtracting the same number from both sides keeps them equal. If \(x = y\), then \(x - z = y - z\).
- Multiplication Property: If two expressions are equal, multiplying both sides by the same number keeps them equal. If \(x = y\), then \(x \times z = y \times z\).
- Division Property: If two expressions are equal, dividing both sides by the same non-zero number keeps them equal. If \(x = y\) and \(z
eq 0\), then \(x \div z = y \div z\).
\[x + 3 = 5 + 3\]
\[x + 3 = 8\]
Transposition of Terms of an Equation
- To move a term from one side of an equation to the other, change the operation sign: a term added on one side becomes subtracted when moved to the other side, and vice versa.
- If a term multiplies the variable on one side (for example \(ax = b\)), moving it to the other side is done by division: \(x = b \div a\) (provided \(a
eq 0\)). - Use transposition to collect like terms and simplify the equation before solving.
Solution:
Transpose 5 to the other side by changing the sign:
\[x = 7 - 5\]
\[x = 2\]
- The solution of an equation is the value of the variable that makes the equation true.
- Always substitute the found value back into the original equation to verify it satisfies both sides.
Verification:
Substitute \(x = 1\) into the left-hand side:
\[3 \times 1 - 5 = 3 - 5 = -2\]
The right-hand side is \(-2\), so LHS = RHS and \(x = 1\) is correct.
Solving Linear Equations in One Variable
- Bring all terms containing the variable to one side and constant terms to the other side by transposition.
- Combine like terms.
- Divide both sides by the coefficient of the variable to obtain the variable's value.
- Finally, verify by substituting the value into the original equation.
Solution:
Transpose \(-y\) to the left side (add \(y\) to both sides):
\[y + y + 3 = 1\]
Simplify the left side by combining like terms:
\[2y + 3 = 1\]
Transpose 3 to the right side (subtract 3 from both sides):
\[2y = 1 - 3\]
\[2y = -2\]
Divide both sides by 2:
\[y = -2 \div 2\]
\[y = -1\]
Verification of the Solution
- Substitute the value obtained for the variable into the original equation.
- Compute the left-hand side (LHS) and right-hand side (RHS) separately.
- If LHS equals RHS, the solution is verified correct.
Solution:
Expand the left-hand side:
\[4x + 8 = 2x + 5\]
Transpose \(2x\) to the left side (subtract \(2x\) from both sides):
\[4x - 2x + 8 = 5\]
Simplify the left side:
\[2x + 8 = 5\]
Transpose 8 to the right side (subtract 8 from both sides):
\[2x = 5 - 8\]
\[2x = -3\]
Divide both sides by 2:
\[x = -3 \div 2\]
\[x = -\frac{3}{2}\]
Verification:
Substitute \(x = -\frac{3}{2}\) into the left-hand side:
\[4\bigl(-\tfrac{3}{2} + 2\bigr) = 4\bigl(\tfrac{1}{2}\bigr) = 2\]
Substitute into the right-hand side:
\[2\bigl(-\tfrac{3}{2}\bigr) + 5 = -3 + 5 = 2\]
Since LHS = RHS, the solution \(x = -\frac{3}{2}\) is correct.
Word Problems
- Read the problem carefully to understand what is asked and which quantities are unknown.
- Choose a variable to represent the unknown quantity.
- Translate the words into an algebraic equation using the chosen variable.
- Solve the equation using the methods for linear equations and check the answer in the context of the problem.
Solution:
Let the number be \(x\).
Form the equation from the sentence:
\[x - 12 = 30\]
Transpose \(-12\) to the right side (add 12 to both sides):
\[x = 30 + 12\]
\[x = 42\]
The number is 42.
Solved Examples
Solution:
Expand the left-hand side:
\[4x + 8 = 2x + 5\]
Transpose \(2x\) to the left side (subtract \(2x\) from both sides):
\[4x - 2x + 8 = 5\]
Simplify:
\[2x + 8 = 5\]
Transpose 8 to the right side (subtract 8 from both sides):
\[2x = 5 - 8\]
\[2x = -3\]
Divide both sides by 2:
\[x = -\frac{3}{2}\]
Verification:
Left side evaluation:
\[4\bigl(-\tfrac{3}{2} + 2\bigr) = 4\bigl(\tfrac{1}{2}\bigr) = 2\]
Right side evaluation:
\[2\bigl(-\tfrac{3}{2}\bigr) + 5 = -3 + 5 = 2\]
Since both sides equal 2, the solution is correct.
Solution:
Let the daughter's age be \(x\) years.
The man's age is \(x + 29\) years.
Form the equation for the sum of ages:
\[x + (x + 29) = 85\]
Simplify the left side by combining like terms:
\[2x + 29 = 85\]
Transpose 29 to the right side (subtract 29 from both sides):
\[2x = 85 - 29\]
\[2x = 56\]
Divide both sides by 2:
\[x = 56 \div 2\]
\[x = 28\]
The daughter's age is 28 years. The man's age is:
\[28 + 29 = 57\]
The man is 57 years old.
FAQs on Chapter Notes: Linear Equations in One Variable
| 1. What is a linear equation in one variable? |  |
| 2. How can I write a verbal statement in algebraic form? | |
| 3. What are the properties of equations that are important for solving linear equations? | |
| 4. How do I verify the solution of a linear equation? | |
| 5. What are some common word problems involving linear equations? | |
