Maharashtra Board Class 8 Maths part 2 Chapter 12 Equations in one variable PDF Download

Equations In One Variable

Let's Recall

In the previous standards we have studied equations in one variable.

The value of the variable which satisfies the given equation is called the solution of the equation.

Solving an equation means to find the solution of the equation.

The equation obtained by performing the same operation on its both sides does not change its solution. Using this property, we obtain new but simpler equations and solve the given equation.

The operations done on both sides of an equation are:

(i) To add the same number to both sides.

(ii) To subtract the same number from both sides.

(iii) To multiply both sides by the same number.

(iv) To divide both sides by the same non zero number.

Fill In The Boxes To Solve The Following Equations

Ex. (1) \(x + 4 = 9\)

\(\therefore x + 4 - \square = 9 - \square\)

\(\therefore x = \square\)

Ex. (2) \(x - 2 = 7\)

\(\therefore x - 2 + \square = 7 + \square\)

\(\therefore x = \square\)

Ex. (3) \(\frac{x}{3} = 4\)

\(\therefore \frac{x}{3} \times \square = 4 \times \square\)

\(\therefore x = \square\)

Ex. (4) \(4x = 24\)

\(\therefore \frac{4x}{\square} = \frac{24}{\square}\)

\(\therefore x = \square\)

Teacher's Note

Solving equations is like finding the missing number in a math puzzle. For example, if you have 5 rupees more and now have 15 rupees total, you can find that you started with 10 rupees.

Exam Trick

Remember: Whatever you do on one side of the equation, do the same thing on the other side. If you add 5 on the left, add 5 on the right too.

Points to Remember

Solution means the value that makes the equation true.
You can add, subtract, multiply, or divide both sides by the same number.
The equation stays equal when you do the same thing to both sides.
Always check your answer by putting it back in the original equation.
Solving means making the variable stand alone on one side.

Let's Learn

Solution Of Equations In One Variable

While solving an equation, sometimes we have to perform many operations on it. We will learn how to find solutions of such equations. Study the following examples.

Ex. (1) Solve The Following Equations

(i) \(2(x - 3) = \frac{3}{5}(x + 4)\)

Solution: Multiplying both sides by 5

\(10(x - 3) = 3(x + 4)\)

\(\therefore 10x - 30 = 3x + 12\)

Adding 30 to both sides

\(\therefore 10x - 30 + 30 = 3x + 12 + 30\)

\(10x = 3x + 42\)

Subtracting 3x from both sides

\(\therefore 10x - 3x = 3x + 42 - 3x\)

\(\therefore 7x = 42\)

Dividing both sides by 7

\(\therefore \frac{7x}{7} = \frac{42}{7}\)

\(\therefore x = 6\)

(ii) \(9x - 4 = 6x + 29\)

Solution: Adding 4 to both sides

\(9x - 4 + 4 = 6x + 29 + 4\)

\(\therefore 9x = 6x + 33\)

Subtracting 6x from both sides

\(\therefore 9x - 6x = 6x + 33 - 6x\)

\(\therefore 3x = 33\)

Dividing both sides by 3

\(\therefore \frac{3x}{3} = \frac{33}{3}\)

\(\therefore x = 11\)

(iii) \(\frac{2}{3} + 5a = 4\)

Solution: Method I

\(\frac{2}{3} + 5a = 4\)

Multiplying each term by 3

\(3 \times \frac{2}{3} + 3 \times 5a = 4 \times 3\)

\(\therefore 2 + 15a = 12\)

\(\therefore 15a = 12 - 2\)

\(\therefore 15a = 10\)

\(\therefore a = \frac{10}{15}\)

\(\therefore a = \frac{2}{3}\)

Method II

Subtracting \(\frac{2}{3}\) from both the sides,

\(\frac{2}{3} + 5a - \frac{2}{3} = 4 - \frac{2}{3}\)

\(\therefore 5a = \frac{12 - 2}{3}\)

\(\therefore 5a = \frac{10}{3}\)

Dividing both sides by 5

\(\frac{5a}{5} = \frac{10}{3} \times \frac{1}{5}\)

\(\therefore a = \frac{2}{3}\)

If A, B, C, D are nonzero expressions such that \(\frac{A}{B} = \frac{C}{D}\) then multiplying both sides by \(B \times D\) we get the equation \(AD = BC\). Using this we will solve examples.

Teacher's Note

When you have fractions in equations, multiply all terms by the denominator to make them easier. For example, if you see \(\frac{x}{2}\), multiply everything by 2 to remove the fraction.

Exam Trick

To solve equations with fractions quickly, multiply the whole equation by the least common multiple of all denominators. This removes all fractions at once.

Points to Remember

Always do the same operation on both sides of the equation.
Move all terms with the variable to one side and numbers to the other side.
Combine like terms before dividing to find the answer.
Multiply by the denominator to remove fractions from equations.
Check your final answer by substituting it back into the original equation.

(iv) \(\frac{(x - 7)}{(x - 2)} = \frac{5}{4}\)

Solution: \(\frac{(x - 7)}{(x - 2)} = \frac{5}{4}\)

\(\therefore 4(x - 7) = 5(x - 2)\)

\(\therefore 4x - 28 = 5x - 10\)

\(\therefore 4x - 5x = -10 + 28\)

\(\therefore -x = 18\)

\(\therefore x = -18\)

(v) \(\frac{8m - 1}{2m + 3} = 2\)

Solution: \(\frac{8m - 1}{2m + 3} = \frac{2}{1}\)

\(\therefore 1(8m - 1) = 2(2m + 3)\)

\(\therefore 8m - 1 = 4m + 6\)

\(\therefore 8m - 4m = 6 + 1\)

\(\therefore 4m = 7\)

\(\therefore m = \frac{7}{4}\)

Teacher's Note

Cross multiplication is very useful when you have two fractions. If \(\frac{a}{b} = \frac{c}{d}\), then \(ad = bc\). This trick saves time in exams.

Exam Trick

When you see a fraction equal to a fraction, use cross multiplication right away. Multiply the top of the left side by the bottom of the right side, and vice versa.

Points to Remember

Cross multiply when you have fractions on both sides of the equation.
Remove brackets first by multiplying the terms inside.
Negative numbers need careful handling when dividing.
Always simplify fractions in your final answer if possible.
Write your working step by step so you can find mistakes easily.

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