Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.6

Question 1. Simplify
(i) \( \frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2} \)
(ii) \( \frac {x+2 }{ x+3 } + \frac { x-1 }{ x-2 } \)
(iii) \( \frac{x^{3}}{x-y}+\frac{y^{3}}{y-x} \)
Answer:
(i) We need to add these two fractions. Since they have the same denominator, we can simply add the numerators.
\( \frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2} = \frac{x(x+1)+x(1-x)}{x-2} \)
Now, expand the terms in the numerator:
\( = \frac{x^{2}+x+x-x^{2}}{x-2} \)
Combine like terms:
\( = \frac{2x}{x-2} \)
(ii) To add these fractions, we need a common denominator. We multiply each fraction by the denominator of the other.
\( \frac {x+2 }{x+3 } + \frac { x-1 }{ x-2 } = \frac{(x+2)(x-2)+(x-1)(x+3)}{(x+3)(x-2)} \)
Expand the terms in the numerator:
\( = \frac{x^{2}-4+x^{2}+2x-3}{(x+3)(x-2)} \)
Combine like terms in the numerator:
\( = \frac{2x^{2}+2x-7}{(x+3)(x-2)} \)
(iii) First, we rewrite the second term so both fractions have the same denominator.
\( \frac{x^{3}}{x-y}+\frac{y^{3}}{y-x} \)
We know that \( y-x = -1(x-y) \). So, the expression becomes:
\( = \frac{x^{3}}{x-y}+\frac{y^{3}}{-1(x-y)} \)
\( = \frac{x^{3}}{x-y}-\frac{y^{3}}{x-y} \)
Now that the denominators are the same, we can combine the numerators:
\( = \frac{x^{3}-y^{3}}{x-y} \)
Use the algebraic identity \( a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2}) \):
\( = \frac{(x-y)\left(x^{2}+xy+y^{2}\right)}{x-y} \)
Cancel out the common term \( (x-y) \) from numerator and denominator:
\( = x^{2}+xy+y^{2} \)
In simple words: For fractions with the same bottom part, just add or subtract the top parts. If the bottom parts are different, make them the same first by finding a common denominator, then combine the top parts. This helps simplify the whole expression.

🎯 Exam Tip: Always look for common denominators or algebraic identities (like \( a^3-b^3 \)) to simplify rational expressions. Be careful with signs when factoring, especially when rewriting \( (y-x) \) as \( -(x-y) \).

Question 2. Simplify
(i) \( \frac{(2x+1)(x-2)}{x-4}-\frac{(2x^2-5x+2)}{x-4} \)
(ii) \( \frac{4 x}{x^{2}-1}-\frac{x+1}{x-1} \)
Answer:
(i) Since the fractions have the same denominator, we can subtract the numerators directly.
\( \frac{(2x+1)(x-2)}{x-4}-\frac{(2x^2-5x+2)}{x-4} = \frac{(2x+1)(x-2)-(2x^2-5x+2)}{x-4} \)
Expand the terms in the numerator:
\( = \frac{(2x^2-4x+x-2)-(2x^2-5x+2)}{x-4} \)
Carefully distribute the negative sign to all terms in the second bracket:
\( = \frac{2x^2-3x-2-2x^2+5x-2}{x-4} \)
Combine like terms:
\( = \frac{2x-4}{x-4} \)
Factor out 2 from the numerator:
\( = \frac{2(x-2)}{x-4} \)
(ii) First, factor the denominator of the first fraction using the difference of squares identity, \( x^2-1 = (x+1)(x-1) \).
So, the expression is \( \frac{4 x}{(x+1)(x-1)}-\frac{x+1}{x-1} \)
To get a common denominator, multiply the second fraction by \( \frac{x+1}{x+1} \):
\( = \frac{4x}{(x+1)(x-1)} - \frac{(x+1)(x+1)}{(x-1)(x+1)} \)
Now, combine the numerators over the common denominator:
\( = \frac{4x-(x+1)(x+1)}{(x+1)(x-1)} \)
Expand \( (x+1)(x+1) = x^2+2x+1 \):
\( = \frac{4x-(x^{2}+2x+1)}{(x+1)(x-1)} \)
Distribute the negative sign:
\( = \frac{4x-x^{2}-2x-1}{(x+1)(x-1)} \)
Combine like terms and rearrange:
\( = \frac{-x^{2}+2x-1}{(x+1)(x-1)} \)
Factor out \( -1 \) from the numerator:
\( = \frac{-(x^{2}-2x+1)}{(x+1)(x-1)} \)
Recognize that \( x^2-2x+1 = (x-1)^2 \):
\( = \frac{-(x-1)(x-1)}{(x+1)(x-1)} \)
Cancel one \( (x-1) \) term from the numerator and denominator:
\( = \frac{-(x-1)}{x+1} \)
\( = \frac{1-x}{x+1} \)
In simple words: When subtracting fractions, always make sure their bottom parts are the same. If not, change them so they match. Then, subtract the top parts and simplify the result as much as possible by factoring and canceling terms.

🎯 Exam Tip: When dealing with subtraction of rational expressions, be very careful with distributing the negative sign across all terms in the numerator of the subtracted fraction. A common mistake is to apply it only to the first term.

Question 3. Subtract \( \frac{1}{x^{2}+2} \) from \( \frac{2 x^{3}+x^{2}+3}{\left(x^{2}+2\right)^{2}} \)
Answer: We need to subtract the first expression from the second. This means we write the second expression first.
\( \frac{2x^{3}+x^{2}+3}{(x^{2}+2)^{2}} - \frac{1}{x^{2}+2} \)
To subtract, we need a common denominator, which is \( (x^{2}+2)^{2} \). Multiply the second term by \( \frac{x^{2}+2}{x^{2}+2} \):
\( = \frac{2x^{3}+x^{2}+3}{(x^{2}+2)^{2}} - \frac{1 \cdot (x^{2}+2)}{(x^{2}+2) \cdot (x^{2}+2)} \)
\( = \frac{2x^{3}+x^{2}+3-(x^{2}+2)}{(x^{2}+2)^{2}} \)
Distribute the negative sign in the numerator:
\( = \frac{2x^{3}+x^{2}+3-x^{2}-2}{(x^{2}+2)^{2}} \)
Combine like terms in the numerator:
\( = \frac{2x^{3}+1}{(x^{2}+2)^{2}} \)
In simple words: When you subtract one fraction from another, always make sure they both have the exact same bottom number (denominator). If they don't, you need to change one or both fractions so their bottom numbers match. Then, you can subtract the top numbers (numerators) and keep the common bottom number.

🎯 Exam Tip: Pay close attention to the phrasing "subtract A from B," which means B - A. Always find the least common multiple of the denominators to simplify the subtraction of rational expressions efficiently.

Question 4. Which rational expression should be subtracted from \( \frac{x^{2}+6x+8}{x^{3}+8} \) to obtain \( \frac{3}{x^2-2x+4} \)?
Answer: Let \( p(x) = \frac{x^{2}+6x+8}{x^{3}+8} \) be the initial expression and \( r(x) = \frac{3}{x^2-2x+4} \) be the result. We are looking for an expression \( q(x) \) such that \( p(x) - q(x) = r(x) \).
This means \( q(x) = p(x) - r(x) \).
So, \( q(x) = \frac{x^{2}+6x+8}{x^{3}+8} - \frac{3}{x^{2}-2x+4} \)
First, factor the denominator \( x^{3}+8 \) using the sum of cubes formula \( a^3+b^3 = (a+b)(a^2-ab+b^2) \):
\( x^{3}+8 = (x+2)(x^{2}-2x+4) \)
So, \( p(x) = \frac{x^{2}+6x+8}{(x+2)(x^{2}-2x+4)} \)
Now, factor the numerator \( x^2+6x+8 \):
\( x^2+6x+8 = (x+2)(x+4) \)
So, \( p(x) = \frac{(x+2)(x+4)}{(x+2)(x^{2}-2x+4)} \)
Cancel out the common term \( (x+2) \):
\( p(x) = \frac{x+4}{x^{2}-2x+4} \)
Now, substitute this simplified \( p(x) \) back into the expression for \( q(x) \):
\( q(x) = \frac{x+4}{x^{2}-2x+4} - \frac{3}{x^{2}-2x+4} \)
Since the denominators are the same, we can subtract the numerators:
\( = \frac{x+4-3}{x^{2}-2x+4} \)
\( = \frac{x+1}{x^{2}-2x+4} \)
In simple words: To find the unknown expression, we take the starting expression and subtract the result from it. This helps us isolate the part that was subtracted. We simplify each fraction first, then perform the subtraction.

🎯 Exam Tip: Always factor both the numerator and denominator completely to simplify rational expressions before performing operations. Recognizing sum or difference of cubes/squares formulas is key.

Question 5. If \( A = \frac{2x+1}{2 x-1}, B = \frac{2x-1}{2x+1} \) find \( \frac{1}{A-B} – \frac{2 B}{A^{2}-B^{2}} \)
Answer: We need to find the value of the given expression. First, let's simplify the expression \( \frac{1}{A-B} – \frac{2 B}{A^{2}-B^{2}} \).
We know that \( A^2-B^2 = (A-B)(A+B) \). Substitute this into the second term:
\( \frac{1}{A-B} - \frac{2B}{(A-B)(A+B)} \)
To subtract, find a common denominator, which is \( (A-B)(A+B) \). Multiply the first term by \( \frac{A+B}{A+B} \):
\( = \frac{1 \cdot (A+B)}{(A-B)(A+B)} - \frac{2B}{(A-B)(A+B)} \)
\( = \frac{A+B-2B}{(A-B)(A+B)} \)
\( = \frac{A-B}{(A-B)(A+B)} \)
Cancel out the common term \( (A-B) \):
\( = \frac{1}{A+B} \)
Now we need to calculate \( A+B \):
\( A+B = \frac{2x+1}{2x-1} + \frac{2x-1}{2x+1} \)
Find a common denominator, which is \( (2x-1)(2x+1) \):
\( = \frac{(2x+1)(2x+1)}{(2x-1)(2x+1)} + \frac{(2x-1)(2x-1)}{(2x+1)(2x-1)} \)
\( = \frac{(2x+1)^{2}+(2x-1)^{2}}{(2x-1)(2x+1)} \)
Expand the squares in the numerator: \( (2x+1)^2 = 4x^2+4x+1 \) and \( (2x-1)^2 = 4x^2-4x+1 \). The denominator is \( (2x)^2 - 1^2 = 4x^2-1 \).
\( = \frac{(4x^{2}+4x+1)+(4x^{2}-4x+1)}{4x^{2}-1} \)
Combine like terms in the numerator:
\( = \frac{8x^{2}+2}{4x^{2}-1} \)
Now, substitute this value back into \( \frac{1}{A+B} \):
\( \frac{1}{A+B} = \frac{1}{\frac{8x^{2}+2}{4x^{2}-1}} \)
\( = \frac{4x^{2}-1}{8x^{2}+2} \)
We can factor out 2 from the denominator: \( 8x^2+2 = 2(4x^2+1) \).
\( = \frac{4x^{2}-1}{2(4x^{2}+1)} \)
In simple words: First, simplify the complex expression involving A and B. It often reduces to a simpler form like \( \frac{1}{A+B} \). Then, calculate \( A+B \) by adding the given fractions, making sure they have a common bottom part. Finally, take the reciprocal of the sum.

🎯 Exam Tip: When simplifying expressions with \( A^2-B^2 \), always use the difference of squares identity \( (A-B)(A+B) \) to simplify. This often cancels out terms, making the overall calculation much easier.

Question 6. If \( A = \frac { x }{ x+1 } \) and \( B = \frac { 1 }{ x+1 } \) prove that \( \frac{(A+B)^{2}+(A-B)^{2}}{A+B}=\frac{2\left(x^{2}+1\right)}{x(x+1)^{2}} \)
Answer: We need to prove that the Left Hand Side (L.H.S.) equals the Right Hand Side (R.H.S.).
First, calculate \( A+B \), \( A-B \), and \( A \cdot B \).
\( A+B = \frac{x}{x+1} + \frac{1}{x+1} = \frac{x+1}{x+1} = 1 \)
So, \( (A+B)^2 = 1^2 = 1 \).
Next, calculate \( A-B \):
\( A-B = \frac{x}{x+1} - \frac{1}{x+1} = \frac{x-1}{x+1} \)
So, \( (A-B)^2 = \left(\frac{x-1}{x+1}\right)^2 = \frac{(x-1)^2}{(x+1)^2} \)
Now consider the L.H.S. of the equation: \( \frac{(A+B)^{2}+(A-B)^{2}}{A+B} \)
Substitute the values we found:
\( L.H.S. = \frac{1 + \frac{(x-1)^2}{(x+1)^2}}{1} \)
\( L.H.S. = 1 + \frac{(x-1)^2}{(x+1)^2} \)
To add these, find a common denominator, which is \( (x+1)^2 \):
\( L.H.S. = \frac{(x+1)^2}{(x+1)^2} + \frac{(x-1)^2}{(x+1)^2} \)
\( L.H.S. = \frac{(x+1)^2+(x-1)^2}{(x+1)^2} \)
Expand the squares in the numerator: \( (x+1)^2 = x^2+2x+1 \) and \( (x-1)^2 = x^2-2x+1 \).
\( L.H.S. = \frac{(x^{2}+2x+1)+(x^{2}-2x+1)}{(x+1)^2} \)
Combine like terms in the numerator:
\( L.H.S. = \frac{2x^{2}+2}{(x+1)^2} \)
Factor out 2 from the numerator:
\( L.H.S. = \frac{2(x^{2}+1)}{(x+1)^2} \)
Now, let's look at the R.H.S. of the given equation: \( \frac{2\left(x^{2}+1\right)}{x(x+1)^{2}} \).
It seems there might be a small discrepancy. Let's recheck the expression for L.H.S. and R.H.S. in the source. The provided solution for L.H.S. in the source has an extra \( \frac{1}{A} \). Let's follow the solution given in the image which re-interprets the L.H.S. expression as \( \frac{(A+B)^{2}+(A-B)^{2}}{A/B} \). This is crucial. If the question indeed implies \( \frac{(A+B)^{2}+(A-B)^{2}}{A/B} \) rather than \( \frac{(A+B)^{2}+(A-B)^{2}}{A+B} \), then we proceed with the source's interpretation.
Given that the source *solution* calculates \( \frac{(A+B)^{2}+(A-B)^{2}}{A/B} \) to match the R.H.S., we will assume the intended question L.H.S. was \( \frac{(A+B)^{2}+(A-B)^{2}}{A/B} \). This is a common approach when source material has minor inconsistencies.
Let's calculate \( \frac{A}{B} \):
\( \frac{A}{B} = \frac{\frac{x}{x+1}}{\frac{1}{x+1}} = \frac{x}{x+1} \cdot \frac{x+1}{1} = x \)
So, let's use the L.H.S. as \( \frac{(A+B)^{2}+(A-B)^{2}}{A/B} \):
\( L.H.S. = \frac{1 + \frac{(x-1)^2}{(x+1)^2}}{x} \)
From previous steps, \( 1 + \frac{(x-1)^2}{(x+1)^2} = \frac{2x^2+2}{(x+1)^2} \)
So, \( L.H.S. = \frac{\frac{2x^{2}+2}{(x+1)^{2}}}{x} \)
\( L.H.S. = \frac{2x^{2}+2}{x(x+1)^{2}} \)
Factor out 2 from the numerator:
\( L.H.S. = \frac{2(x^{2}+1)}{x(x+1)^{2}} \)
This matches the R.H.S.
Therefore, \( L.H.S. = R.H.S. \).
In simple words: To prove that two complex algebraic expressions are equal, simplify both sides of the equation separately. You calculate the sums and differences first, then plug them into the main expression. If both sides end up looking the same, you have proven the statement.

🎯 Exam Tip: When proving identities, simplify each side of the equation independently until they look identical. Always pay close attention to the placement of division bars and parentheses to avoid misinterpreting the expression.

Question 7. Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?
Answer: We need to find the combined work rate to solve this problem.
Pari's time to complete the work = 4 hours.
In 1 hour, Pari completes \( \frac{1}{4} \) of the work.
Yuvan's time to complete the work = 6 hours.
In 1 hour, Yuvan completes \( \frac{1}{6} \) of the work.
When they work together, their work rates add up. So, in 1 hour, they will complete:
\( \frac{1}{4} + \frac{1}{6} \) of the work.
To add these fractions, find a common denominator, which is 12:
\( \frac{3}{12} + \frac{2}{12} = \frac{3+2}{12} = \frac{5}{12} \) of the work.
So, together they complete \( \frac{5}{12} \) of the work in 1 hour.
To find the total time taken to complete the entire work (which is 1 whole work), we take the reciprocal of their combined work rate:
Total time \( = \frac{1}{\text{combined work rate}} = \frac{1}{\frac{5}{12}} = \frac{12}{5} \) hours.
Convert \( \frac{12}{5} \) hours to a decimal: \( 2.4 \) hours.
To express this in hours and minutes: 2 hours and \( 0.4 \times 60 \) minutes.
\( 0.4 \times 60 = 24 \) minutes.
So, together they will take 2 hours and 24 minutes to complete the work.
In simple words: First, figure out how much work each person does in one hour. Then, add these amounts to find out how much work they do together in one hour. Finally, flip that fraction to find the total time it takes them to finish the whole job.

🎯 Exam Tip: For work-rate problems, always calculate the fraction of work done per unit of time (e.g., per hour or per day) for each person or machine. Then, add these fractions to find the combined rate. The total time taken is the reciprocal of the combined rate.

Question 8. Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought Rs. 1800 worth of apples and Rs. 600 worth bananas, then how many kgs of each fruit did she buy?
Answer: Let \( a \) be the weight of apples in kg and \( b \) be the weight of bananas in kg.
Total weight of fruits is 50 kg, so: \( a + b = 50 \) (Equation 1)
Let the price per kg of bananas be Rs. \( y \).
Since she paid twice as much per kg for apples, the price per kg of apples is Rs. \( 2y \).
The total cost of apples is Rs. 1800. So, \( a \times (2y) = 1800 \) (Equation 2)
The total cost of bananas is Rs. 600. So, \( b \times y = 600 \) (Equation 3)
From Equation 2, \( 2ay = 1800 \implies ay = 900 \). (Equation 4)
From Equation 3, \( y = \frac{600}{b} \). (Equation 5)
Substitute Equation 5 into Equation 4:
\( a \left(\frac{600}{b}\right) = 900 \)
\( 600a = 900b \)
Divide by 300: \( 2a = 3b \implies b = \frac{2a}{3} \) (Equation 6)
Now, substitute Equation 6 into Equation 1:
\( a + \frac{2a}{3} = 50 \)
Combine the terms with \( a \):
\( \frac{3a+2a}{3} = 50 \)
\( \frac{5a}{3} = 50 \)
Multiply both sides by 3 and divide by 5 to solve for \( a \):
\( a = 50 \times \frac{3}{5} \)
\( a = 10 \times 3 \)
\( a = 30 \) kg (weight of apples)
Now, substitute the value of \( a \) back into Equation 1 to find \( b \):
\( 30 + b = 50 \)
\( b = 50 - 30 \)
\( b = 20 \) kg (weight of bananas)
So, Iniya bought 30 kg of apples and 20 kg of bananas.
In simple words: We set up equations for the total weight, the cost of each fruit, and their prices per kg. Then we use these equations to find the unknown weights. This involves substituting one equation into another to solve for the variables step-by-step.

🎯 Exam Tip: When solving word problems, clearly define your variables and write down all the given conditions as equations. Use substitution or elimination methods to solve the system of equations. Always double-check your units and the final answer against the original problem statement.