OP Malhotra Class 9 Maths Solutions Chapter 5 Simultaneous Linear Equations in Two Variable Exercise 5 (A)

Question 1. If (5, k) is a solution of the equation \( 2x + y - 7 = 0 \) find the value of k.
Answer: The given equation is \( 2x + y - 7 = 0 \).
Since \( (5, k) \) is a solution to this equation, we can substitute \( x = 5 \) and \( y = k \) into the equation.
\( 2(5) + k - 7 = 0 \)
\( 10 + k - 7 = 0 \)
\( 3 + k = 0 \)
\( \implies k = -3 \)
So, the value of \( k \) is -3. This means the point \( (5, -3) \) lies on the line represented by the equation.
In simple words: When you put \( x = 5 \) and \( y = k \) into the equation, it makes the equation true. Just solve for \( k \) after putting in \( x = 5 \).

🎯 Exam Tip: When a point is a solution to an equation, its coordinates can be directly substituted into the equation to find any unknown values.

Solve the following simultaneous equations:

Question 2. \( x + y = 5, x - y = 3 \)
Answer: We have the following system of equations:
\( x + y = 5 \) ... (i)
\( x - y = 3 \) ... (ii)
We can solve this system by adding the two equations together. Adding them will eliminate the \( y \) term.
Adding (i) and (ii):
\( (x + y) + (x - y) = 5 + 3 \)
\( 2x = 8 \)
\( \implies x = \frac{8}{2} \)
\( \implies x = 4 \)
Now, substitute the value of \( x = 4 \) into equation (i):
\( 4 + y = 5 \)
\( \implies y = 5 - 4 \)
\( \implies y = 1 \)
Thus, the solution to the simultaneous equations is \( x = 4 \) and \( y = 1 \). This pair of values satisfies both equations at the same time.
In simple words: Add the two equations together to get rid of \( y \). Solve for \( x \). Then put \( x \) back into one of the original equations to find \( y \).

🎯 Exam Tip: For equations like these, adding or subtracting can quickly eliminate one variable. Always check your answer by substituting both \( x \) and \( y \) values into the other original equation.

Question 3. \( y = 2x - 6, y = 0 \)
Answer: We have the following system of equations:
\( y = 2x - 6 \) ... (i)
\( y = 0 \) ... (ii)
Since we know that \( y = 0 \) from equation (ii), we can directly substitute this value into equation (i). This is a very straightforward substitution.
Substitute \( y = 0 \) into equation (i):
\( 0 = 2x - 6 \)
Now, we need to solve for \( x \):
\( 6 = 2x \)
\( \implies x = \frac{6}{2} \)
\( \implies x = 3 \)
Therefore, the solution to this system of equations is \( x = 3 \) and \( y = 0 \). This means the line \( y = 2x - 6 \) crosses the x-axis at \( x = 3 \).
In simple words: Since \( y \) is already given as 0, just put 0 in place of \( y \) in the first equation. Then, solve to find what \( x \) is.

🎯 Exam Tip: When one variable is already expressed in terms of a constant or a simple expression, direct substitution is the most efficient method.

Question 4. \( p = 2q - 1, q = 5 - 3p \)
Answer: We have the following system of equations:
\( p = 2q - 1 \) ... (i)
\( q = 5 - 3p \) ... (ii)
We will use the substitution method here. Substitute the expression for \( p \) from equation (i) into equation (ii). This will give us an equation with only one variable, \( q \).
Substitute \( p = 2q - 1 \) into equation (ii):
\( q = 5 - 3(2q - 1) \)
Now, simplify and solve for \( q \):
\( q = 5 - 6q + 3 \)
\( q = 8 - 6q \)
\( q + 6q = 8 \)
\( 7q = 8 \)
\( \implies q = \frac{8}{7} \)
Now that we have the value of \( q \), substitute it back into equation (i) to find \( p \):
\( p = 2\left(\frac{8}{7}\right) - 1 \)
\( p = \frac{16}{7} - 1 \)
\( p = \frac{16 - 7}{7} \)
\( p = \frac{9}{7} \)
So, the solution is \( p = \frac{9}{7} \) and \( q = \frac{8}{7} \). These values satisfy both equations simultaneously.
In simple words: Take the value of \( p \) from the first equation and put it into the second equation. Solve for \( q \). Then use that \( q \) value in the first equation to find \( p \).

🎯 Exam Tip: Substitution is a good method when one variable is already isolated or easily isolated in one of the equations. Be careful with distributing negative signs, like \( -3(2q-1) \).

Question 5. \( 9x + 4y = 5, 4x - 5y = 9 \)
Answer: We have the following system of equations:
\( 9x + 4y = 5 \) ... (i)
\( 4x - 5y = 9 \) ... (ii)
We will use the elimination method. To eliminate \( y \), we need to make the coefficients of \( y \) equal and opposite. Multiply equation (i) by 5 and equation (ii) by 4.
Multiplying equation (i) by 5:
\( 5(9x + 4y) = 5(5) \)
\( \implies 45x + 20y = 25 \) ... (iii)
Multiplying equation (ii) by 4:
\( 4(4x - 5y) = 4(9) \)
\( \implies 16x - 20y = 36 \) ... (iv)
Now, add equation (iii) and equation (iv) to eliminate \( y \):
\( (45x + 20y) + (16x - 20y) = 25 + 36 \)
\( 45x + 16x = 61 \)
\( 61x = 61 \)
\( \implies x = \frac{61}{61} \)
\( \implies x = 1 \)
Now, substitute the value of \( x = 1 \) into equation (i) to find \( y \):
\( 9(1) + 4y = 5 \)
\( 9 + 4y = 5 \)
\( 4y = 5 - 9 \)
\( 4y = -4 \)
\( \implies y = \frac{-4}{4} \)
\( \implies y = -1 \)
So, the solution to the simultaneous equations is \( x = 1 \) and \( y = -1 \). The elimination method is very effective when coefficients can be easily made the same.
In simple words: Make the \( y \) numbers in both equations the same size but with opposite signs by multiplying the whole equations. Add them together to get rid of \( y \). Solve for \( x \). Then put \( x \) back into one original equation to find \( y \).

🎯 Exam Tip: The key to the elimination method is choosing which variable to eliminate and finding the least common multiple of its coefficients to simplify calculations.

Question 6. \( v \perp 2v - 5, 2v - v - 5 \)
Answer: [No solution provided for this question in the source.]

🎯 Exam Tip: Always double-check that you have extracted the question text exactly as it appears in the source, as sometimes OCR errors can lead to unclear questions.

Question 7. \( 3x - 7y + 10 = 0, y - 2x - 3 = 0 \)
Answer: We have the following system of equations:
From \( 3x - 7y + 10 = 0 \), we get \( 3x - 7y = -10 \) ... (i)
From \( y - 2x - 3 = 0 \), we get \( y = 2x + 3 \) ... (ii)
We will use the substitution method. Substitute the expression for \( y \) from equation (ii) into equation (i). This will help us solve for \( x \) first.
Substitute \( y = 2x + 3 \) into equation (i):
\( 3x - 7(2x + 3) = -10 \)
Now, simplify and solve for \( x \):
\( 3x - 14x - 21 = -10 \)
\( -11x - 21 = -10 \)
\( -11x = -10 + 21 \)
\( -11x = 11 \)
\( \implies x = \frac{11}{-11} \)
\( \implies x = -1 \)
Now that we have \( x = -1 \), substitute it back into equation (ii) to find \( y \):
\( y = 2(-1) + 3 \)
\( y = -2 + 3 \)
\( y = 1 \)
So, the solution is \( x = -1 \) and \( y = 1 \). This method is clear and helps manage the signs carefully.
In simple words: Rearrange the second equation to get \( y \) alone. Put that expression for \( y \) into the first equation. Solve for \( x \), then use that \( x \) to find \( y \).

🎯 Exam Tip: Always rearrange equations into a standard form (e.g., \( Ax + By = C \)) before starting to solve, as it reduces errors. Be careful when distributing negative signs during substitution.

Question 8. \( 20u - 30v = 13, 10v - 10u = -5 \)
Answer: We have the following system of equations:
\( 20u - 30v = 13 \) ... (i)
Rearranging the second equation \( 10v - 10u = -5 \) to a standard form \( Au + Bv = C \):
\( -10u + 10v = -5 \)
Multiplying by -1 to make \( u \) positive: \( 10u - 10v = 5 \) ... (ii)
We will use the elimination method. To eliminate \( u \), multiply equation (ii) by 2.
Multiplying equation (ii) by 2:
\( 2(10u - 10v) = 2(5) \)
\( \implies 20u - 20v = 10 \) ... (iii)
Now, subtract equation (iii) from equation (i):
\( (20u - 30v) - (20u - 20v) = 13 - 10 \)
\( 20u - 30v - 20u + 20v = 3 \)
\( -10v = 3 \)
\( \implies v = -\frac{3}{10} \)
Now, substitute the value of \( v = -\frac{3}{10} \) into equation (ii) to find \( u \):
\( 10u - 10\left(-\frac{3}{10}\right) = 5 \)
\( 10u + 3 = 5 \)
\( 10u = 5 - 3 \)
\( 10u = 2 \)
\( \implies u = \frac{2}{10} \)
\( \implies u = \frac{1}{5} \)
So, the solution is \( u = \frac{1}{5} \) and \( v = -\frac{3}{10} \). Always try to reorder equations for consistency before solving.
In simple words: First, rewrite the second equation so \( u \) comes before \( v \). Then, multiply the second equation by 2. Subtract this new equation from the first one to find \( v \). Finally, use \( v \) in the second original equation to find \( u \).

🎯 Exam Tip: Always align variables in the same order (e.g., \( u \) then \( v \)) across all equations before performing addition or subtraction, as this helps prevent calculation mistakes.

Question 9. \( 2x - 3y = 1.3, y - x = 0.5 \)
Answer: We have the following system of equations:
\( 2x - 3y = 1.3 \) ... (i)
\( y - x = 0.5 \) ... (ii)
From equation (ii), we can easily express \( y \) in terms of \( x \):
\( y = 0.5 + x \)
Now, substitute this expression for \( y \) into equation (i). This is the substitution method.
Substitute \( y = 0.5 + x \) into equation (i):
\( 2x - 3(0.5 + x) = 1.3 \)
Simplify and solve for \( x \):
\( 2x - 1.5 - 3x = 1.3 \)
\( -x - 1.5 = 1.3 \)
\( -x = 1.3 + 1.5 \)
\( -x = 2.8 \)
\( \implies x = -2.8 \)
Now, substitute the value of \( x = -2.8 \) back into the rearranged equation for \( y \):
\( y = 0.5 + (-2.8) \)
\( y = 0.5 - 2.8 \)
\( y = -2.3 \)
So, the solution is \( x = -2.8 \) and \( y = -2.3 \). Working with decimals is similar to integers, just be careful with placement.
In simple words: Get \( y \) by itself from the second equation. Put that into the first equation to find \( x \). Then use that \( x \) to find \( y \).

🎯 Exam Tip: When dealing with decimals, it's often helpful to multiply the entire equation by a power of 10 to clear the decimals, turning them into integers for easier calculation. However, direct decimal calculations are also valid.

Question 10. \( 11x + 15y + 23 = 0, 7x - 2y - 20 = 0 \)
Answer: We have the following system of equations:
\( 11x + 15y = -23 \) ... (i)
\( 7x - 2y = 20 \) ... (ii)
We will use the elimination method. To eliminate \( y \), we need the coefficients of \( y \) to be the same magnitude but opposite signs. Multiply equation (i) by 2 and equation (ii) by 15.
Multiplying equation (i) by 2:
\( 2(11x + 15y) = 2(-23) \)
\( \implies 22x + 30y = -46 \) ... (iii)
Multiplying equation (ii) by 15:
\( 15(7x - 2y) = 15(20) \)
\( \implies 105x - 30y = 300 \) ... (iv)
Now, add equation (iii) and equation (iv) to eliminate \( y \):
\( (22x + 30y) + (105x - 30y) = -46 + 300 \)
\( 22x + 105x = 254 \)
\( 127x = 254 \)
\( \implies x = \frac{254}{127} \)
\( \implies x = 2 \)
Now, substitute the value of \( x = 2 \) into equation (ii) to find \( y \):
\( 7(2) - 2y = 20 \)
\( 14 - 2y = 20 \)
\( -2y = 20 - 14 \)
\( -2y = 6 \)
\( \implies y = \frac{6}{-2} \)
\( \implies y = -3 \)
So, the solution to the simultaneous equations is \( x = 2 \) and \( y = -3 \). This elimination process is robust for any coefficients.
In simple words: Multiply the first equation by 2 and the second equation by 15 to make the \( y \) numbers the same but opposite. Add the two new equations to find \( x \). Then put \( x \) back into one of the original equations to find \( y \).

🎯 Exam Tip: Always make sure to bring the constant term to the right side of the equation (i.e., \( Ax + By = C \)) before starting the elimination or substitution process.

Question 11. \( 3 - (x - 5) = y + 2, 2(x + y) = 4 - 3y \)
Answer: First, let's simplify both equations into the standard form \( Ax + By = C \).
For the first equation: \( 3 - (x - 5) = y + 2 \)
\( 3 - x + 5 = y + 2 \)
\( 8 - x = y + 2 \)
\( -x - y = 2 - 8 \)
\( -x - y = -6 \)
\( \implies x + y = 6 \) ... (i)
For the second equation: \( 2(x + y) = 4 - 3y \)
\( 2x + 2y = 4 - 3y \)
\( 2x + 2y + 3y = 4 \)
\( \implies 2x + 5y = 4 \) ... (ii)
Now we have a simplified system:
\( x + y = 6 \)
\( 2x + 5y = 4 \)
From equation (i), we can express \( x \) in terms of \( y \): \( x = 6 - y \).
Substitute \( x = 6 - y \) into equation (ii):
\( 2(6 - y) + 5y = 4 \)
\( 12 - 2y + 5y = 4 \)
\( 12 + 3y = 4 \)
\( 3y = 4 - 12 \)
\( 3y = -8 \)
\( \implies y = -\frac{8}{3} \)
Now, substitute the value of \( y = -\frac{8}{3} \) back into \( x = 6 - y \):
\( x = 6 - \left(-\frac{8}{3}\right) \)
\( x = 6 + \frac{8}{3} \)
\( x = \frac{18}{3} + \frac{8}{3} \)
\( x = \frac{18 + 8}{3} \)
\( x = \frac{26}{3} \)
So, the solution is \( x = \frac{26}{3} \) and \( y = -\frac{8}{3} \). Simplifying first makes the rest of the steps much easier.
In simple words: First, clean up both equations to make them simple like "number \( x \) plus number \( y \) equals a constant". Then, get \( x \) alone from the first simple equation and put it into the second one to find \( y \). Finally, use that \( y \) to find \( x \).

🎯 Exam Tip: Always simplify and rearrange the given equations into the standard linear form \( Ax + By = C \) before attempting to solve them. This reduces complexity and helps avoid mistakes.

Question 12.
(a) \( \frac{x}{2}+\frac{y}{4}=6, \frac{x}{5}-\frac{y}{2}=0 \)
(b) \( \frac{x}{4}-3=\frac{y}{6}, \frac{1}{2}x-y=-2 \)
Answer:
(a) We have the system:
\( \frac{x}{2} + \frac{y}{4} = 6 \) ... (i)
\( \frac{x}{5} - \frac{y}{2} = 0 \) ... (ii)
To eliminate fractions, multiply equation (i) by the LCM of 2 and 4 (which is 4):
\( 4\left(\frac{x}{2}\right) + 4\left(\frac{y}{4}\right) = 4(6) \)
\( \implies 2x + y = 24 \) ... (iii)
Multiply equation (ii) by the LCM of 5 and 2 (which is 10):
\( 10\left(\frac{x}{5}\right) - 10\left(\frac{y}{2}\right) = 10(0) \)
\( \implies 2x - 5y = 0 \) ... (iv)
Now we have a simpler system:
\( 2x + y = 24 \)
\( 2x - 5y = 0 \)
Subtract equation (iv) from equation (iii) to eliminate \( x \):
\( (2x + y) - (2x - 5y) = 24 - 0 \)
\( 2x + y - 2x + 5y = 24 \)
\( 6y = 24 \)
\( \implies y = \frac{24}{6} \)
\( \implies y = 4 \)
Substitute \( y = 4 \) into equation (iii):
\( 2x + 4 = 24 \)
\( 2x = 24 - 4 \)
\( 2x = 20 \)
\( \implies x = \frac{20}{2} \)
\( \implies x = 10 \)
So, for part (a), \( x = 10 \) and \( y = 4 \).

(b) We have the system:
\( \frac{x}{4} - 3 = \frac{y}{6} \)
\( \frac{1}{2}x - y = -2 \)
First, rearrange the first equation to standard form. Multiply by LCM of 4 and 6 (which is 12):
\( 12\left(\frac{x}{4}\right) - 12(3) = 12\left(\frac{y}{6}\right) \)
\( 3x - 36 = 2y \)
\( \implies 3x - 2y = 36 \) ... (v)
For the second equation, rearrange to standard form:
\( \frac{1}{2}x - y = -2 \)
\( \implies x - 2y = -4 \) ... (vi) (Multiplying by 2)
Now we have a simpler system:
\( 3x - 2y = 36 \)
\( x - 2y = -4 \)
Subtract equation (vi) from equation (v) to eliminate \( y \):
\( (3x - 2y) - (x - 2y) = 36 - (-4) \)
\( 3x - 2y - x + 2y = 36 + 4 \)
\( 2x = 40 \)
\( \implies x = \frac{40}{2} \)
\( \implies x = 20 \)
Substitute \( x = 20 \) into equation (vi):
\( 20 - 2y = -4 \)
\( -2y = -4 - 20 \)
\( -2y = -24 \)
\( \implies y = \frac{-24}{-2} \)
\( \implies y = 12 \)
So, for part (b), \( x = 20 \) and \( y = 12 \). Always clear fractions first to make solving easier and prevent calculation errors.
In simple words: For both parts, first get rid of the fractions by multiplying each equation by a number that clears all the denominators. This makes the equations much simpler to work with. Then, solve the simple equations using either adding, subtracting, or substituting.

🎯 Exam Tip: Always clear denominators in fractional equations by multiplying by the Least Common Multiple (LCM) of all denominators. This converts them into integer-coefficient equations, simplifying calculations significantly.

Question 13.
(a) \( \frac{7}{x}+\frac{8}{y}=2, \frac{2}{x}+\frac{12}{y} = 20 \)
(b) \( \frac{1}{7x} + \frac{1}{6y}=3, \frac{1}{2x}-\frac{1}{3y} = 5 \)
Answer:
(a) We have the system:
\( \frac{7}{x} + \frac{8}{y} = 2 \) ... (i)
\( \frac{2}{x} + \frac{12}{y} = 20 \) ... (ii)
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). The system becomes:
\( 7u + 8v = 2 \) ... (iii)
\( 2u + 12v = 20 \) ... (iv)
We will use the elimination method. To eliminate \( u \), multiply equation (iii) by 2 and equation (iv) by 7.
Multiplying equation (iii) by 2:
\( 2(7u + 8v) = 2(2) \)
\( \implies 14u + 16v = 4 \) ... (v)
Multiplying equation (iv) by 7:
\( 7(2u + 12v) = 7(20) \)
\( \implies 14u + 84v = 140 \) ... (vi)
Subtract equation (v) from equation (vi):
\( (14u + 84v) - (14u + 16v) = 140 - 4 \)
\( 68v = 136 \)
\( \implies v = \frac{136}{68} \)
\( \implies v = 2 \)
Now, substitute \( v = 2 \) into equation (iii):
\( 7u + 8(2) = 2 \)
\( 7u + 16 = 2 \)
\( 7u = 2 - 16 \)
\( 7u = -14 \)
\( \implies u = \frac{-14}{7} \)
\( \implies u = -2 \)
Now, we find \( x \) and \( y \) using \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \):
\( u = -2 \implies \frac{1}{x} = -2 \implies x = -\frac{1}{2} \)
\( v = 2 \implies \frac{1}{y} = 2 \implies y = \frac{1}{2} \)
So, for part (a), \( x = -\frac{1}{2} \) and \( y = \frac{1}{2} \).

(b) We have the system:
\( \frac{1}{7x} + \frac{1}{6y} = 3 \) ... (vii)
\( \frac{1}{2x} - \frac{1}{3y} = 5 \) ... (viii)
Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). The system becomes:
\( \frac{1}{7}u + \frac{1}{6}v = 3 \) ... (ix)
\( \frac{1}{2}u - \frac{1}{3}v = 5 \) ... (x)
To clear fractions in (ix), multiply by LCM of 7 and 6 (which is 42):
\( 42\left(\frac{1}{7}u\right) + 42\left(\frac{1}{6}v\right) = 42(3) \)
\( 6u + 7v = 126 \) ... (xi)
To clear fractions in (x), multiply by LCM of 2 and 3 (which is 6):
\( 6\left(\frac{1}{2}u\right) - 6\left(\frac{1}{3}v\right) = 6(5) \)
\( 3u - 2v = 30 \) ... (xii)
Now we have a simpler system:
\( 6u + 7v = 126 \)
\( 3u - 2v = 30 \)
To eliminate \( u \), multiply equation (xii) by 2:
\( 2(3u - 2v) = 2(30) \)
\( \implies 6u - 4v = 60 \) ... (xiii)
Subtract equation (xiii) from equation (xi):
\( (6u + 7v) - (6u - 4v) = 126 - 60 \)
\( 6u + 7v - 6u + 4v = 66 \)
\( 11v = 66 \)
\( \implies v = \frac{66}{11} \)
\( \implies v = 6 \)
Now, substitute \( v = 6 \) into equation (xii):
\( 3u - 2(6) = 30 \)
\( 3u - 12 = 30 \)
\( 3u = 30 + 12 \)
\( 3u = 42 \)
\( \implies u = \frac{42}{3} \)
\( \implies u = 14 \)
Finally, find \( x \) and \( y \):
\( u = 14 \implies \frac{1}{x} = 14 \implies x = \frac{1}{14} \)
\( v = 6 \implies \frac{1}{y} = 6 \implies y = \frac{1}{6} \)
So, for part (b), \( x = \frac{1}{14} \) and \( y = \frac{1}{6} \). This substitution method is crucial for rational equations.
In simple words: For equations where \( x \) and \( y \) are in the denominator, first pretend that \( 1/x \) and \( 1/y \) are new simple letters (like \( u \) and \( v \)). Solve for these new letters using your usual methods. Once you find \( u \) and \( v \), just flip them over to get the real \( x \) and \( y \).

🎯 Exam Tip: When variables appear in the denominator, introduce new variables (e.g., \( u = 1/x, v = 1/y \)) to transform the equations into a linear system, which is much easier to solve. Remember to substitute back to find the original variables at the end.

Question 14.
(a) \( 65x - 33y = 97, 33x - 65y = 1 \)
(b) \( 23x + 31y = 77, 31x + 23y = 85 \)
Answer:
(a) We have the system:
\( 65x - 33y = 97 \) ... (i)
\( 33x - 65y = 1 \) ... (ii)
This is a case of symmetric coefficients (where coefficients of \( x \) and \( y \) swap between equations). We solve by adding and subtracting the equations.
Adding (i) and (ii):
\( (65x - 33y) + (33x - 65y) = 97 + 1 \)
\( 98x - 98y = 98 \)
Divide by 98:
\( x - y = 1 \) ... (iii)
Subtracting (ii) from (i):
\( (65x - 33y) - (33x - 65y) = 97 - 1 \)
\( 65x - 33y - 33x + 65y = 96 \)
\( 32x + 32y = 96 \)
Divide by 32:
\( x + y = 3 \) ... (iv)
Now we have a simpler system:
\( x - y = 1 \)
\( x + y = 3 \)
Add (iii) and (iv):
\( (x - y) + (x + y) = 1 + 3 \)
\( 2x = 4 \)
\( \implies x = \frac{4}{2} \)
\( \implies x = 2 \)
Subtract (iii) from (iv):
\( (x + y) - (x - y) = 3 - 1 \)
\( x + y - x + y = 2 \)
\( 2y = 2 \)
\( \implies y = \frac{2}{2} \)
\( \implies y = 1 \)
So, for part (a), \( x = 2 \) and \( y = 1 \).

(b) We have the system:
\( 23x + 31y = 77 \) ... (v)
\( 31x + 23y = 85 \) ... (vi)
This is another case of symmetric coefficients.
Adding (v) and (vi):
\( (23x + 31y) + (31x + 23y) = 77 + 85 \)
\( 54x + 54y = 162 \)
Divide by 54:
\( x + y = 3 \) ... (vii)
Subtracting (v) from (vi):
\( (31x + 23y) - (23x + 31y) = 85 - 77 \)
\( 31x + 23y - 23x - 31y = 8 \)
\( 8x - 8y = 8 \)
Divide by 8:
\( x - y = 1 \) ... (viii)
Now we have a simpler system:
\( x + y = 3 \)
\( x - y = 1 \)
Add (vii) and (viii):
\( (x + y) + (x - y) = 3 + 1 \)
\( 2x = 4 \)
\( \implies x = \frac{4}{2} \)
\( \implies x = 2 \)
Subtract (viii) from (vii):
\( (x + y) - (x - y) = 3 - 1 \)
\( x + y - x + y = 2 \)
\( 2y = 2 \)
\( \implies y = \frac{2}{2} \)
\( \implies y = 1 \)
So, for part (b), \( x = 2 \) and \( y = 1 \). This method is a shortcut for these special types of systems.
In simple words: For equations where the numbers in front of \( x \) and \( y \) swap places, first add the two equations together, then subtract them. This will give you two much simpler equations. Solve these two new simple equations to find \( x \) and \( y \).

🎯 Exam Tip: For symmetric coefficient equations (e.g., \( ax + by = c \) and \( bx + ay = d \)), adding and subtracting the original equations is the most efficient method to get two new simpler linear equations, which can then be solved easily.

Question 15.
(a) \( 4x + \frac { 6 }{ y } = 15, 6x - \frac { 8 }{ y } = 14; y \ne 0 \)
Find p, if \( y = px - 2 \)
(b) \( \frac { 4 }{ x } + 5y = 7, \frac { 3 }{ x } + 4y = 5; x \ne 0 \)
Answer:
(a) We have the system:
\( 4x + \frac{6}{y} = 15 \) ... (i)
\( 6x - \frac{8}{y} = 14 \) ... (ii)
Let \( v = \frac{1}{y} \). The system becomes:
\( 4x + 6v = 15 \) ... (iii)
\( 6x - 8v = 14 \) ... (iv)
To eliminate \( v \), multiply equation (iii) by 4 and equation (iv) by 3.
Multiplying equation (iii) by 4:
\( 4(4x + 6v) = 4(15) \)
\( \implies 16x + 24v = 60 \) ... (v)
Multiplying equation (iv) by 3:
\( 3(6x - 8v) = 3(14) \)
\( \implies 18x - 24v = 42 \) ... (vi)
Add equation (v) and equation (vi) to eliminate \( v \):
\( (16x + 24v) + (18x - 24v) = 60 + 42 \)
\( 34x = 102 \)
\( \implies x = \frac{102}{34} \)
\( \implies x = 3 \)
Now, substitute \( x = 3 \) into equation (i) to find \( y \):
\( 4(3) + \frac{6}{y} = 15 \)
\( 12 + \frac{6}{y} = 15 \)
\( \frac{6}{y} = 15 - 12 \)
\( \frac{6}{y} = 3 \)
\( \implies 3y = 6 \)
\( \implies y = \frac{6}{3} \)
\( \implies y = 2 \)
So, \( x = 3 \) and \( y = 2 \).
Now, we need to find \( p \) using the relation \( y = px - 2 \):
Substitute \( x = 3 \) and \( y = 2 \):
\( 2 = p(3) - 2 \)
\( 2 = 3p - 2 \)
\( 2 + 2 = 3p \)
\( 4 = 3p \)
\( \implies p = \frac{4}{3} \)
Thus, the value of \( p \) is \( \frac{4}{3} \).

(b) We have the system:
\( \frac{4}{x} + 5y = 7 \) ... (vii)
\( \frac{3}{x} + 4y = 5 \) ... (viii)
Let \( u = \frac{1}{x} \). The system becomes:
\( 4u + 5y = 7 \) ... (ix)
\( 3u + 4y = 5 \) ... (x)
To eliminate \( u \), multiply equation (ix) by 3 and equation (x) by 4.
Multiplying equation (ix) by 3:
\( 3(4u + 5y) = 3(7) \)
\( \implies 12u + 15y = 21 \) ... (xi)
Multiplying equation (x) by 4:
\( 4(3u + 4y) = 4(5) \)
\( \implies 12u + 16y = 20 \) ... (xii)
Subtract equation (xi) from equation (xii):
\( (12u + 16y) - (12u + 15y) = 20 - 21 \)
\( 12u + 16y - 12u - 15y = -1 \)
\( y = -1 \)
Now, substitute \( y = -1 \) into equation (x) to find \( u \):
\( 3u + 4(-1) = 5 \)
\( 3u - 4 = 5 \)
\( 3u = 5 + 4 \)
\( 3u = 9 \)
\( \implies u = \frac{9}{3} \)
\( \implies u = 3 \)
Finally, find \( x \) using \( u = \frac{1}{x} \):
\( u = 3 \implies \frac{1}{x} = 3 \implies x = \frac{1}{3} \)
So, for part (b), \( x = \frac{1}{3} \) and \( y = -1 \). This approach effectively turns complex equations into simpler linear forms.
In simple words: For part (a), solve the first two equations for \( x \) and \( y \) by making \( 1/y \) a new letter, then substitute \( x \) and \( y \) into \( y = px - 2 \) to find \( p \). For part (b), make \( 1/x \) a new letter, then solve the new equations for that letter and \( y \), and finally find the real \( x \).

🎯 Exam Tip: For equations with variables in the denominator, simplify by using substitution (e.g., \( u = 1/x \), \( v = 1/y \)) to convert them into standard linear form. Always remember to solve for the original variables at the very end.

Question 16.
(a) \( \frac{6}{x+y}=\frac{7}{x-y}+3, \frac{1}{2(x+y)} = \frac{1}{3(x-y)}, \text{ where } x + y \ne 0, x - y \ne 0 \).
(b) \( \frac{44}{x+y}+\frac{30}{x-y}=10, \frac{55}{x+y}+\frac{40}{x-y} = 13, \text{ where } x + y \ne 0, x - y \ne 0 \).
Answer:
(a) We have the system:
\( \frac{6}{x+y} = \frac{7}{x-y} + 3 \)
\( \frac{1}{2(x+y)} = \frac{1}{3(x-y)} \)
Let \( a = x+y \) and \( b = x-y \). The equations become:
\( \frac{6}{a} = \frac{7}{b} + 3 \)
\( \frac{1}{2a} = \frac{1}{3b} \)
From the first transformed equation, rearrange to get standard form:
\( \frac{6}{a} - \frac{7}{b} = 3 \) ... (i)
From the second transformed equation, cross-multiply:
\( 3b = 2a \)
\( \implies a = \frac{3}{2}b \) ... (ii)
Now substitute \( a = \frac{3}{2}b \) into equation (i):
\( \frac{6}{\frac{3}{2}b} - \frac{7}{b} = 3 \)
\( \frac{12}{3b} - \frac{7}{b} = 3 \)
\( \frac{4}{b} - \frac{7}{b} = 3 \)
\( -\frac{3}{b} = 3 \)
\( \implies 3b = -3 \)
\( \implies b = -1 \)
Now substitute \( b = -1 \) back into \( a = \frac{3}{2}b \):
\( a = \frac{3}{2}(-1) \)
\( \implies a = -\frac{3}{2} \)
Now we have \( x+y = a \) and \( x-y = b \). So:
\( x+y = -\frac{3}{2} \) ... (iii)
\( x-y = -1 \) ... (iv)
Add (iii) and (iv):
\( (x+y) + (x-y) = -\frac{3}{2} + (-1) \)
\( 2x = -\frac{3}{2} - \frac{2}{2} \)
\( 2x = -\frac{5}{2} \)
\( \implies x = -\frac{5}{4} \)
Subtract (iv) from (iii):
\( (x+y) - (x-y) = -\frac{3}{2} - (-1) \)
\( 2y = -\frac{3}{2} + 1 \)
\( 2y = -\frac{3}{2} + \frac{2}{2} \)
\( 2y = -\frac{1}{2} \)
\( \implies y = -\frac{1}{4} \)
So, for part (a), \( x = -\frac{5}{4} \) and \( y = -\frac{1}{4} \).

(b) We have the system:
\( \frac{44}{x+y} + \frac{30}{x-y} = 10 \)
\( \frac{55}{x+y} + \frac{40}{x-y} = 13 \)
Let \( u = \frac{1}{x+y} \) and \( v = \frac{1}{x-y} \). The system becomes:
\( 44u + 30v = 10 \) ... (v)
\( 55u + 40v = 13 \) ... (vi)
To eliminate \( v \), multiply equation (v) by 4 and equation (vi) by 3.
Multiplying equation (v) by 4:
\( 4(44u + 30v) = 4(10) \)
\( \implies 176u + 120v = 40 \) ... (vii)
Multiplying equation (vi) by 3:
\( 3(55u + 40v) = 3(13) \)
\( \implies 165u + 120v = 39 \) ... (viii)
Subtract equation (viii) from equation (vii):
\( (176u + 120v) - (165u + 120v) = 40 - 39 \)
\( 11u = 1 \)
\( \implies u = \frac{1}{11} \)
Now, substitute \( u = \frac{1}{11} \) into equation (v):
\( 44\left(\frac{1}{11}\right) + 30v = 10 \)
\( 4 + 30v = 10 \)
\( 30v = 10 - 4 \)
\( 30v = 6 \)
\( \implies v = \frac{6}{30} \)
\( \implies v = \frac{1}{5} \)
Now we have \( x+y = \frac{1}{u} \) and \( x-y = \frac{1}{v} \). So:
\( x+y = \frac{1}{1/11} = 11 \) ... (ix)
\( x-y = \frac{1}{1/5} = 5 \) ... (x)
Add (ix) and (x):
\( (x+y) + (x-y) = 11 + 5 \)
\( 2x = 16 \)
\( \implies x = \frac{16}{2} \)
\( \implies x = 8 \)
Subtract (x) from (ix):
\( (x+y) - (x-y) = 11 - 5 \)
\( 2y = 6 \)
\( \implies y = \frac{6}{2} \)
\( \implies y = 3 \)
So, for part (b), \( x = 8 \) and \( y = 3 \). This multi-step substitution is a powerful tool.
In simple words: For both parts, use new letters for \( x+y \) and \( x-y \) (or their reciprocals). Solve the new, simpler equations for these letters. Once you have their values, set up two more simple equations to solve for the original \( x \) and \( y \).

🎯 Exam Tip: For equations of the form \( \frac{A}{ax+by} + \frac{B}{cx+dy} = E \), always use substitution like \( u = \frac{1}{ax+by} \) and \( v = \frac{1}{cx+dy} \) to convert them into a linear system. Remember to perform a second substitution to find \( x \) and \( y \).

Question 17.
(i) \( 2 (3u - v) = 5uv, 2 (u + 3v) = 5uv \)
(ii) \( 3 (a + 3b) = 11ab, 3 (2a + b) = 7ab \)
Answer:
(i) We have the system:
\( 2(3u - v) = 5uv \)
\( 2(u + 3v) = 5uv \)
First, expand and rearrange each equation:
\( 6u - 2v = 5uv \)
\( 2u + 6v = 5uv \)
Since \( u, v \ne 0 \) (otherwise \( 5uv = 0 \), which leads to trivial solutions like \( u=0, v=0 \)), we can divide both equations by \( uv \). This transforms them into linear equations in terms of \( \frac{1}{v} \) and \( \frac{1}{u} \).
Dividing the first equation by \( uv \):
\( \frac{6u}{uv} - \frac{2v}{uv} = \frac{5uv}{uv} \)
\( \implies \frac{6}{v} - \frac{2}{u} = 5 \) ... (i)
Dividing the second equation by \( uv \):
\( \frac{2u}{uv} + \frac{6v}{uv} = \frac{5uv}{uv} \)
\( \implies \frac{2}{v} + \frac{6}{u} = 5 \) ... (ii)
Let \( x = \frac{1}{v} \) and \( y = \frac{1}{u} \). The system becomes:
\( 6x - 2y = 5 \) ... (iii)
\( 2x + 6y = 5 \) ... (iv)
To eliminate \( y \), multiply equation (iii) by 3:
\( 3(6x - 2y) = 3(5) \)
\( \implies 18x - 6y = 15 \) ... (v)
Add equation (v) and equation (iv):
\( (18x - 6y) + (2x + 6y) = 15 + 5 \)
\( 20x = 20 \)
\( \implies x = 1 \)
Now, substitute \( x = 1 \) into equation (iii):
\( 6(1) - 2y = 5 \)
\( 6 - 2y = 5 \)
\( -2y = 5 - 6 \)
\( -2y = -1 \)
\( \implies y = \frac{1}{2} \)
Now, find \( u \) and \( v \) using \( x = \frac{1}{v} \) and \( y = \frac{1}{u} \):
\( x = 1 \implies \frac{1}{v} = 1 \implies v = 1 \)
\( y = \frac{1}{2} \implies \frac{1}{u} = \frac{1}{2} \implies u = 2 \)
So, for part (i), \( u = 2 \) and \( v = 1 \).

(ii) We have the system:
\( 3(a + 3b) = 11ab \)
\( 3(2a + b) = 7ab \)
First, expand and rearrange each equation:
\( 3a + 9b = 11ab \)
\( 6a + 3b = 7ab \)
Since \( a, b \ne 0 \), we can divide both equations by \( ab \). This is a common technique for this type of problem.
Dividing the first equation by \( ab \):
\( \frac{3a}{ab} + \frac{9b}{ab} = \frac{11ab}{ab} \)
\( \implies \frac{3}{b} + \frac{9}{a} = 11 \) ... (vi)
Dividing the second equation by \( ab \):
\( \frac{6a}{ab} + \frac{3b}{ab} = \frac{7ab}{ab} \)
\( \implies \frac{6}{b} + \frac{3}{a} = 7 \) ... (vii)
Let \( x = \frac{1}{b} \) and \( y = \frac{1}{a} \). The system becomes:
\( 3x + 9y = 11 \) ... (viii)
\( 6x + 3y = 7 \) ... (ix)
To eliminate \( x \), multiply equation (viii) by 2:
\( 2(3x + 9y) = 2(11) \)
\( \implies 6x + 18y = 22 \) ... (x)
Subtract equation (ix) from equation (x):
\( (6x + 18y) - (6x + 3y) = 22 - 7 \)
\( 15y = 15 \)
\( \implies y = 1 \)
Now, substitute \( y = 1 \) into equation (ix):
\( 6x + 3(1) = 7 \)
\( 6x + 3 = 7 \)
\( 6x = 7 - 3 \)
\( 6x = 4 \)
\( \implies x = \frac{4}{6} \)
\( \implies x = \frac{2}{3} \)
Now, find \( a \) and \( b \) using \( x = \frac{1}{b} \) and \( y = \frac{1}{a} \):
\( x = \frac{2}{3} \implies \frac{1}{b} = \frac{2}{3} \implies b = \frac{3}{2} \)
\( y = 1 \implies \frac{1}{a} = 1 \implies a = 1 \)
So, for part (ii), \( a = 1 \) and \( b = \frac{3}{2} \). Dividing by the product of variables simplifies the problem significantly.
In simple words: For both parts, first expand the equations. Then, divide all parts of each equation by \( uv \) or \( ab \) (the product of the variables). This turns them into simpler equations with \( 1/u \) and \( 1/v \) (or \( 1/a \) and \( 1/b \)). Solve for these new terms, then flip them over to find the actual values of \( u, v \) (or \( a, b \)).

🎯 Exam Tip: When equations involve products of variables (e.g., \( uv \) or \( ab \)), dividing the entire equation by the product of the variables is a standard technique to transform them into linear equations solvable by substitution methods.

Question 18.
\[ \frac{3x-6}{4}+3-\frac{5y-4}{2}=\frac{5y}{2} \]
\[ \frac{y-x}{4}+\frac{x}{8}-\frac{7x-5y}{3}=y-2x \]
Answer:
First, let's simplify the first equation:
\[ \frac{3x-6}{4}+3-\frac{5y-4}{2}=\frac{5y}{2} \]
Multiply all terms by 4 to clear the denominators:
\[ (3x-6) + 12 - 2(5y-4) = 2(5y) \]
\[ 3x - 6 + 12 - 10y + 8 = 10y \]
\[ 3x + 14 = 20y \]
\[ 3x - 20y = -14 \quad \text{... (i)} \]
Now, let's simplify the second equation:
\[ \frac{y-x}{4}+\frac{x}{8}-\frac{7x-5y}{3}=y-2x \]
Multiply all terms by 24 (the L.C.M. of 4, 8, 3) to clear the denominators:
\[ 6(y-x) + 3x - 8(7x-5y) = 24(y-2x) \]
\[ 6y - 6x + 3x - 56x + 40y = 24y - 48x \]
Group similar terms:
\[ (-6x+3x-56x+48x) + (6y+40y-24y) = 0 \]
\[ -11x + 22y = 0 \]
\[ 22y = 11x \]
\[ x = \frac{22y}{11} \]
\[ x = 2y \quad \text{... (ii)} \]
Substitute the value of \( x \) from equation (ii) into equation (i):
\[ 3(2y) - 20y = -14 \]
\[ 6y - 20y = -14 \]
\[ -14y = -14 \]
\[ y = \frac{-14}{-14} \]
\[ y = 1 \]
Now substitute the value of \( y \) back into equation (ii) to find \( x \):
\[ x = 2(1) \]
\[ x = 2 \]
Therefore, the solution to the system of equations is \( x = 2 \) and \( y = 1 \). This method of substitution helps solve complex systems by reducing the number of variables in one equation.
In simple words: First, we make both equations simpler by getting rid of the fractions. Then, we use one equation to find out what \( x \) is in terms of \( y \). We put that into the other equation to find \( y \). Finally, we use the value of \( y \) to find \( x \).

🎯 Exam Tip: When dealing with fractional equations, always multiply by the L.C.M. of the denominators to eliminate fractions first. This simplifies the equations significantly before you attempt to solve them.

Question 19.
\[ x - y = 0.9, \frac{11}{2(x+y)} = 1 \]
Answer:
Let's write the first equation as:
\[ x - y = 0.9 \quad \text{... (i)} \]
Now, simplify the second equation:
\[ \frac{11}{2(x+y)} = 1 \]
Multiply both sides by \( 2(x+y) \):
\[ 11 = 2(x+y) \]
\[ x + y = \frac{11}{2} \]
\[ x + y = 5.5 \quad \text{... (ii)} \]
Now we have a system of two simple linear equations:
1. \( x - y = 0.9 \)
2. \( x + y = 5.5 \)
Add equation (i) and equation (ii):
\[ (x - y) + (x + y) = 0.9 + 5.5 \]
\[ x - y + x + y = 6.4 \]
\[ 2x = 6.4 \]
\[ x = \frac{6.4}{2} \]
\[ x = 3.2 \]
Substitute the value of \( x \) into equation (i):
\[ 3.2 - y = 0.9 \]
\[ -y = 0.9 - 3.2 \]
\[ -y = -2.3 \]
\[ y = 2.3 \]
So, the solution is \( x = 3.2 \) and \( y = 2.3 \). Converting decimals to fractions can sometimes make calculations easier, especially when dealing with multiplication or division.
In simple words: We have two equations. One is about \( x \) minus \( y \), and the other is about \( x \) plus \( y \). We add the two equations together to get rid of \( y \) and find \( x \). Then we use that \( x \) to find \( y \).

🎯 Exam Tip: For equations involving \( x-y \) and \( x+y \), adding or subtracting the two equations is usually the fastest way to solve for the individual variables.

Question 20.
\[ \frac{x}{2}+y=0.8, \frac{7}{x+\frac{y}{2}} = 10 \]
Answer:
First, simplify the given equations:
Equation 1:
\[ \frac{x}{2}+y=0.8 \]
Change the decimal to a fraction: \( 0.8 = \frac{8}{10} = \frac{4}{5} \)
\[ \frac{x}{2}+y=\frac{4}{5} \]
Multiply by 10 (L.C.M. of 2 and 5) to clear denominators:
\[ 5x + 10y = 8 \quad \text{... (i)} \]
Equation 2:
\[ \frac{7}{x+\frac{y}{2}} = 10 \]
Multiply both sides by \( x+\frac{y}{2} \):
\[ 7 = 10 \left( x+\frac{y}{2} \right) \]
\[ 7 = 10x + 5y \]
\[ 10x + 5y = 7 \quad \text{... (ii)} \]
Now we have a system of two linear equations:
1. \( 5x + 10y = 8 \)
2. \( 10x + 5y = 7 \)
To solve this, multiply equation (i) by 2:
\[ 2(5x + 10y) = 2(8) \]
\[ 10x + 20y = 16 \quad \text{... (iii)} \]
Subtract equation (ii) from equation (iii):
\[ (10x + 20y) - (10x + 5y) = 16 - 7 \]
\[ 10x + 20y - 10x - 5y = 9 \]
\[ 15y = 9 \]
\[ y = \frac{9}{15} \]
\[ y = \frac{3}{5} \]
Now substitute \( y = \frac{3}{5} \) into equation (i):
\[ 5x + 10\left(\frac{3}{5}\right) = 8 \]
\[ 5x + 2 \times 3 = 8 \]
\[ 5x + 6 = 8 \]
\[ 5x = 8 - 6 \]
\[ 5x = 2 \]
\[ x = \frac{2}{5} \]
So, the solution is \( x = \frac{2}{5} \) and \( y = \frac{3}{5} \). Always remember to simplify expressions before solving for variables, as it makes the process much smoother.
In simple words: First, we change the equations into a simpler form without fractions or decimals. Then we make the number in front of \( x \) (or \( y \)) the same in both equations. We subtract one equation from the other to find one variable, and then use that to find the other variable.

🎯 Exam Tip: When decimals are involved, convert them to fractions to avoid calculation errors, especially when multiplying or dividing. Always check your solution by plugging \( x \) and \( y \) back into the original equations.

Question 21. Solve the system of equations: \( 2x + y = 35 \), \( 3x + 4y = 65 \). Then find the value of \( \frac{x}{y} \).
Answer:
Given the equations:
1. \( 2x + y = 35 \quad \text{... (i)} \)
2. \( 3x + 4y = 65 \quad \text{... (ii)} \)
To eliminate \( y \), multiply equation (i) by 4:
\[ 4(2x + y) = 4(35) \]
\[ 8x + 4y = 140 \quad \text{... (iii)} \]
Now, subtract equation (ii) from equation (iii):
\[ (8x + 4y) - (3x + 4y) = 140 - 65 \]
\[ 8x + 4y - 3x - 4y = 75 \]
\[ 5x = 75 \]
\[ x = \frac{75}{5} \]
\[ x = 15 \]
Substitute the value of \( x \) into equation (i):
\[ 2(15) + y = 35 \]
\[ 30 + y = 35 \]
\[ y = 35 - 30 \]
\[ y = 5 \]
So, \( x = 15 \) and \( y = 5 \).
Now, we need to find the value of \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{15}{5} \]
\[ \frac{x}{y} = 3 \]
The solution provides values for \( x \) and \( y \) which can then be used to calculate other expressions, making simultaneous equations very versatile.
In simple words: We have two math puzzles to solve at once. We multiply the first puzzle by 4 to make the \( y \) part match the second puzzle. Then we subtract the puzzles to find \( x \). After we know \( x \), we use it to find \( y \). Finally, we divide \( x \) by \( y \) to get the final answer.

🎯 Exam Tip: When asked to find an expression like \( \frac{x}{y} \) after solving a system, ensure you calculate the values of \( x \) and \( y \) accurately first, and then perform the final calculation.

Question 21. (a) Solve the system of equations: \( x - y = a - b, ax - by = a^2 + b^2 \).
Answer:
Given the equations:
1. \( x - y = a - b \quad \text{... (i)} \)
2. \( ax - by = a^2 + b^2 \quad \text{... (ii)} \)
From equation (i), we can express \( x \) in terms of \( y \), \( a \), and \( b \):
\[ x = y + a - b \]
Substitute this expression for \( x \) into equation (ii):
\[ a(y + a - b) - by = a^2 + b^2 \]
\[ ay + a^2 - ab - by = a^2 + b^2 \]
Group terms with \( y \) and move other terms to the right side:
\[ ay - by = a^2 + b^2 - a^2 + ab \]
\[ y(a - b) = b^2 + ab \]
\[ y(a - b) = b(b + a) \]
Assuming \( a \neq b \), we can divide by \( (a - b) \):
\[ y = \frac{b(a+b)}{a-b} \]
Now, substitute the value of \( y \) back into equation (i):
\[ x - \frac{b(a+b)}{a-b} = a - b \]
\[ x = a - b + \frac{b(a+b)}{a-b} \]
Find a common denominator:
\[ x = \frac{(a-b)(a-b) + b(a+b)}{a-b} \]
\[ x = \frac{a^2 - 2ab + b^2 + ab + b^2}{a-b} \]
\[ x = \frac{a^2 - ab + 2b^2}{a-b} \]
So, \( x = \frac{a^2 - ab + 2b^2}{a-b} \) and \( y = \frac{b(a+b)}{a-b} \). In algebra, careful expansion and grouping of terms are crucial for solving systems of equations with parameters.
In simple words: We are solving two equations where the answers for \( x \) and \( y \) will have \( a \) and \( b \) in them. We use the first equation to write \( x \) as something with \( y \), \( a \), and \( b \). Then we put this into the second equation to find \( y \). Finally, we use the value of \( y \) to find \( x \).

🎯 Exam Tip: When working with literal coefficients (like \( a \) and \( b \)), treat them as numbers when performing algebraic operations. Be meticulous with distributing terms and combining like terms.

Question 21. (b) Solve the system of equations: \( ax + by = a - b, bx - ay = a + b \).
Answer:
Given the equations:
1. \( ax + by = a - b \quad \text{... (i)} \)
2. \( bx - ay = a + b \quad \text{... (ii)} \)
To eliminate \( y \), multiply equation (i) by \( a \) and equation (ii) by \( b \):
Multiply (i) by \( a \):
\[ a(ax + by) = a(a - b) \]
\[ a^2x + aby = a^2 - ab \quad \text{... (iii)} \]
Multiply (ii) by \( b \):
\[ b(bx - ay) = b(a + b) \]
\[ b^2x - aby = ab + b^2 \quad \text{... (iv)} \]
Now, add equation (iii) and equation (iv):
\[ (a^2x + aby) + (b^2x - aby) = (a^2 - ab) + (ab + b^2) \]
\[ a^2x + aby + b^2x - aby = a^2 - ab + ab + b^2 \]
\[ a^2x + b^2x = a^2 + b^2 \]
Factor out \( x \):
\[ x(a^2 + b^2) = a^2 + b^2 \]
Assuming \( a^2 + b^2 \neq 0 \) (which is true unless \( a=0 \) and \( b=0 \)), we can divide by \( (a^2 + b^2) \):
\[ x = \frac{a^2 + b^2}{a^2 + b^2} \]
\[ x = 1 \]
Substitute the value of \( x = 1 \) into equation (i):
\[ a(1) + by = a - b \]
\[ a + by = a - b \]
\[ by = a - b - a \]
\[ by = -b \]
Assuming \( b \neq 0 \), we can divide by \( b \):
\[ y = \frac{-b}{b} \]
\[ y = -1 \]
Thus, the solution is \( x = 1 \) and \( y = -1 \). This method of eliminating a variable by multiplying equations by constants is highly effective.
In simple words: We have two equations with \( x \), \( y \), \( a \), and \( b \). To solve them, we make the \( y \) parts cancel out by multiplying the equations by \( a \) and \( b \) separately. After adding them, we find \( x \). Then, we use the value of \( x \) to easily find \( y \).

🎯 Exam Tip: This type of problem often leads to a common factor like \( (a^2 + b^2) \) that simplifies nicely. Look for opportunities to add or subtract equations to eliminate variables, especially when coefficients are related.

Question 22. Solve the system of equations: \( 3 - 2(3x + 4y) = x \), \( \frac{x-3}{4}-\frac{y-4}{5}=\frac{1}{10} \).
Answer:
First, simplify the given equations:
Equation 1:
\[ 3 - 2(3x + 4y) = x \]
\[ 3 - 6x - 8y = x \]
Move all \( x \) and \( y \) terms to one side and constants to the other:
\[ 3 = x + 6x + 8y \]
\[ 7x + 8y = 3 \quad \text{... (i)} \]
Equation 2:
\[ \frac{x-3}{4}-\frac{y-4}{5}=\frac{1}{10} \]
Multiply by 20 (L.C.M. of 4, 5, and 10) to clear the denominators:
\[ 5(x-3) - 4(y-4) = 2(1) \]
\[ 5x - 15 - 4y + 16 = 2 \]
\[ 5x - 4y + 1 = 2 \]
\[ 5x - 4y = 2 - 1 \]
\[ 5x - 4y = 1 \quad \text{... (ii)} \]
Now we have a system of two linear equations:
1. \( 7x + 8y = 3 \)
2. \( 5x - 4y = 1 \)
To eliminate \( y \), multiply equation (ii) by 2:
\[ 2(5x - 4y) = 2(1) \]
\[ 10x - 8y = 2 \quad \text{... (iii)} \]
Add equation (i) and equation (iii):
\[ (7x + 8y) + (10x - 8y) = 3 + 2 \]
\[ 7x + 8y + 10x - 8y = 5 \]
\[ 17x = 5 \]
\[ x = \frac{5}{17} \]
Substitute the value of \( x \) into equation (i):
\[ 7\left(\frac{5}{17}\right) + 8y = 3 \]
\[ \frac{35}{17} + 8y = 3 \]
\[ 8y = 3 - \frac{35}{17} \]
\[ 8y = \frac{3 \times 17 - 35}{17} \]
\[ 8y = \frac{51 - 35}{17} \]
\[ 8y = \frac{16}{17} \]
\[ y = \frac{16}{17 \times 8} \]
\[ y = \frac{2}{17} \]
So, the solution is \( x = \frac{5}{17} \) and \( y = \frac{2}{17} \). It's important to be careful with signs when distributing and combining terms, especially after multiplying through by a negative number.
In simple words: First, we tidy up both equations by getting rid of brackets and fractions. This gives us two simpler equations. Then, we multiply one equation so that we can cancel out the \( y \) parts when we add the equations together. This helps us find \( x \). Once we have \( x \), we use it to find \( y \).

🎯 Exam Tip: Double-check calculations when combining fractions, especially when subtracting. Always keep track of positive and negative signs, as a small error can propagate through the entire solution.

Question 23. Can the following system of equations hold simultaneously? \( \frac{3}{x} + 4y = 7 \), \( \frac{-2}{x} + 7y = 5 \), \( 5x + \frac{4}{y} = 9 \). If yes, find x and y.
Answer:
Given the three equations:
1. \( \frac{3}{x} + 4y = 7 \quad \text{... (i)} \)
2. \( \frac{-2}{x} + 7y = 5 \quad \text{... (ii)} \)
3. \( 5x + \frac{4}{y} = 9 \quad \text{... (iii)} \)
Let's try to solve the first two equations for \( x \) and \( y \). To eliminate \( \frac{1}{x} \), multiply equation (i) by 2 and equation (ii) by 3:
Multiply (i) by 2:
\[ 2\left(\frac{3}{x} + 4y\right) = 2(7) \]
\[ \frac{6}{x} + 8y = 14 \quad \text{... (iv)} \]
Multiply (ii) by 3:
\[ 3\left(\frac{-2}{x} + 7y\right) = 3(5) \]
\[ \frac{-6}{x} + 21y = 15 \quad \text{... (v)} \]
Add equation (iv) and equation (v):
\[ \left(\frac{6}{x} + 8y\right) + \left(\frac{-6}{x} + 21y\right) = 14 + 15 \]
\[ \frac{6}{x} + 8y - \frac{6}{x} + 21y = 29 \]
\[ 29y = 29 \]
\[ y = \frac{29}{29} \]
\[ y = 1 \]
Substitute \( y = 1 \) into equation (i):
\[ \frac{3}{x} + 4(1) = 7 \]
\[ \frac{3}{x} + 4 = 7 \]
\[ \frac{3}{x} = 7 - 4 \]
\[ \frac{3}{x} = 3 \]
\[ x = \frac{3}{3} \]
\[ x = 1 \]
So, from the first two equations, we found \( x = 1 \) and \( y = 1 \).
Now, we need to check if these values also satisfy the third equation (iii):
\[ 5x + \frac{4}{y} = 9 \]
Substitute \( x = 1 \) and \( y = 1 \):
\[ 5(1) + \frac{4}{1} = 9 \]
\[ 5 + 4 = 9 \]
\[ 9 = 9 \]
Since the values satisfy all three equations, the system of equations can hold simultaneously. It's crucial to check consistency across all equations when dealing with more than two.
In simple words: We have three math puzzles. We solve the first two puzzles together to find \( x \) and \( y \). Then, we check if these same \( x \) and \( y \) values also work for the third puzzle. If they do, then all three puzzles can be solved at the same time, and we found the solution!

🎯 Exam Tip: When given more than two equations, solve any pair of equations first. Once you find the values for the variables, substitute them into the remaining equation(s) to verify if the system is consistent (i.e., if they can hold simultaneously).

Question 24. If the following three equations hold simultaneously for x and y, find p. \( 3x - 2y = 6 \), \( \frac{x}{3}-\frac{y}{6}=\frac{1}{2} \), \( x - py = 6 \).
Answer:
Given the three equations:
1. \( 3x - 2y = 6 \quad \text{... (i)} \)
2. \( \frac{x}{3}-\frac{y}{6}=\frac{1}{2} \quad \text{... (ii)} \)
3. \( x - py = 6 \quad \text{... (iii)} \)
First, simplify equation (ii) by multiplying by 6 (L.C.M. of 3, 6, 2):
\[ 6\left(\frac{x}{3}\right) - 6\left(\frac{y}{6}\right) = 6\left(\frac{1}{2}\right) \]
\[ 2x - y = 3 \quad \text{... (iv)} \]
Now, let's solve equation (i) and equation (iv) simultaneously for \( x \) and \( y \).
1. \( 3x - 2y = 6 \)
2. \( 2x - y = 3 \)
Multiply equation (iv) by 2 to make the coefficient of \( y \) match equation (i):
\[ 2(2x - y) = 2(3) \]
\[ 4x - 2y = 6 \quad \text{... (v)} \]
Subtract equation (i) from equation (v):
\[ (4x - 2y) - (3x - 2y) = 6 - 6 \]
\[ 4x - 2y - 3x + 2y = 0 \]
\[ x = 0 \]
Substitute \( x = 0 \) into equation (iv):
\[ 2(0) - y = 3 \]
\[ 0 - y = 3 \]
\[ -y = 3 \]
\[ y = -3 \]
So, the solution from the first two equations is \( x = 0 \) and \( y = -3 \).
Since these values must also satisfy the third equation (iii) for the system to hold simultaneously, substitute \( x = 0 \) and \( y = -3 \) into equation (iii):
\[ x - py = 6 \]
\[ 0 - p(-3) = 6 \]
\[ 3p = 6 \]
\[ p = \frac{6}{3} \]
\[ p = 2 \]
Therefore, the value of \( p \) is 2. This demonstrates how a consistent solution from a subset of equations can be used to determine unknown parameters in other equations.
In simple words: We have three equations, and we need to find the value of \( p \). First, we make the first two equations simpler and solve them to find \( x \) and \( y \). Then, we use these values of \( x \) and \( y \) in the third equation. This helps us work out what \( p \) must be.

🎯 Exam Tip: For problems involving an unknown parameter in one of the equations, first solve the subset of equations that only contain the known variables. Then, substitute these solved values into the equation with the parameter to find its value.

Question 25. The sides of an equilateral triangle are \( (6x + 33y) \) cm, \( (8x + 9y - 5) \) cm and \( (10x + 12y - 8) \) cm respectively. Find the length of each side.
Answer:
In an equilateral triangle, all three sides are equal in length.
So, we can set up two equations by equating the side lengths:
1. \( 6x + 33y = 8x + 9y - 5 \)
Rearrange the terms to form a linear equation:
\[ 8x - 6x + 9y - 33y = 5 \]
\[ 2x - 24y = 5 \quad \text{... (i)} \]
2. \( 8x + 9y - 5 = 10x + 12y - 8 \)
Rearrange the terms:
\[ 10x - 8x + 12y - 9y = -5 + 8 \]
\[ 2x + 3y = 3 \quad \text{... (ii)} \]
Now we have a system of two linear equations:
1. \( 2x - 24y = 5 \)
2. \( 2x + 3y = 3 \)
Subtract equation (ii) from equation (i) to eliminate \( x \):
\[ (2x - 24y) - (2x + 3y) = 5 - 3 \]
\[ 2x - 24y - 2x - 3y = 2 \]
\[ -27y = 2 \]
\[ y = \frac{-2}{27} \]
Substitute the value of \( y \) into equation (ii):
\[ 2x + 3\left(\frac{-2}{27}\right) = 3 \]
\[ 2x - \frac{6}{27} = 3 \]
\[ 2x - \frac{2}{9} = 3 \]
\[ 2x = 3 + \frac{2}{9} \]
\[ 2x = \frac{3 \times 9 + 2}{9} \]
\[ 2x = \frac{27 + 2}{9} \]
\[ 2x = \frac{29}{9} \]
\[ x = \frac{29}{9 \times 2} \]
\[ x = \frac{29}{18} \]
So, \( x = \frac{29}{18} \) and \( y = \frac{-2}{27} \).
Finally, find the length of each side using one of the expressions for a side (e.g., \( 2x + 3y \), which we know equals 3):
Side length \( = 2x + 3y = 3 \).
Therefore, the length of each side of the equilateral triangle is 3 cm. Defining relationships between unknown quantities using equations is a fundamental aspect of problem-solving in geometry.
In simple words: Since all sides of an equilateral triangle are the same length, we set the given expressions for the sides equal to each other to make two equations. We solve these two equations to find the values of \( x \) and \( y \). Once we have \( x \) and \( y \), we put them back into any of the side expressions to find the actual length of one (and thus all) sides.

🎯 Exam Tip: For word problems, translate the given information into mathematical equations first. Always check the properties of the geometric figure (like an equilateral triangle having equal sides) to set up the correct relationships.