GSEB Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.6

Question 1. Solve the following equations.
(i) \( \frac { 8x - 3 }{ 3x } = 2 \)
(ii) \( \frac { 9x }{ 7 - 6x } = 15 \)
(iii) \( \frac { z }{ z + 15 } = \frac { 4 }{ 9 } \)
(iv) \( \frac { 3y + 4 }{ 2 - 6y } = \frac { -2 }{ 5 } \)
(v) \( \frac { 7y + 4 }{ y + 2 } = \frac { -4 }{ 3 } \)
Answer:
(i) We are given the equation: \( \frac { 8x - 3 }{ 3x } = 2 \)
To remove the fraction, we multiply both sides of the equation by \( 3x \).
\( \frac { 8x - 3 }{ 3x } \times 3x = 2 \times 3x \)
This simplifies to \( 8x - 3 = 6x \).
Next, we rearrange the terms. We move \( -3 \) to the right side and \( 6x \) to the left side.
\( 8x - 6x = 3 \)
\( 2x = 3 \)
Finally, we divide both sides by 2 to find the value of \( x \).
\( x = \frac { 3 }{ 2 } \)
(ii) We begin with the equation: \( \frac { 9x }{ 7 - 6x } = 15 \)
To eliminate the denominator, we multiply both sides by \( (7 - 6x) \).
\( \frac { 9x }{ 7 - 6x } \times (7 - 6x) = 15 \times (7 - 6x) \)
Simplifying, we get \( 9x = 105 - 90x \).
Now, we want to gather all terms containing \( x \) on one side. We transpose \( -90x \) to the left-hand side.
\( 9x + 90x = 105 \)
\( 99x = 105 \)
To isolate \( x \), we divide both sides by 99.
\( x = \frac { 105 }{ 99 } \)
This fraction can be simplified by dividing both the numerator and denominator by 3, giving us \( x = \frac { 35 }{ 33 } \).
(iii) We have the equation: \( \frac { z }{ z + 15 } = \frac { 4 }{ 9 } \)
We can solve this by using cross-multiplication.
\( 9 \times z = 4 \times (z + 15) \)
\( \implies \) \( 9z = 4z + 60 \)
To collect terms with \( z \), we move \( 4z \) from the right to the left side.
\( 9z - 4z = 60 \)
\( 5z = 60 \)
Finally, we divide both sides by 5 to find \( z \).
\( z = \frac { 60 }{ 5 } \)
\( \implies \) \( z = 12 \)
(iv) The given equation is: \( \frac { 3y + 4 }{ 2 - 6y } = \frac { -2 }{ 5 } \)
We will use cross-multiplication to solve this.
\( 5 \times (3y + 4) = -2 \times (2 - 6y) \)
Expanding both sides, we get \( 15y + 20 = -4 + 12y \).
To group the \( y \) terms, we transpose \( 12y \) to the left side and the constant \( 20 \) to the right side.
\( 15y - 12y = -4 - 20 \)
Simplifying both sides, we have \( 3y = -24 \).
To find \( y \), we divide both sides by 3.
\( y = \frac { -24 }{ 3 } \)
\( \implies \) \( y = -8 \)
(v) We are presented with the equation: \( \frac { 7y + 4 }{ y + 2 } = \frac { -4 }{ 3 } \)
We can solve this by cross-multiplication.
\( 3 \times (7y + 4) = -4 \times (y + 2) \)
Expanding the terms, we get \( 21y + 12 = -4y - 8 \).
Next, we rearrange the equation by moving \( -4y \) to the left side and \( 12 \) to the right side.
\( 21y + 4y = -8 - 12 \)
Combining like terms, we have \( 25y = -20 \).
Finally, we divide both sides by 25 to find the value of \( y \).
\( y = \frac { -20 }{ 25 } \)
\( \implies \) \( y = \frac { -4 }{ 5 } \)
In simple words: To solve these equations, first get rid of any fractions by multiplying. Then, move all the terms with the unknown letter to one side and all the numbers to the other. Lastly, divide to find the value of the letter.

Exam Tip: Always double-check your calculations, especially when dealing with negative numbers and transposing terms. Simplify fractions to their lowest form for the final answer.

Question 2. The ages of Hari and Harry are in the ratio 5: 7. Four years from now the ratio of their ages will be 3: 4. Find their present ages.
Answer:
Let Hari's current age be \( 5x \) years and Harry's current age be \( 7x \) years, based on their given ratio.
After four years, Hari's age will be \( (5x + 4) \) years, and Harry's age will be \( (7x + 4) \) years.
According to the problem's condition, the ratio of their ages after four years will be 3:4.
So, we can write the equation as \( \frac { 5x + 4 }{ 7x + 4 } = \frac { 3 }{ 4 } \).
By using cross-multiplication, we get:
\( 4(5x + 4) = 3(7x + 4) \)
Expanding both sides, this becomes \( 20x + 16 = 21x + 12 \).
Now, we transpose the terms: move \( 21x \) to the left-hand side and \( 16 \) to the right-hand side.
\( 20x - 21x = 12 - 16 \)
\( -x = -4 \)
\( \implies \) \( x = 4 \).
Therefore, Hari's present age is \( 5 \times 4 = 20 \) years.
And Harry's present age is \( 7 \times 4 = 28 \) years.
In simple words: We set up the current ages using a ratio and then adjust for ages after four years. Forming an equation with the new ratio allows us to solve for 'x' and find their actual current ages.

Exam Tip: When solving age problems with ratios, represent the current ages with a variable (e.g., 5x, 7x). Remember to correctly add or subtract years for future or past ages before setting up the final ratio equation.

Question 3. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \( \frac { 3 }{ 2 } \). Find the rational number.
Answer:
Let the numerator of the rational number be \( x \).
Based on the problem statement, the denominator is 8 greater than the numerator, so the denominator is \( x + 8 \).
The original rational number is \( \frac { x }{ x + 8 } \).
If the numerator is increased by 17, the new numerator becomes \( x + 17 \).
If the denominator is decreased by 1, the new denominator becomes \( (x + 8) - 1 = x + 7 \).
The new rational number formed is \( \frac { x + 17 }{ x + 7 } \).
According to the condition given in the problem, this new number is equal to \( \frac { 3 }{ 2 } \).
So, we set up the equation: \( \frac { x + 17 }{ x + 7 } = \frac { 3 }{ 2 } \).
To solve this, we use cross-multiplication:
\( 2 \times (x + 17) = 3 \times (x + 7) \)
Expanding both sides, we get \( 2x + 34 = 3x + 21 \).
Next, we transpose the terms: move \( 3x \) to the left side and \( 34 \) to the right side.
\( 2x - 3x = 21 - 34 \)
\( -x = -13 \)
\( \implies \) \( x = 13 \).
So, the numerator is \( 13 \).
The denominator is \( x + 8 = 13 + 8 = 21 \).
Therefore, the rational number is \( \frac { 13 }{ 21 } \).
In simple words: We represent the unknown numerator with 'x'. Then we write the denominator and the new numerator/denominator based on the given conditions. Equating the new fraction to 3/2 lets us solve for 'x' and find the original rational number.

Exam Tip: Clearly define your variables (numerator, denominator) at the beginning. Pay close attention to how the numerator and denominator change (increased, decreased) before forming the new fraction.