RBSE Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.2

Chapter 4 Linear Equations In Two Variables Ex 4.2

Solve each of the following systems of simultaneous linear equations by the method of substitution. [Q1. to Q6.]

Question 1. 2x + 3y = 9
3x + 4y = 5
Answer: The given equations are:
\( 2x + 3y = 9 \) ...(i)
\( 3x + 4y = 5 \) ...(ii)
From equation (i), we can write:
\( 2x = 9 - 3y \)
\( x = \frac{9-3y}{2} \) ...(iii)
Now, substitute this value of \(x\) from equation (iii) into equation (ii):
\( 3 \left( \frac{9-3y}{2} \right) + 4y = 5 \)
To remove the fraction, multiply the entire equation by 2:
\( 3(9-3y) + 2(4y) = 2(5) \)
\( 27 - 9y + 8y = 10 \)
Combine the \(y\) terms:
\( 27 - y = 10 \)
Now, isolate \(y\):
\( -y = 10 - 27 \)
\( -y = -17 \)
\( y = 17 \)
Next, substitute the value of \(y = 17\) back into equation (iii) to find \(x\):
\( x = \frac{9-3(17)}{2} \)
\( x = \frac{9-51}{2} \)
\( x = \frac{-42}{2} \)
\( x = -21 \)
Thus, the solution for the system of equations is \(x = -21\) and \(y = 17\). Solving systems of linear equations helps find points where two lines intersect on a graph.
In simple words: First, we change one equation to find \(x\) in terms of \(y\). Then we put this \(x\) into the second equation to find \(y\). Finally, we use the value of \(y\) to find \(x\).

🎯 Exam Tip: Always verify your solution by substituting the \(x\) and \(y\) values back into both original equations to ensure they both hold true. This catches errors.

Question 2. Solve the following system of equations:
\( x + 2y = -1 \)
\( 2x - 3y = 12 \)
Answer: The given equations are:
\( x + 2y = -1 \) ...(i)
\( 2x - 3y = 12 \) ...(ii)
From equation (i), express \(x\) in terms of \(y\):
\( x = -1 - 2y \) ...(iii)
Substitute this value of \(x\) from equation (iii) into equation (ii):
\( 2(-1 - 2y) - 3y = 12 \)
Distribute the 2:
\( -2 - 4y - 3y = 12 \)
Combine the \(y\) terms:
\( -2 - 7y = 12 \)
Add 2 to both sides:
\( -7y = 12 + 2 \)
\( -7y = 14 \)
Divide by -7:
\( y = \frac{14}{-7} \)
\( y = -2 \)
Now, substitute the value of \(y = -2\) back into equation (iii) to find \(x\):
\( x = -1 - 2(-2) \)
\( x = -1 + 4 \)
\( x = 3 \)
Thus, the required solution for the system of equations is \(x = 3\) and \(y = -2\). This method is particularly useful when one of the variables has a coefficient of 1 or -1, making it easy to isolate.
In simple words: We changed the first equation to get \(x\) by itself. Then we put that \(x\) into the second equation to find \(y\). Finally, we used \(y\) to find the correct \(x\).

🎯 Exam Tip: When dealing with negative numbers, be extra careful with signs, especially during multiplication and subtraction, to avoid common calculation errors.

Question 3. 3x + 2y = 11
2x + 3y = 4
Answer: The given equations are:
\( 3x + 2y = 11 \) ...(i)
\( 2x + 3y = 4 \) ...(ii)
From equation (i), express \(2y\) in terms of \(x\):
\( 2y = 11 - 3x \)
\( y = \frac{11 - 3x}{2} \) ...(iii)
Substitute this value of \(y\) from equation (iii) into equation (ii):
\( 2x + 3 \left( \frac{11 - 3x}{2} \right) = 4 \)
Multiply the entire equation by 2 to clear the denominator:
\( 2(2x) + 3(11 - 3x) = 2(4) \)
\( 4x + 33 - 9x = 8 \)
Combine the \(x\) terms:
\( -5x + 33 = 8 \)
Subtract 33 from both sides:
\( -5x = 8 - 33 \)
\( -5x = -25 \)
Divide by -5:
\( x = \frac{-25}{-5} \)
\( x = 5 \)
Now, substitute the value of \(x = 5\) back into equation (iii) to find \(y\):
\( y = \frac{11 - 3(5)}{2} \)
\( y = \frac{11 - 15}{2} \)
\( y = \frac{-4}{2} \)
\( y = -2 \)
Hence, the required solution is \(x = 5\) and \(y = -2\). This process helps find the specific point where the two lines represented by these equations cross.
In simple words: We took the first equation and solved it for \(y\). Then we put this expression for \(y\) into the second equation. This helped us find \(x\), and then we used \(x\) to find \(y\).

🎯 Exam Tip: When solving equations with fractions, multiplying by the least common multiple of the denominators is often the quickest way to clear fractions and simplify calculations.

Question 4. 8x + 5y = 9
3x + 2y = 4
Answer: The given equations are:
\( 8x + 5y = 9 \) ...(i)
\( 3x + 2y = 4 \) ...(ii)
From equation (ii), express \(2y\) in terms of \(x\):
\( 2y = 4 - 3x \)
\( y = \frac{4 - 3x}{2} \) ...(iii)
Substitute this value of \(y\) from equation (iii) into equation (i):
\( 8x + 5 \left( \frac{4 - 3x}{2} \right) = 9 \)
Multiply the entire equation by 2 to clear the denominator:
\( 2(8x) + 5(4 - 3x) = 2(9) \)
\( 16x + 20 - 15x = 18 \)
Combine the \(x\) terms:
\( (16x - 15x) + 20 = 18 \)
\( x + 20 = 18 \)
Subtract 20 from both sides:
\( x = 18 - 20 \)
\( x = -2 \)
Now, substitute the value of \(x = -2\) back into equation (iii) to find \(y\):
\( y = \frac{4 - 3(-2)}{2} \)
\( y = \frac{4 + 6}{2} \)
\( y = \frac{10}{2} \)
\( y = 5 \)
Thus, the required solution is \(x = -2\) and \(y = 5\). This demonstrates how substituting an expression for one variable can simplify the problem to a single-variable equation.
In simple words: We rearranged the second equation to get \(y\) by itself. Then we put that expression for \(y\) into the first equation to solve for \(x\). Once we had \(x\), we found \(y\).

🎯 Exam Tip: When choosing which variable to isolate, look for variables with smaller coefficients or those that are already nearly isolated to minimize calculation steps and potential errors.

Question 5. 4x - 5y = 39
2x - 7y = 51
Answer: The given equations are:
\( 4x - 5y = 39 \) ...(i)
\( 2x - 7y = 51 \) ...(ii)
From equation (ii), express \(2x\) in terms of \(y\):
\( 2x - 51 = 7y \)
\( 7y = 2x - 51 \)
\( y = \frac{2x - 51}{7} \) ...(iii)
Substitute this value of \(y\) from equation (iii) into equation (i):
\( 4x - 5 \left( \frac{2x - 51}{7} \right) = 39 \)
Multiply the entire equation by 7 to clear the denominator:
\( 7(4x) - 5(2x - 51) = 7(39) \)
\( 28x - 10x + 255 = 273 \)
Combine the \(x\) terms:
\( 18x + 255 = 273 \)
Subtract 255 from both sides:
\( 18x = 273 - 255 \)
\( 18x = 18 \)
Divide by 18:
\( x = \frac{18}{18} \)
\( x = 1 \)
Now, substitute the value of \(x = 1\) back into equation (iii) to find \(y\):
\( y = \frac{2(1) - 51}{7} \)
\( y = \frac{2 - 51}{7} \)
\( y = \frac{-49}{7} \)
\( y = -7 \)
Thus, the required solution is \(x = 1\) and \(y = -7\). The substitution method helps transform a system of two equations into a single, solvable equation.
In simple words: We rearranged the second equation to get \(y\) by itself. Then we put that expression for \(y\) into the first equation to solve for \(x\). After finding \(x\), we easily found \(y\).

🎯 Exam Tip: Always double-check your arithmetic, especially when dealing with multiplication of fractions and combining terms with different signs, as these are common sources of errors.

Question 6. 5x - 2y = 19
3x + y = 18
Answer: The given equations are:
\( 5x - 2y = 19 \) ...(i)
\( 3x + y = 18 \) ...(ii)
From equation (ii), express \(y\) in terms of \(x\):
\( y = 18 - 3x \) ...(iii)
Substitute this value of \(y\) from equation (iii) into equation (i):
\( 5x - 2(18 - 3x) = 19 \)
Distribute the -2:
\( 5x - 36 + 6x = 19 \)
Combine the \(x\) terms:
\( (5x + 6x) - 36 = 19 \)
\( 11x - 36 = 19 \)
Add 36 to both sides:
\( 11x = 19 + 36 \)
\( 11x = 55 \)
Divide by 11:
\( x = \frac{55}{11} \)
\( x = 5 \)
Now, substitute the value of \(x = 5\) back into equation (iii) to find \(y\):
\( y = 18 - 3(5) \)
\( y = 18 - 15 \)
\( y = 3 \)
Hence, the required solution is \(x = 5\) and \(y = 3\). Isolating a variable that has a coefficient of 1 or -1 simplifies the substitution process significantly.
In simple words: We quickly got \(y\) by itself from the second equation. Then we put this expression for \(y\) into the first equation to find \(x\). Once \(x\) was known, finding \(y\) was simple.

🎯 Exam Tip: Look for the easiest variable to isolate. If a variable has a coefficient of 1 or -1, isolating it will prevent fractions and simplify calculations.

Question 7. Solve the following system of equations:
\( 2x + y = 13 \)
\( 5x - 3y = 16 \)
Answer: The given equations are:
\( 2x + y = 13 \) ...(i)
\( 5x - 3y = 16 \) ...(ii)
To solve using the elimination method, we can make the coefficients of \(y\) opposite. Multiply equation (i) by 3:
\( 3(2x + y) = 3(13) \)
\( 6x + 3y = 39 \) ...(iii)
Now, add equation (iii) and equation (ii):
\( (6x + 3y) + (5x - 3y) = 39 + 16 \)
\( 6x + 5x + 3y - 3y = 55 \)
\( 11x = 55 \)
Divide by 11:
\( x = \frac{55}{11} \)
\( x = 5 \)
Now, substitute the value of \(x = 5\) back into equation (i) to find \(y\):
\( 2(5) + y = 13 \)
\( 10 + y = 13 \)
Subtract 10 from both sides:
\( y = 13 - 10 \)
\( y = 3 \)
Hence, the required solution is \(x = 5\) and \(y = 3\). This method is efficient when coefficients can be easily made additive inverses.
In simple words: We multiplied the first equation so that the \(y\) terms would cancel out when added to the second equation. This gave us \(x\), and then we used \(x\) to find \(y\).

🎯 Exam Tip: In the elimination method, aim to multiply equations by numbers that will make one pair of variable coefficients identical but with opposite signs, allowing them to cancel out when added.

Question 8. 0.4x + 0.3y = 1.7
0.7x - 0.2y = 0.8
Answer: The given equations are:
\( 0.4x + 0.3y = 1.7 \) ...(i)
\( 0.7x - 0.2y = 0.8 \) ...(ii)
To simplify, we can multiply both equations by 10 to remove decimals:
\( 4x + 3y = 17 \) ...(iii)
\( 7x - 2y = 8 \) ...(iv)
To use the elimination method, we can make the coefficients of \(y\) opposite. Multiply equation (iii) by 2 and equation (iv) by 3:
Multiply (iii) by 2: \( 2(4x + 3y) = 2(17) \)
\( 8x + 6y = 34 \) ...(v)
Multiply (iv) by 3: \( 3(7x - 2y) = 3(8) \)
\( 21x - 6y = 24 \) ...(vi)
Now, add equation (v) and equation (vi):
\( (8x + 6y) + (21x - 6y) = 34 + 24 \)
\( 8x + 21x + 6y - 6y = 58 \)
\( 29x = 58 \)
Divide by 29:
\( x = \frac{58}{29} \)
\( x = 2 \)
Now, substitute the value of \(x = 2\) back into equation (iii) to find \(y\):
\( 4(2) + 3y = 17 \)
\( 8 + 3y = 17 \)
Subtract 8 from both sides:
\( 3y = 17 - 8 \)
\( 3y = 9 \)
Divide by 3:
\( y = \frac{9}{3} \)
\( y = 3 \)
Hence, the required solution is \(x = 2\) and \(y = 3\). Clearing decimals at the start simplifies the entire calculation process.
In simple words: We first removed the decimals by multiplying by 10. Then, we made the \(y\) numbers match so they would cancel out when we added the two equations. This gave us \(x\), and then we found \(y\).

🎯 Exam Tip: When dealing with decimal coefficients, convert them into whole numbers by multiplying the entire equation by a power of 10. This makes calculations much simpler and reduces errors.

Question 9. \( \frac{x}{7} + \frac{y}{3} = 5 \)
\( \frac{x}{2} - \frac{y}{9} = 6 \)
Answer: The given equations are:
\( \frac{x}{7} + \frac{y}{3} = 5 \) ...(i)
\( \frac{x}{2} - \frac{y}{9} = 6 \) ...(ii)
First, simplify each equation by finding a common denominator to clear the fractions:
For equation (i): LCM of 7 and 3 is 21. Multiply by 21:
\( 21 \left( \frac{x}{7} \right) + 21 \left( \frac{y}{3} \right) = 21(5) \)
\( 3x + 7y = 105 \) ...(iii)
For equation (ii): LCM of 2 and 9 is 18. Multiply by 18:
\( 18 \left( \frac{x}{2} \right) - 18 \left( \frac{y}{9} \right) = 18(6) \)
\( 9x - 2y = 108 \) ...(iv)
Now we have a system of simplified equations:
\( 3x + 7y = 105 \)
\( 9x - 2y = 108 \)
From (iii), we can express \(y\) in terms of \(x\):
\( 7y = 105 - 3x \)
\( y = \frac{105 - 3x}{7} \)
Substituting this into equation (iv):
\( 9x - 2 \left( \frac{105 - 3x}{7} \right) = 108 \)
Multiply by 7 to clear the fraction:
\( 63x - 2(105 - 3x) = 756 \)
\( 63x - 210 + 6x = 756 \)
\( 69x - 210 = 756 \)
\( 69x = 756 + 210 \)
\( 69x = 966 \)
\( x = \frac{966}{69} \)
\( x = 14 \)
Now, substitute \(x = 14\) back into equation (iii) (or the simplified form to find \(y\)):
\( 3(14) + 7y = 105 \)
\( 42 + 7y = 105 \)
\( 7y = 105 - 42 \)
\( 7y = 63 \)
\( y = \frac{63}{7} \)
\( y = 9 \)
Hence, the required solution is \(x = 14\) and \(y = 9\). Eliminating fractions early simplifies all subsequent calculations, making the solution process smoother.
In simple words: First, we got rid of the fractions in both equations. This gave us simpler equations. Then we used the substitution method: we solved one equation for \(y\) and put that into the other equation to find \(x\). Finally, we used \(x\) to find \(y\).

🎯 Exam Tip: Always clear fractions from equations as the first step by multiplying by the Least Common Multiple (LCM) of the denominators. This makes the equations easier to work with.

Question 10. 11x + 15y = -23
7x - 2y = 20
Answer: The given equations are:
\( 11x + 15y = -23 \) ...(i)
\( 7x - 2y = 20 \) ...(ii)
To solve using the elimination method, we can make the coefficients of \(y\) opposite. Multiply equation (i) by 2 and equation (ii) by 15:
Multiply (i) by 2: \( 2(11x + 15y) = 2(-23) \)
\( 22x + 30y = -46 \) ...(iii)
Multiply (ii) by 15: \( 15(7x - 2y) = 15(20) \)
\( 105x - 30y = 300 \) ...(iv)
Now, add equation (iii) and equation (iv):
\( (22x + 30y) + (105x - 30y) = -46 + 300 \)
\( 22x + 105x + 30y - 30y = 254 \)
\( 127x = 254 \)
Divide by 127:
\( x = \frac{254}{127} \)
\( x = 2 \)
Now, substitute the value of \(x = 2\) back into equation (ii) to find \(y\):
\( 7(2) - 2y = 20 \)
\( 14 - 2y = 20 \)
Subtract 14 from both sides:
\( -2y = 20 - 14 \)
\( -2y = 6 \)
Divide by -2:
\( y = \frac{6}{-2} \)
\( y = -3 \)
Hence, the required solution is \(x = 2\) and \(y = -3\). Choosing to eliminate \(y\) here was effective because 15 and 2 have a small LCM, simplifying the multiplications.
In simple words: We multiplied the first equation by 2 and the second by 15 to make the \(y\) numbers opposites. When we added them, the \(y\) terms canceled, and we found \(x\). Then we used \(x\) to find \(y\).

🎯 Exam Tip: When using elimination, choose the variable to eliminate based on the LCM of its coefficients. A smaller LCM means smaller multipliers, which reduces the chance of calculation errors.

Question 11. Solve the following system of equations:
\( 3x - 7y + 10 = 0 \)
\( y - 2x = 3 \)
Answer: The given equations are:
\( 3x - 7y + 10 = 0 \)
\( 3x - 7y = -10 \) ...(i)
\( y - 2x = 3 \)
\( -2x + y = 3 \) ...(ii)
To solve using the elimination method, we can make the coefficients of \(y\) opposite. Multiply equation (ii) by 7:
\( 7(-2x + y) = 7(3) \)
\( -14x + 7y = 21 \) ...(iii)
Now, add equation (i) and equation (iii):
\( (3x - 7y) + (-14x + 7y) = -10 + 21 \)
\( 3x - 14x - 7y + 7y = 11 \)
\( -11x = 11 \)
Divide by -11:
\( x = \frac{11}{-11} \)
\( x = -1 \)
Now, substitute the value of \(x = -1\) back into equation (ii) to find \(y\):
\( y - 2(-1) = 3 \)
\( y + 2 = 3 \)
Subtract 2 from both sides:
\( y = 3 - 2 \)
\( y = 1 \)
Hence, the required solution is \(x = -1\) and \(y = 1\). Rearranging equations into the standard form \(Ax + By = C\) often simplifies the application of elimination.
In simple words: First, we put both equations into a standard form. Then, we multiplied the second equation to make the \(y\) terms cancel out when added to the first equation. This helped us find \(x\), and then we used \(x\) to find \(y\).

🎯 Exam Tip: Always rearrange equations into the standard form \(Ax + By = C\) before applying elimination or substitution methods. This makes the steps clearer and reduces confusion.

Question 12. \( x + 2y = \frac{3}{2} \)
\( 2x + y = \frac{3}{2} \)
Answer: The given equations are:
\( x + 2y = \frac{3}{2} \) ...(i)
\( 2x + y = \frac{3}{2} \) ...(ii)
To solve using the elimination method, we can make the coefficients of \(y\) match. Multiply equation (ii) by 2:
\( 2(2x + y) = 2 \left( \frac{3}{2} \right) \)
\( 4x + 2y = 3 \) ...(iii)
Now, subtract equation (i) from equation (iii):
\( (4x + 2y) - (x + 2y) = 3 - \frac{3}{2} \)
\( 4x - x + 2y - 2y = \frac{6}{2} - \frac{3}{2} \)
\( 3x = \frac{3}{2} \)
Divide by 3:
\( x = \frac{3}{2} \times \frac{1}{3} \)
\( x = \frac{1}{2} \)
Now, substitute the value of \(x = \frac{1}{2}\) back into equation (i) to find \(y\):
\( \frac{1}{2} + 2y = \frac{3}{2} \)
Subtract \( \frac{1}{2} \) from both sides:
\( 2y = \frac{3}{2} - \frac{1}{2} \)
\( 2y = \frac{2}{2} \)
\( 2y = 1 \)
Divide by 2:
\( y = \frac{1}{2} \)
Hence, the required solution is \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\). This shows how elimination works effectively even with fractional constants, especially when coefficients align well.
In simple words: We multiplied the second equation by 2 to make the \(y\) parts the same. Then we subtracted the first equation from the new second equation to make \(y\) disappear, which helped us find \(x\). Finally, we used \(x\) to find \(y\).

🎯 Exam Tip: When equations have fractional constants, it's often best to work with them directly rather than converting to decimals to maintain precision and avoid rounding errors.

Solve each of the following (Q13. to Q15.)

Question 13. 8v - 3u = 5uv
6v - 5u = 2uv
Answer: The given equations are:
\( 8v - 3u = 5uv \) ...(i)
\( 6v - 5u = 2uv \) ...(ii)
Assuming \(u \neq 0\) and \(v \neq 0\), divide both equations by \(uv\):
From (i): \( \frac{8v}{uv} - \frac{3u}{uv} = \frac{5uv}{uv} \)
\( \frac{8}{u} - \frac{3}{v} = 5 \) ...(iii)
From (ii): \( \frac{6v}{uv} - \frac{5u}{uv} = \frac{2uv}{uv} \)
\( \frac{6}{u} - \frac{5}{v} = 2 \) ...(iv)
Let \( A = \frac{1}{u} \) and \( B = \frac{1}{v} \). The equations become a linear system:
\( 8A - 3B = 5 \) ...(v)
\( 6A - 5B = 2 \) ...(vi)
To solve by elimination, multiply equation (v) by 5 and equation (vi) by 3:
Multiply (v) by 5: \( 5(8A - 3B) = 5(5) \)
\( 40A - 15B = 25 \) ...(vii)
Multiply (vi) by 3: \( 3(6A - 5B) = 3(2) \)
\( 18A - 15B = 6 \) ...(viii)
Subtract equation (viii) from equation (vii):
\( (40A - 15B) - (18A - 15B) = 25 - 6 \)
\( 40A - 18A - 15B + 15B = 19 \)
\( 22A = 19 \)
\( A = \frac{19}{22} \)
Now, substitute \( A = \frac{19}{22} \) back into equation (v):
\( 8 \left( \frac{19}{22} \right) - 3B = 5 \)
\( \frac{4 \times 19}{11} - 3B = 5 \)
\( \frac{76}{11} - 3B = 5 \)
\( -3B = 5 - \frac{76}{11} \)
\( -3B = \frac{55 - 76}{11} \)
\( -3B = \frac{-21}{11} \)
\( B = \frac{-21}{11 \times (-3)} \)
\( B = \frac{7}{11} \)
Now, convert back to \(u\) and \(v\):
Since \( A = \frac{1}{u} \), \( u = \frac{1}{A} = \frac{1}{\frac{19}{22}} = \frac{22}{19} \)
Since \( B = \frac{1}{v} \), \( v = \frac{1}{B} = \frac{1}{\frac{7}{11}} = \frac{11}{7} \)
Thus, the required solution is \( u = \frac{22}{19} \) and \( v = \frac{11}{7} \). These types of equations are first converted into a standard linear form before solving.
In simple words: We first divided everything by \(uv\) to make the equations simpler. Then we changed \(1/u\) to \(A\) and \(1/v\) to \(B\) to get normal equations. We solved these for \(A\) and \(B\), and then used them to find the original \(u\) and \(v\).

🎯 Exam Tip: For equations with variables in the denominator (like \(uv\)), divide by the variable product to convert them into linear form using reciprocal substitutions like \(A = 1/u\) and \(B = 1/v\).

Question 14. \( \frac{1}{2x} + \frac{1}{y} = -1 \)
\( \frac{1}{x} + \frac{1}{2y} = 8 \)
Answer: The given equations are:
\( \frac{1}{2x} + \frac{1}{y} = -1 \) ...(i)
\( \frac{1}{x} + \frac{1}{2y} = 8 \) ...(ii)
Let \( A = \frac{1}{x} \) and \( B = \frac{1}{y} \). The equations become a linear system:
\( \frac{1}{2}A + B = -1 \) ...(iii)
\( A + \frac{1}{2}B = 8 \) ...(iv)
Multiply equation (iii) by 2 and equation (iv) by 2 to clear the fractions:
From (iii): \( A + 2B = -2 \) ...(v)
From (iv): \( 2A + B = 16 \) ...(vi)
To solve by elimination, multiply equation (vi) by 2:
\( 2(2A + B) = 2(16) \)
\( 4A + 2B = 32 \) ...(vii)
Subtract equation (v) from equation (vii):
\( (4A + 2B) - (A + 2B) = 32 - (-2) \)
\( 4A - A + 2B - 2B = 32 + 2 \)
\( 3A = 34 \)
\( A = \frac{34}{3} \)
Now, substitute \( A = \frac{34}{3} \) back into equation (vi):
\( 2 \left( \frac{34}{3} \right) + B = 16 \)
\( \frac{68}{3} + B = 16 \)
\( B = 16 - \frac{68}{3} \)
\( B = \frac{48 - 68}{3} \)
\( B = \frac{-20}{3} \)
Now, convert back to \(x\) and \(y\):
Since \( A = \frac{1}{x} \), \( x = \frac{1}{A} = \frac{1}{\frac{34}{3}} = \frac{3}{34} \)
Since \( B = \frac{1}{y} \), \( y = \frac{1}{B} = \frac{1}{\frac{-20}{3}} = \frac{-3}{20} \)
Thus, the required solution is \( x = \frac{3}{34} \) and \( y = -\frac{3}{20} \). This substitution method effectively transforms a complex system into a simpler, solvable one.
In simple words: We changed \(1/x\) to \(A\) and \(1/y\) to \(B\) to make the equations straight lines. Then we cleared any remaining fractions and used elimination to find \(A\) and \(B\). Finally, we converted back to find the original \(x\) and \(y\).

🎯 Exam Tip: Always introduce new variables for the reciprocal terms (\(1/x\), \(1/y\)) to convert non-linear equations into a solvable linear system. This is a standard technique.

Question 15. \( \frac{5}{x+y} - \frac{2}{x-y} = -1 \)
\( \frac{15}{x+y} + \frac{7}{x-y} = 10 \)
Answer: The given equations are:
\( \frac{5}{x+y} - \frac{2}{x-y} = -1 \) ...(i)
\( \frac{15}{x+y} + \frac{7}{x-y} = 10 \) ...(ii)
Let \( A = \frac{1}{x+y} \) and \( B = \frac{1}{x-y} \). The equations become a linear system:
\( 5A - 2B = -1 \) ...(iii)
\( 15A + 7B = 10 \) ...(iv)
To solve by elimination, multiply equation (iii) by 3:
\( 3(5A - 2B) = 3(-1) \)
\( 15A - 6B = -3 \) ...(v)
Subtract equation (v) from equation (iv):
\( (15A + 7B) - (15A - 6B) = 10 - (-3) \)
\( 15A - 15A + 7B + 6B = 10 + 3 \)
\( 13B = 13 \)
\( B = \frac{13}{13} \)
\( B = 1 \)
Now, substitute \( B = 1 \) back into equation (iii):
\( 5A - 2(1) = -1 \)
\( 5A - 2 = -1 \)
\( 5A = -1 + 2 \)
\( 5A = 1 \)
\( A = \frac{1}{5} \)
Now, convert back to \(x+y\) and \(x-y\):
Since \( A = \frac{1}{x+y} \), \( \frac{1}{x+y} = \frac{1}{5} \)
\( x+y = 5 \) ...(vi)
Since \( B = \frac{1}{x-y} \), \( \frac{1}{x-y} = 1 \)
\( x-y = 1 \) ...(vii)
Now, we have a new system of linear equations in terms of \(x\) and \(y\). Add equation (vi) and equation (vii):
\( (x+y) + (x-y) = 5 + 1 \)
\( x+x+y-y = 6 \)
\( 2x = 6 \)
\( x = \frac{6}{2} \)
\( x = 3 \)
Substitute \( x = 3 \) back into equation (vi):
\( 3 + y = 5 \)
\( y = 5 - 3 \)
\( y = 2 \)
Hence, the required solution is \(x = 3\) and \(y = 2\). This problem involves a two-step substitution, first for the compound terms and then for the individual variables.
In simple words: We replaced the complex parts \(1/(x+y)\) and \(1/(x-y)\) with new letters \(A\) and \(B\). We solved for \(A\) and \(B\) using elimination. Then, we used \(A\) and \(B\) to create two simpler equations for \(x\) and \(y\), which we then solved to find \(x\) and \(y\).

🎯 Exam Tip: For systems involving expressions like \( (x+y) \) or \( (x-y) \) in the denominator, substitute them with new variables (e.g., A and B) to transform the system into a linear form. This makes it much easier to solve.