Maharashtra Board Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.4 Solutions

Question 1. Solve the following simultaneous equations.
(i) \( \frac{2}{x} - \frac{3}{y} = 15 \); \( \frac{8}{x} + \frac{5}{y} = 77 \)
(ii) \( \frac{10}{x+y} + \frac{2}{x-y} = 4 \); \( \frac{15}{x+y} - \frac{5}{x-y} = -2 \)
(iii) \( \frac{27}{x-2} + \frac{31}{y+3} = 85 \); \( \frac{31}{x-2} + \frac{27}{y+3} = 89 \)
(iv) \( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \); \( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} \)
Answer:
These non-linear equations are solved by converting them into a linear form using suitable substitutions.

(i) Given equations:
\( \frac{2}{x} - \frac{3}{y} = 15 \) — (1)
\( \frac{8}{x} + \frac{5}{y} = 77 \) — (2)
Substituting \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \) in equations (1) and (2), we get:
\( 2a - 3b = 15 \) — (3)
\( 8a + 5b = 77 \) — (4)
Multiplying equation (3) by 4, we get:
\( 8a - 12b = 60 \) — (5)
Subtracting equation (5) from equation (4), we get:
\( (8a + 5b) - (8a - 12b) = 77 - 60 \)
\( 17b = 17 \)
\( \implies b = 1 \)
Substituting \( b = 1 \) in equation (3), we get:
\( 2a - 3(1) = 15 \)
\( 2a - 3 = 15 \)
\( 2a = 18 \)
\( \implies a = 9 \)
Resubstituting the values of \( a \) and \( b \):
\( a = \frac{1}{x} \)
\( 9 = \frac{1}{x} \)
\( \implies x = \frac{1}{9} \)
\( b = \frac{1}{y} \)
\( 1 = \frac{1}{y} \)
\( \implies y = 1 \)
Therefore, the solution is \( (x, y) = \left(\frac{1}{9}, 1\right) \).

(ii) Given equations:
\( \frac{10}{x+y} + \frac{2}{x-y} = 4 \) — (1)
\( \frac{15}{x+y} - \frac{5}{x-y} = -2 \) — (2)
Substituting \( \frac{1}{x+y} = a \) and \( \frac{1}{x-y} = b \) in equations (1) and (2), we get:
\( 10a + 2b = 4 \) — (3)
\( 15a - 5b = -2 \) — (4)
Multiplying equation (3) by 5 and equation (4) by 2, we get:
\( 50a + 10b = 20 \) — (5)
\( 30a - 10b = -4 \) — (6)
Adding equations (5) and (6), we get:
\( 80a = 16 \)
\( \implies a = \frac{16}{80} = \frac{1}{5} \)
Substituting \( a = \frac{1}{5} \) in equation (3), we get:
\( 10\left(\frac{1}{5}\right) + 2b = 4 \)
\( 2 + 2b = 4 \)
\( 2b = 2 \)
\( \implies b = 1 \)
Resubstituting the values of \( a \) and \( b \):
\( \frac{1}{x+y} = \frac{1}{5} \)
\( \implies x + y = 5 \) — (7)
\( \frac{1}{x-y} = 1 \)
\( \implies x - y = 1 \) — (8)
Adding equations (7) and (8), we get:
\( 2x = 6 \)
\( \implies x = 3 \)
Substituting \( x = 3 \) in equation (7), we get:
\( 3 + y = 5 \)
\( \implies y = 2 \)
Therefore, the solution is \( (x, y) = (3, 2) \).

(iii) Given equations:
\( \frac{27}{x-2} + \frac{31}{y+3} = 85 \) — (1)
\( \frac{31}{x-2} + \frac{27}{y+3} = 89 \) — (2)
Substituting \( \frac{1}{x-2} = a \) and \( \frac{1}{y+3} = b \) in equations (1) and (2), we get:
\( 27a + 31b = 85 \) — (3)
\( 31a + 27b = 89 \) — (4)
Adding equations (3) and (4), we get:
\( 58a + 58b = 174 \)
Dividing by 58 on both sides:
\( \implies a + b = 3 \) — (5)
Subtracting equation (3) from equation (4), we get:
\( 4a - 4b = 4 \)
Dividing by 4 on both sides:
\( \implies a - b = 1 \) — (6)
Adding equations (5) and (6), we get:
\( 2a = 4 \)
\( \implies a = 2 \)
Substituting \( a = 2 \) in equation (5), we get:
\( 2 + b = 3 \)
\( \implies b = 1 \)
Resubstituting the values of \( a \) and \( b \):
\( \frac{1}{x-2} = 2 \)
\( \implies x - 2 = \frac{1}{2} \)
\( \implies x = 2 + 0.5 = 2.5 = \frac{5}{2} \)
\( \frac{1}{y+3} = 1 \)
\( \implies y + 3 = 1 \)
\( \implies y = 1 - 3 = -2 \)
Therefore, the solution is \( (x, y) = \left(\frac{5}{2}, -2\right) \).

(iv) Given equations:
\( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \) — (1)
\( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} \) — (2)
Substituting \( \frac{1}{3x+y} = a \) and \( \frac{1}{3x-y} = b \) in equations (1) and (2), we get:
\( a + b = \frac{3}{4} \)
Multiplying by 4:
\( \implies 4a + 4b = 3 \) — (3)
\( \frac{a}{2} - \frac{b}{2} = -\frac{1}{8} \)
Multiplying by 8:
\( \implies 4a - 4b = -1 \) — (4)
Adding equations (3) and (4), we get:
\( 8a = 2 \)
\( \implies a = \frac{2}{8} = \frac{1}{4} \)
Subtracting equation (4) from equation (3), we get:
\( 8b = 4 \)
\( \implies b = \frac{4}{8} = \frac{1}{2} \)
Resubstituting the values of \( a \) and \( b \):
\( \frac{1}{3x+y} = \frac{1}{4} \)
\( \implies 3x + y = 4 \) — (5)
\( \frac{1}{3x-y} = \frac{1}{2} \)
\( \implies 3x - y = 2 \) — (6)
Adding equations (5) and (6), we get:
\( 6x = 6 \)
\( \implies x = 1 \)
Substituting \( x = 1 \) in equation (5), we get:
\( 3(1) + y = 4 \)
\( 3 + y = 4 \)
\( \implies y = 1 \)
Therefore, the solution is \( (x, y) = (1, 1) \).
In simple words: To solve these complex equations, we temporarily replace the fractional parts with simpler variables like a and b. Once we solve for these new variables, we substitute them back to find the final values of x and y.

🎯 Exam Tip: Always remember to substitute the original variables back at the end of the problem to find the final values of x and y, as forgetting this step is a very common mistake.

Question 1. Solve the following simultaneous equations:
(i) \( \frac{2}{x} - \frac{3}{y} = 15 \) ; \( \frac{8}{x} + \frac{5}{y} = 77 \)
(ii) \( \frac{10}{x+y} + \frac{2}{x-y} = 4 \) ; \( \frac{15}{x+y} - \frac{5}{x-y} = -2 \)
(iii) \( \frac{27}{x-2} + \frac{31}{y+3} = 85 \) ; \( \frac{31}{x-2} + \frac{27}{y+3} = 89 \)
(iv) \( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \) ; \( \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8} \)
Answer:
(i) The given simultaneous equations are:
\( \frac{2}{x} - \frac{3}{y} = 15 \) ...(i)
\( \frac{8}{x} + \frac{5}{y} = 77 \) ...(ii)
Let \( \frac{1}{x} = p \) and \( \frac{1}{y} = q \)
\( \therefore \) Equations (i) and (ii) become
\( 2p - 3q = 15 \) ...(iii)
\( 8p + 5q = 77 \) ...(iv)
Multiplying equation (iii) by 4, we get
\( 8p - 12q = 60 \) ...(v)
Subtracting equation (v) from (iv), we get
\( (8p + 5q) - (8p - 12q) = 77 - 60 \)

\( \implies 17q = 17 \)

\( \implies q = 1 \)
Substituting \( q = 1 \) in equation (iii), we get
\( 2p - 3(1) = 15 \)

\( \implies 2p - 3 = 15 \)

\( \implies 2p = 18 \)

\( \implies p = 9 \)
Now, substituting the values of \( p \) and \( q \) back to find \( x \) and \( y \):
\( p = \frac{1}{x} \)

\( \implies 9 = \frac{1}{x} \)

\( \implies x = \frac{1}{9} \)
And,
\( q = \frac{1}{y} \)

\( \implies 1 = \frac{1}{y} \)

\( \implies y = 1 \)
Therefore, the solution of the given simultaneous equations is \( (x, y) = \left(\frac{1}{9}, 1\right) \). This systematic method of substitution helps simplify complex rational equations into standard linear forms.
In simple words: When variables are in the bottom part of a fraction, we replace them with simpler letters like p and q to make them easy to solve. Once we find the values of p and q, we flip them back to get the final answers for x and y.

🎯 Exam Tip: Always remember to substitute the original variables back at the end of the solution to find the final values of x and y, as leaving the answer in terms of p and q will result in a loss of marks.

Question 1. Solve the following simultaneous equations:
(i) \( \frac{2}{x} - \frac{3}{y} = 15 \), \( \frac{8}{x} + \frac{5}{y} = 77 \)
(ii) \( \frac{10}{x+y} + \frac{2}{x-y} = 4 \), \( \frac{15}{x+y} - \frac{5}{x-y} = -2 \)
Answer:
(i) The given simultaneous equations are:
\( \frac{2}{x} - \frac{3}{y} = 15 \) ...(i)
\( \frac{8}{x} + \frac{5}{y} = 77 \) ...(ii)
Let \( \frac{1}{x} = p \) and \( \frac{1}{y} = q \).
Equations (i) and (ii) become:
\( 2p - 3q = 15 \) ...(iii)
\( 8p + 5q = 77 \) ...(iv)
Multiplying equation (iii) by 4, we get:
\( 8p - 12q = 60 \) ...(v)
Subtracting equation (v) from equation (iv):
\[ \begin{array}{rcc} 8p + 5q & = & 77 \\ 8p - 12q & = & 60 \\ (-) \quad (+) & & (-) \\ \hline 17q & = & 17 \end{array} \]
\( \implies q = \frac{17}{17} = 1 \)
Substituting \( q = 1 \) in equation (iii), we get:
\( 2p - 3(1) = 15 \)
\( \implies 2p - 3 = 15 \)
\( \implies 2p = 15 + 3 = 18 \)
\( \implies p = \frac{18}{2} = 9 \)
\( \implies (p, q) = (9, 1) \)
Resubstituting the values of \( p \) and \( q \), we get:
\( 9 = \frac{1}{x} \) and \( 1 = \frac{1}{y} \)
\( \implies x = \frac{1}{9} \) and \( y = 1 \)
\( \implies (x, y) = \left(\frac{1}{9}, 1\right) \) is the solution of the given simultaneous equations.

(ii) The given simultaneous equations are:
\( \frac{10}{x+y} + \frac{2}{x-y} = 4 \) ...(i)
\( \frac{15}{x+y} - \frac{5}{x-y} = -2 \) ...(ii)
Let \( \frac{1}{x+y} = p \) and \( \frac{1}{x-y} = q \).
Equations (i) and (ii) become:
\( 10p + 2q = 4 \)
\( \implies 5p + q = 2 \) ...(iii) [Dividing both sides by 2]
\( 15p - 5q = -2 \) ...(iv)
Multiplying equation (iii) by 5, we get:
\( 25p + 5q = 10 \) ...(v)
Adding equations (iv) and (v), we get:
\[ \begin{array}{rcc} 15p - 5q & = & -2 \\ +\quad 25p + 5q & = & 10 \\ \hline 40p & = & 8 \end{array} \]
\( \implies p = \frac{8}{40} = \frac{1}{5} \)
Substituting \( p = \frac{1}{5} \) in equation (iii), we get:
\( 5\left(\frac{1}{5}\right) + q = 2 \)
\( \implies 1 + q = 2 \)
\( \implies q = 1 \)
Resubstituting the values of \( p \) and \( q \), we get:
\( \frac{1}{x+y} = \frac{1}{5} \)
\( \implies x + y = 5 \) ...(vi)
And,
\( \frac{1}{x-y} = 1 \)
\( \implies x - y = 1 \) ...(vii)
Adding equations (vi) and (vii), we get:
\( 2x = 6 \)
\( \implies x = 3 \)
Substituting \( x = 3 \) in equation (vi), we get:
\( 3 + y = 5 \)
\( \implies y = 2 \)
\( \implies (x, y) = (3, 2) \) is the solution of the given simultaneous equations.
In simple words: When variables are in the denominator, we temporarily replace them with simpler letters like p and q to make them easy to solve. Once we find the values of p and q, we substitute them back to find the final values of x and y.

🎯 Exam Tip: Always remember to substitute back the original variables (x and y) at the end of the problem, as solving only for p and q is a common incomplete step where students lose marks.

Question (ii) Solve the simultaneous equations (continued).
Answer: Substituting \( p = \frac{1}{5} \) in equation (iii), we get:
\( 5\left(\frac{1}{5}\right) + q = 2 \)
\( \implies 1 + q = 2 \)
\( \implies q = 2 - 1 = 1 \)
\( \implies (p, q) = \left(\frac{1}{5}, 1\right) \)
Resubstituting the values of \( p \) and \( q \), we get:
\( \frac{1}{5} = \frac{1}{x+y} \) and \( 1 = \frac{1}{x-y} \)
\( \implies x + y = 5 \) ...(vi)
and \( x - y = 1 \) ...(vii)
Adding equations (vi) and (vii), we get:
\( (x + y) + (x - y) = 5 + 1 \)
\( \implies 2x = 6 \)
\( \implies x = \frac{6}{2} = 3 \)
Substituting \( x = 3 \) in equation (vi), we get:
\( 3 + y = 5 \)
\( \implies y = 5 - 3 = 2 \)
This systematic process of substitution helps us break down complex equations into simpler linear forms.
\( \implies (x, y) = (3, 2) \) is the solution of the given simultaneous equations.
In simple words: We find the values of our temporary variables first, then substitute them back to get the final values of x and y. This makes solving complex equations much easier.

🎯 Exam Tip: Always double-check your final values of x and y by substituting them back into the original equations to ensure they satisfy both.

Question (iii) Solve the simultaneous equations:
\( \frac{27}{x-2} + \frac{31}{y+3} = 85 \) and \( \frac{31}{x-2} + \frac{27}{y+3} = 89 \)
Answer: The given simultaneous equations are:
\( \frac{27}{x-2} + \frac{31}{y+3} = 85 \) ...(i)
\( \frac{31}{x-2} + \frac{27}{y+3} = 89 \) ...(ii)
Let \( \frac{1}{x-2} = p \) and \( \frac{1}{y+3} = q \)
\( \implies \) Equations (i) and (ii) become:
\( 27p + 31q = 85 \) ...(iii)
\( 31p + 27q = 89 \) ...(iv)
Adding equations (iii) and (iv), we get:
\( (27p + 31q) + (31p + 27q) = 85 + 89 \)
\( \implies 58p + 58q = 174 \)
Dividing both sides by 58, we get:
\( \implies p + q = 3 \) ...(v)
This simplification helps us convert a complex system into a much more manageable linear form.
In simple words: We replace the complicated fractions with simpler letters like p and q to make the equations easier to add and solve.

🎯 Exam Tip: When the coefficients of the variables are interchanged (like 27 and 31 here), adding and subtracting the equations is the fastest way to simplify them.

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Question 1. Solve the following simultaneous equations:
\( \frac{27}{x-2} + \frac{31}{y+3} = 85 \)
\( \frac{31}{x-2} + \frac{27}{y+3} = 89 \)
Answer:
Let \( \frac{1}{x-2} = p \) and \( \frac{1}{y+3} = q \).
The equations become:
\( 27p + 31q = 85 \) ... (iii)
\( 31p + 27q = 89 \) ... (iv)
Adding equations (iii) and (iv), we get:
\( 27p + 31q = 85 \)
\( + \; 31p + 27q = 89 \)
___________________
\( 58p + 58q = 174 \)
\( \therefore p + q = \frac{174}{58} \) ... [Dividing both sides by 58]
\( \dots p + q = 3 \) ... (v)

Subtracting equation (iv) from (iii), we get:
\( 27p + 31q = 85 \)
\( - \; (31p + 27q = 89) \)
___________________
\( -4p + 4q = -4 \)
\( \dots p - q = \frac{-4}{-4} \) ... [Dividing both sides by -4]
\( \dots p - q = 1 \) ... (vi)

Adding equations (v) and (vi), we get:
\( p + q = 3 \)
\( + \; p - q = 1 \)
___________________
\( 2p = 4 \)
\( \therefore p = \frac{4}{2} = 2 \)

Substituting \( p = 2 \) in equation (v), we get:
\( 2 + q = 3 \)
\( \therefore q = 3 - 2 = 1 \)
\( \therefore (p, q) = (2, 1) \)

Resubstituting the values of \( p \) and \( q \), we get:
\( 2 = \frac{1}{x-2} \) and \( 1 = \frac{1}{y+3} \)
\( \therefore 2(x-2) = 1 \) and \( y+3 = 1 \)
\( \therefore 2x-4 = 1 \) and \( y = 1-3 \)
\( \therefore 2x = 1+4 \) and \( y = -2 \)
\( \therefore 2x = 5 \) and \( y = -2 \)
\( \therefore x = \frac{5}{2} \) and \( y = -2 \)

\( \therefore (x, y) = \left(\frac{5}{2}, -2\right) \) is the solution of the given simultaneous equations.
In simple words: To solve these equations, we first simplify them by replacing the complicated fraction parts with simpler letters like p and q. After finding the values of p and q, we substitute them back to find the final values of x and y.

🎯 Exam Tip: When substituting back the values of p and q, be extremely careful with cross-multiplication and signs to avoid simple calculation errors.

Question (iv) Solve the following simultaneous equations:
\( \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \)
\( \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8} \)
Answer:
The given simultaneous equations are:
\( \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \)     ...(i)
\( \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8} \)     ...(ii)

Let \( \frac{1}{3x + y} = p \) and \( \frac{1}{3x - y} = q \)

Therefore, equations (i) and (ii) become:
\( p + q = \frac{3}{4} \)     ...(iii)
\( \frac{1}{2}p - \frac{1}{2}q = -\frac{1}{8} \)     ...(iv)

Multiplying equation (iv) by 2, we get:
\( p - q = -\frac{1}{4} \)     ...(v)

Adding equations (iii) and (v), we get:
\( p + q = \frac{3}{4} \)
\( + \quad p - q = -\frac{1}{4} \)
__________________
\( 2p = \frac{2}{4} \)

\( \implies 2p = \frac{1}{2} \)

\( \implies p = \frac{1}{4} \)
In simple words: We substitute the complex fraction terms with simpler variables \(p\) and \(q\) to make the equations easier to solve. By adding the two simplified equations, we eliminate \(q\) and easily find that \(p\) equals \(1/4\).

🎯 Exam Tip: When dealing with equations containing variables in the denominator, always use substitution to convert them into standard linear equations first to avoid calculation errors.

Question. Solve the simultaneous equations (Continuation).
Answer: Substituting \( p = \frac{1}{4} \) in equation (iii), we get:
\( \frac{1}{4} + q = \frac{3}{4} \)
\( \therefore q = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
\( \dots (p, q) = \left(\frac{1}{4}, \frac{1}{2}\right) \)
Resubstituting the values of \( p \) and \( q \), we get:
\( \frac{1}{4} = \frac{1}{3x + y} \) and \( \frac{1}{2} = \frac{1}{3x - y} \)
\( \therefore 3x + y = 4 \) ...(vi)
and \( 3x - y = 2 \) ...(vii)
Adding equations (vi) and (vii), we get:
\( (3x + y) + (3x - y) = 4 + 2 \)
\( \implies 6x = 6 \)
\( \implies x = \frac{6}{6} = 1 \)
Substituting \( x = 1 \) in equation (vi), we get:
\( 3(1) + y = 4 \)
\( \dots 3 + y = 4 \)
\( \therefore y = 4 - 3 = 1 \)
\( \therefore (x, y) = (1, 1) \) is the solution of the given simultaneous equations.
In simple words: We find the values of p and q first, then substitute them back to get two simpler equations in x and y. Solving those gives us the final answer.

🎯 Exam Tip: When resubstituting variables, double-check your signs to avoid simple calculation errors that can affect the final values of x and y.

Question 1. Complete the following table. (Textbook pg. no. 16)
Answer:

In simple words: These equations have variables in the denominator, which means their degree is -1, not 1. Therefore, they are not linear equations.

🎯 Exam Tip: Remember that a linear equation must have variables with a power of 1. If the variables are in the denominator, the equation is non-linear.

Question 2. In the above table the equations are not linear. Can you convert the equations into linear equations? (Textbook pg. no. 17)
Answer: Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions. This mathematical technique simplifies complex non-linear systems into standard solvable forms.

Steps for solving equations reducible to a pair of linear equations:
• Step 1: Select suitable variables other than those which are in the equations.
• Step 2: Replace the given variables with new variables such that the given equations become linear equations in two variables.
• Step 3: Solve the new simultaneous equations and find the values of the new variables.
• Step 4: By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.
In simple words: We can change complicated equations into simpler linear ones by replacing the tricky parts with new letters. Once we solve the simple equations, we swap the original letters back to get the final answer.

🎯 Exam Tip: Always remember to perform the final back-substitution step to find the values of the original variables, as this is a very common place where students lose marks.

Question 3. To solve given equations fill the below boxes suitably. (Text book pg.no. 19)

Question 3. Solve the following simultaneous equations: \( \frac{5}{x-1} + \frac{1}{y-2} = 2 \); \( \frac{6}{x-1} - \frac{3}{y-2} = 1 \)
Answer:
• Given equations: \( \frac{5}{x-1} + \frac{1}{y-2} = 2 \) and \( \frac{6}{x-1} - \frac{3}{y-2} = 1 \)
• Replacing \( \left(\frac{1}{x-1}\right) \) by \( m \) and \( \left(\frac{1}{y-2}\right) \) by \( n \)
• New equations: \( 5m + n = 2 \) and \( 6m - 3n = 1 \)
• On solving: \( m = \frac{1}{3} \), \( n = \frac{1}{3} \)
• Replacing \( m, n \) by their original values: \( \frac{1}{x-1} = \frac{1}{3} \) and \( \frac{1}{y-2} = \frac{1}{3} \)
• On solving: \( x-1 = 3 \), \( y-2 = 3 \)
\( \implies x = 4 \), \( y = 5 \)
• Therefore, \( (x, y) = (4, 5) \) is the solution of the given simultaneous equations.
In simple words: We substitute the fraction terms with simpler variables like m and n, solve the basic equations, and then substitute back to find the final values of x and y.

🎯 Exam Tip: Always substitute back the original variables at the end of the solution to find the values of x and y, as forgetting this step is a common way to lose marks.

Question 4. The examples on textbook pg. no. 17 and 18 obtained by transformation are solved by elimination method. If you solve these equations by graphical method and by Cramer's rule will you get the same answers? Solve and check it. (Textbook pg. no. 18)
(i) \( \frac{4}{x} + \frac{5}{y} = 7 \); \( \frac{3}{x} + \frac{4}{y} = 5 \)
(ii) \( \frac{4}{x-y} + \frac{1}{x+y} = 3 \); \( \frac{2}{x-y} - \frac{3}{x+y} = 5 \)
Answer: Yes, we will get the exact same answers regardless of the method used (Elimination, Graphical, or Cramer's Rule) because the solution to a consistent system of linear equations is unique.

Verification for sub-part (i):
Equations: \( \frac{4}{x} + \frac{5}{y} = 7 \) and \( \frac{3}{x} + \frac{4}{y} = 5 \)
Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \). The equations become:
\( 4a + 5b = 7 \) ---(1)
\( 3a + 4b = 5 \) ---(2)

1. By Cramer's Rule:
\( D = \begin{vmatrix} 4 & 5 \\ 3 & 4 \end{vmatrix} = (4 \times 4) - (5 \times 3) = 16 - 15 = 1 \)
\( D_a = \begin{vmatrix} 7 & 5 \\ 5 & 4 \end{vmatrix} = (7 \times 4) - (5 \times 5) = 28 - 25 = 3 \)
\( D_b = \begin{vmatrix} 4 & 7 \\ 3 & 5 \end{vmatrix} = (4 \times 5) - (7 \times 3) = 20 - 21 = -1 \)
\( \implies a = \frac{D_a}{D} = \frac{3}{1} = 3 \)
\( \implies b = \frac{D_b}{D} = \frac{-1}{1} = -1 \)
Since \( a = \frac{1}{x} \)
\( \implies x = \frac{1}{3} \)
Since \( b = \frac{1}{y} \)
\( \implies y = -1 \)
So, \( (x, y) = \left(\frac{1}{3}, -1\right) \).

2. By Graphical Method:
To plot \( 4a + 5b = 7 \) and \( 3a + 4b = 5 \):
For \( 4a + 5b = 7 \): If \( a = 3 \), \( b = -1 \); if \( a = -2 \), \( b = 3 \).
For \( 3a + 4b = 5 \): If \( a = 3 \), \( b = -1 \); if \( a = -1 \), \( b = 2 \).
The two lines intersect at the point \( (a, b) = (3, -1) \).
Substituting back, we get \( (x, y) = \left(\frac{1}{3}, -1\right) \).

Verification for sub-part (ii):
Equations: \( \frac{4}{x-y} + \frac{1}{x+y} = 3 \) and \( \frac{2}{x-y} - \frac{3}{x+y} = 5 \)
Let \( \frac{1}{x-y} = a \) and \( \frac{1}{x+y} = b \). The equations become:
\( 4a + b = 3 \) ---(1)
\( 2a - 3b = 5 \) ---(2)

1. By Cramer's Rule:
\( D = \begin{vmatrix} 4 & 1 \\ 2 & -3 \end{vmatrix} = (4 \times -3) - (1 \times 2) = -12 - 2 = -14 \)
\( D_a = \begin{vmatrix} 3 & 1 \\ 5 & -3 \end{vmatrix} = (3 \times -3) - (1 \times 5) = -9 - 5 = -14 \)
\( D_b = \begin{vmatrix} 4 & 3 \\ 2 & 5 \end{vmatrix} = (4 \times 5) - (3 \times 2) = 20 - 6 = 14 \)
\( \implies a = \frac{D_a}{D} = \frac{-14}{-14} = 1 \)
\( \implies b = \frac{D_b}{D} = \frac{14}{-14} = -1 \)
Since \( a = \frac{1}{x-y} \)
\( \implies x-y = 1 \) ---(3)
Since \( b = \frac{1}{x+y} \)
\( \implies x+y = -1 \) ---(4)
Solving (3) and (4):
Adding them: \( 2x = 0 \)
\( \implies x = 0 \)
Substituting \( x = 0 \) in (4): \( 0 + y = -1 \)
\( \implies y = -1 \)
So, \( (x, y) = (0, -1) \).

2. By Graphical Method:
Plotting \( 4a + b = 3 \) and \( 2a - 3b = 5 \):
The two lines intersect at \( (a, b) = (1, -1) \).
This gives the simultaneous linear equations \( x-y = 1 \) and \( x+y = -1 \).
Plotting these two lines, they intersect at \( (x, y) = (0, -1) \).
Thus, both methods yield the exact same solutions.
In simple words: No matter which mathematical method you use—whether you draw graphs, use determinants (Cramer's Rule), or eliminate variables—the final answer for the unknown values will always be the same because there is only one correct solution.

🎯 Exam Tip: When solving using Cramer's Rule, double-check the signs of your determinants (\( D, D_x, D_y \)) as a single sign error will lead to an incorrect final coordinate.

Question (ii) Solve the following simultaneous equations:
\( \frac{4}{x-y} + \frac{1}{x+y} = 3 \)
\( \frac{2}{x-y} - \frac{3}{x+y} = 5 \)
Answer:
The given simultaneous equations are:
\( \frac{4}{x-y} + \frac{1}{x+y} = 3 \quad \dots \text{(i)} \)
\( \frac{2}{x-y} - \frac{3}{x+y} = 5 \quad \dots \text{(ii)} \)

Let \( \frac{1}{x-y} = p \) and \( \frac{1}{x+y} = q \).

\( \therefore \) Equations (i) and (ii) become:
\( 4p + q = 3 \quad \dots \text{(iii)} \)
\( 2p - 3q = 5 \quad \dots \text{(iv)} \)

Graphical method:
For equation \( 4p + q = 3 \):
\( \therefore q = 3 - 4p \)

For equation \( 2p - 3q = 5 \):
\( \therefore 3q = 2p - 5 \)
\( \therefore q = \frac{2p-5}{3} \)

Graph Plotting Details:

• Scale: On both axes, \( 1\text{ cm} = 1\text{ unit} \)
• Line 1 \( (4p + q = 3) \): Passes through points \( (0, 3) \), \( (0.5, 1) \), \( (1, -1) \), and \( (2, -5) \).
• Line 2 \( (2p - 3q = 5) \): Passes through points \( (-2, -3) \), \( (1, -1) \), \( (4, 1) \), and \( (7, 3) \).
• Intersection Point: The two lines intersect at point \( (1, -1) \).

Since the intersection point is \( (1, -1) \), we have:
\( p = 1 \) and \( q = -1 \)

Now, substituting the values of \( p \) and \( q \) back into our assumptions:
\( \frac{1}{x-y} = 1 \)
\( \implies x - y = 1 \quad \dots \text{(v)} \)

\( \frac{1}{x+y} = -1 \)
\( \implies x + y = -1 \quad \dots \text{(vi)} \)

Adding equations (v) and (vi):
\( (x - y) + (x + y) = 1 + (-1) \)
\( \implies 2x = 0 \)
\( \implies x = 0 \)

Substituting \( x = 0 \) in equation (v):
\( 0 - y = 1 \)
\( \implies y = -1 \)

Thus, the final solution is \( (x, y) = (0, -1) \).
In simple words: We temporarily replace the fractions with simpler letters \( p \) and \( q \) to make them easy to graph. By plotting both lines, we find they cross at \( p = 1 \) and \( q = -1 \), which we then use to solve for our original variables \( x = 0 \) and \( y = -1 \).

🎯 Exam Tip: Always label your axes, write down the scale in the top right corner of your graph paper, and clearly mark the coordinates of the intersection point to secure full marks.

Question 1. Solve the following simultaneous equations: \( \frac{4}{x-y} + \frac{1}{x+y} = 3 \) and \( \frac{2}{x-y} - \frac{3}{x+y} = 5 \)
Answer: Let \( \frac{1}{x-y} = p \) and \( \frac{1}{x+y} = q \).
The equations become:
\( 4p + q = 3 \) --- (i)
\( 2p - 3q = 5 \) --- (ii)
Solving equations (i) and (ii), we get:
\( p = 1 \) and \( q = -1 \) is the solution of the simultaneous equations \( 4p + q = 3 \) and \( 2p - 3q = 5 \).
Re-substituting the values of \( p \) and \( q \), we get:
\( \frac{1}{x-y} = 1 \) and \( \frac{1}{x+y} = -1 \)
\( \implies x - y = 1 \)
\( \implies x + y = -1 \)
Now, let us find the coordinates to plot these two linear equations on a graph.
For \( x - y = 1 \)
\( \implies y = x - 1 \)

For \( x + y = -1 \)
\( \implies y = -1 - x \)

Plotting these points on a coordinate plane, we draw the lines representing \( x - y = 1 \) and \( x + y = -1 \).
The two lines intersect at point \( (0, -1) \).
\( \implies x = 0 \) and \( y = -1 \) is the solution of the simultaneous equations \( x - y = 1 \) and \( x + y = -1 \).
\( \implies (x, y) = (0, -1) \) is the solution of the given simultaneous equations. This graphical intersection confirms our algebraic substitution perfectly.
In simple words: To solve these complex equations, we temporarily replace the fraction terms with simpler variables like p and q. After finding their values, we convert them back to x and y equations and plot them on a graph to find where they cross.

🎯 Exam Tip: Always write the scale on the top right corner of your graph paper and clearly label the intersection point with its coordinates to secure full marks.