A statement of equality of two algebraic expressions, which involve one or more unknown quantities is known as an equation. If there are two unknown quantities then equation is called linear equation in two variables. A linear equation is an equation which involves linear polynomials. A value of the variable which makes the two sides of the equation equal is called the solution of the equation. Same quantity can be added/subtracted to/from both the sides of an equation without changing the equality. Both the sides of an equation can be multiplied/divided by the same non-zero number without changing the equality. Note :- To find value of variables in any equation we required number of equation equal to number of variables in equation.
Graph Of Linear Equation ax + by + c = 0 in Two Variables, where a ≠ 0, b ≠ 0
Step I: Obtain the linear equation, let the equation be ax + by + c = 0. Step II: Express y in terms of x to obtain \(y=-\left( \frac{ax+c}{b} \right)\) Step III: Give any two values to x and calculate the corresponding values of y from the expression in step II to obtain two solutions, say (α1, β1) and (α2, β2). If possible take values of x as integers in such a manner that the corresponding values of y are also integers. Step IV: Plot points (α1, β1) and (α2, β2) on a graph paper. Step V: Join the points marked in step IV to obtain a line. The line obtained is the graph of the equation ax + by + c = 0.
Graph Of Linear Equation ax + by + c = 0 in Two Variables Example Problems With Solutions
Example 1: Draw the graph of the equation y – x = 2. Solution: We have, y – x = 2 ⇒ y = x + 2 When x = 1, y = 1 + 2 = 3 When x = 3, y = 3 + 2 = 5 Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation. Plotting the points (1, 3) and (3, 5) on the graph paper and drawing a line joining them, we obtain the graph of the line represented by the given equation as shown in Fig.
Example 2: Draw a graph of the line x – 2y = 3. From the graph, find the coordinates of the point when (i) x = – 5 (ii) y = 0. Solution: We have x – 2y = 3 ⇒ \(y=\frac { x-3 }{ 2 }\) When x = 1, y = = –1 When x = –1, y = = –2 Plotting points (1, –1) & (–1, –2) on graph paper & joining them, we get straight line as shown in fig. This line is required graph of equation x – 2y = 3. To find the coordinates of the point when x = –5, we draw a line parallel to y-axis and passing through (–5, 0). This line meets the graph of x – 2y = 3 at a point from which we draw a line parallel to x-axis which crosses y-axis at y = –4. So, the coordinates of the required point are (–5, –4). Since y = 0 on x-axis. So, the required point is the point where the line meets x-axis. From the graph the coordinates of such point are (3, 0). Hence, required points are (–5, –4) and (3, 0).
Example 3: Draw the graph of (i) x – 7y = – 42 (ii) x – 3y = 6 (iii) x – y + 1 = 0 (iv) 3x + 2y = 12 Solution: Note: (i) The graph of any linear equation is a line and every solution of equations lies on the graph of that equation. (ii) If a point (a, b) is not on the line then this point is not a solution of given equation. ∵ (2, 9) and (4, 1) are on the line ∴ These two points are solution of given equation But (1, 10) and (7, –4) are not on the line so these two are not solutions.
Example 3: If (9/2, 6) is lies on graph of 4x + ky = 12 then find value of k. Solution: Note: (1) Equation of x-axis is y = 0 and any point in ordered pair form which is on the x axis is (±a, 0). (2) Equation of y axis is x = 0 and any point on y axis is (0, ±b) (3) Graph of line x = ±a is parallel to y axis (4) Graph of line y = ±b is parallel to x axis Concurrent lines: Three or more lines are called concurrent if all lines passes through a common point. These all lines a, b, c, d, e are passes through O. ∴ These are concurrent lines Note: From a point there are infinite lines can pass, so we can find (or make) infinite equations of lines which passes through a given point.
Example 4: Find five equations of lines which passes through (3, –5). Solution: x + y = –2, x – y = 8, 2x + y = 1, 2x – y = 11, 2x + 3y + 9 = 0
By the method of elimination by substitution, only those equations can be solved, which have unique solution. But the method of cross multiplication discussed below is applicable in all the cases; whether the system has a unique solution, no solution or infinitely many solutions. Let us solve the following system of equations a1x + b1y + c1 = 0 ….(1) a2x + b2y + c2 = 0 ….(2) Multiplying equation (1) by b2 and equation (2) by b1, we get a1b2x + b1b2y + b2c1 = 0 ….(3) a2b1x + b1b2y + b1c2 = 0 ….(4) Subtracting equation (4) from equation (3), we get (a1b2 – a2b1) x + (b2c1 – b1c2) = 0 \(\Rightarrow x=\frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\) \(\left[ {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}\ne 0\text{ and }\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}} \right]\) \(\text{Similarly, }y=\frac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\) These values of x and y can also be written as \(\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\)
Cross Multiplication Method Examples
Example 1: Solve the following system of equations by cross-multiplication method. 2x + 3y + 8 = 0 4x + 5y + 14 = 0 Sol. The given system of equations is 2x + 3y + 8 = 0 4x + 5y + 14 = 0 By cross-multiplication, we get \(\Rightarrow \frac{x}{3\times 14-5\times 8}=\frac{x}{3\times 14-5\times 8}=\frac{1}{2\times 5-4\times 3}\) \(\Rightarrow \frac{x}{42-40}=\frac{-y}{28-32}=\frac{1}{10-12} \) \(\Rightarrow \frac{x}{2}=\frac{-y}{-4}=\frac{1}{-2} \) \(\Rightarrow \frac { x }{ 2 }\) = \(\frac { -1 }{ 2 }\) ⇒ x = – 1 \(\Rightarrow \frac { -y }{ -4 }\) = \(\frac { -1 }{ 2 }\) ⇒ y = – 2 Hence, the solution is x = – 1, y = – 2 We can verify the solution.
Example 2: Solve the follownig system of equations by the method of cross-multiplication. 2x – 6y + 10 = 0 3x – 7y + 13 = 0 Sol. The given system of equations is 2x – 6y + 10 = 0 ….(1) 3x – 7y + 13 = 0 ….(2) Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{-6\times 13-(-7)\times 10}=\frac{-y}{2\times 13-3\times 10}=\frac{1}{2\times (-7)-3\times (-6)} \) \(\Rightarrow \frac{x}{78+70}=\frac{-y}{26-30}=\frac{1}{-14+18} \) \(\Rightarrow \frac{x}{-8}=\frac{-y}{-4}=\frac{1}{4} \) \(\Rightarrow \frac { x }{ -8 }\) = \(\frac { 1 }{ 4 }\) ⇒ x = – 2 \(\Rightarrow \frac { -y }{ -4 }\) = \(\frac { 1 }{ 4 }\) ⇒ y = 1 Hence, the solution is x = – 2, y = 1 Example 3: Solve the following system of equations by the method of cross-multiplication. 11x + 15y = – 23; 7x – 2y = 20 Sol. The given system of equations is 11x + 15y + 23 = 0 7x – 2y – 20 = 0 Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{15\times (-20)-(-2)\times 23}=\frac{-y}{11\times (-20)-7\times 23}=\frac{1}{11\times (-2)-7\times 15}\) \(\Rightarrow \frac{x}{-300+46}=\frac{-y}{-220-161}=\frac{1}{-22-105} \) \(\Rightarrow \frac{x}{-254}=\frac{-y}{-381}=\frac{1}{-127} \) \(\Rightarrow \frac{x}{-254}=\frac{1}{-127}\Rightarrow x=2 \) \(\text{and}\frac{-y}{-381}=\frac{1}{-127}\Rightarrow \text{y}=\text{ }-3 \) Hence, x = 2, y = – 3 is the required solution.
Example 4: Solve the following system of equations by cross-multiplication method. ax + by = a – b; bx – ay = a + b Sol. Rewriting the given system of equations, we get ax + by – (a – b) = 0 bx – ay – (a + b) = 0 Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{b\times \{-(a+b)\}-(-a)\times \{-(a-b)\}}=\frac{-y}{-a(a+b)+b(a-b)}=\frac{1}{-{{a}^{2}}-{{b}^{2}}} \) \(\Rightarrow \frac{x}{-ab-{{b}^{2}}-{{a}^{2}}+ab}=\frac{-y}{-{{a}^{2}}-ab+ab-{{b}^{2}}}=\frac{1}{-({{a}^{2}}+{{b}^{2}})} \) \(\Rightarrow \frac{x}{-({{a}^{2}}+{{b}^{2}})}=\frac{-y}{-({{a}^{2}}+{{b}^{2}})}=\frac{1}{-({{a}^{2}}+{{b}^{2}})} \) \(\Rightarrow \frac{x}{-({{a}^{2}}+{{b}^{2}})}\frac{1}{-({{a}^{2}}+{{b}^{2}})}\Rightarrow x=1 \) \(and\text{ }\frac{-y}{-({{a}^{2}}+{{b}^{2}})}\frac{1}{-({{a}^{2}}+{{b}^{2}})}\Rightarrow y=-1 \)
Example 5: Solve the following system of equations by cross-multiplication method. x + y = a – b; ax – by = a2 + b2 Sol. The given system of equations can be rewritten as: x + y – (a – b) = 0 ax – by – (a2 + b2) = 0 Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{-({{a}^{2}}+{{b}^{2}})-(-b)\times \{-(a-b)\}}=\frac{-y}{-({{a}^{2}}+{{b}^{2}})-a\times \{-(a-b)\}}=\frac{1}{-b-a} \) \(\Rightarrow \frac{x}{-({{a}^{2}}+{{b}^{2}})-b(a-b)}=\frac{-y}{-({{a}^{2}}+{{b}^{2}})+a(a-b)}=\frac{1}{-(b+a)} \) \(\Rightarrow \frac{x}{-{{a}^{2}}-{{b}^{2}}-ab+{{b}^{2}}}=\frac{-y}{-{{a}^{2}}-{{b}^{2}}+{{a}^{2}}-ab}=\frac{1}{-(a+b)} \) \(\Rightarrow \frac{x}{-a(a+b)}=\frac{-y}{-b(a+b)}=\frac{1}{-(a+b)} \) \(\Rightarrow \frac{x}{-a(a+b)}=\frac{1}{-(a+b)}\Rightarrow x=a \) \(and\text{ }\frac{-y}{-b(a+b)}=\frac{1}{-(a+b)}\Rightarrow y=-b \)
Example 6: Solve the following system of equations by the method of cross-multiplication: \(\frac{x}{a}+\frac{y}{b}=a+b \) ; \(\frac{x}{{{a}^{2}}}+\frac{y}{{{b}^{2}}}=2 \) Sol: The given system of equations is rewritten as: \(\frac{x}{a}+\frac{y}{b}-\left( a+b \right)\) ….(1) \(\frac{x}{{{a}^{2}}}+\frac{y}{{{b}^{2}}}-2 \) ….(2) Multiplying equation (1) by ab, we get bx + ay – ab (a + b) = 0 ….(3) Multiplying equation (2) by a2 b2, we get b2x + a2y – 2a2b2 = 0 ….(4) Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{-2{{a}^{3}}{{b}^{2}}+{{a}^{3}}b(a+b)}=\frac{-y}{-2{{a}^{2}}{{b}^{3}}+a{{b}^{3}}(a+b)}=\frac{1}{{{a}^{2}}b-a{{b}^{2}}} \) \(\Rightarrow \frac{x}{-2{{a}^{3}}{{b}^{2}}+{{a}^{4}}b+{{a}^{3}}{{b}^{2}}}=\frac{y}{-2{{a}^{2}}{{b}^{3}}+{{a}^{2}}{{b}^{3}}+a{{b}^{4}}}=\frac{1}{ab(a-b)} \) \(\Rightarrow \frac{x}{{{a}^{4}}b-{{a}^{3}}{{b}^{2}}}=\frac{-y}{a{{b}^{4}}-{{a}^{2}}{{b}^{3}}}=\frac{1}{ab(a-b)} \) \(\Rightarrow \frac{x}{{{a}^{3}}b(a-b)}=\frac{y}{a{{b}^{3}}(a-b)}=\frac{1}{ab(a-b)} \) \(\Rightarrow \frac{x}{{{a}^{3}}b(a-b)}=\frac{1}{ab(a-b)} \) \(\Rightarrow x=\frac{{{a}^{3}}b(a-b)}{ab(a-b)}={{a}^{2}} \) \(And\text{ }\frac{y}{a{{b}^{3}}(a-b)}=\frac{1}{ab(a-b)} \) \(\Rightarrow y=\frac{a{{b}^{3}}(a-b)}{ab(a-b)}={{b}^{2}} \) Hence, the solution x = a2, y = b2
Example 7: Solve the following system of equations by cross-multiplication method – ax + by = 1; bx + ay = \(\frac{{{(a+b)}^{2}}}{{{a}^{2}}+{{b}^{2}}}-1\) Sol: The given system of equations can be written as ax + by – 1 = 0 ….(1) \(bx+ay=\frac{{{(a+b)}^{2}}}{{{a}^{2}}+{{b}^{2}}}-1 \) \(\Rightarrow bx+ay=\frac{{{a}^{2}}+2ab+{{b}^{2}}-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \) \(\Rightarrow bx+ay=\frac{2ab}{{{a}^{2}}+{{b}^{2}}} \) \(\Rightarrow bx+ay-\frac{2ab}{{{a}^{2}}+{{b}^{2}}}=0 \) ….. (2) Rewritting the equations (1) and (2), we have ax + by – 1 = 0 \(\Rightarrow bx+ay-\frac{2ab}{{{a}^{2}}+{{b}^{2}}}=0 \) Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{b\times \left( \frac{-2ab}{{{a}^{2}}+{{b}^{2}}} \right)-a\times (-1)}=\frac{-y}{a\times \left( \frac{-2ab}{{{a}^{2}}+{{b}^{2}}} \right)-b\times (-1)}=\frac{1}{a\times a-b\times b} \) \(\Rightarrow \frac{x}{-\frac{2a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}+a}=\frac{-y}{\frac{-2{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}}+b}=\frac{1}{{{a}^{2}}-{{b}^{2}}} \) \(\Rightarrow \frac{x}{\frac{-2a{{b}^{2}}+{{a}^{3}}+a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}=\frac{-y}{\frac{-2{{a}^{2}}b+{{a}^{2}}b+{{b}^{3}}}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}} \) \(\Rightarrow \frac{x}{\frac{a({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{-y}{\frac{b({{b}^{2}}-{{a}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}} \) \( \Rightarrow \frac{x}{\frac{a({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}}\Rightarrow x=\frac{a}{{{a}^{2}}+{{b}^{2}}} \) \(and\text{ }\frac{-y}{\frac{b({{b}^{2}}-{{a}^{2}})}{{{a}^{2}}+{{b}^{2}}}}=\frac{1}{{{a}^{2}}-{{b}^{2}}}\Rightarrow y=\frac{b}{{{a}^{2}}+{{b}^{2}}} \) Hence, the solution is \(x=\frac{a}{{{a}^{2}}+{{b}^{2}}},y=\frac{b}{{{a}^{2}}+{{b}^{2}}} \)
Example 8: Solve the following system of equations in x and y by cross-multiplication method (a – b) x + (a + b) y = a2 – 2ab – b2 (a + b) (x + y) = a2 + b2 Sol: The given system of equations can be rewritten as : (a – b) x + (a +b) y – (a2 – 2ab – b2) = 0 (a + b) x + (a + b) y – (a2 + b2) = 0 Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{(a+b)\times \{-({{a}^{2}}+{{b}^{2}})\}-(a+b)\times \{-({{a}^{2}}-2ab-{{b}^{2}})\}}=\frac{-y}{(a-b)\times \{-({{a}^{2}}+{{b}^{2}})\}-(a+b)\times \{-({{a}^{2}}-2ab-{{b}^{2}})\}}=\frac{1}{(a-b)\times (a+b)-(a+b)\times (a+b)} \) \(\Rightarrow \frac{x}{-(a+b)({{a}^{2}}+{{b}^{2}})+(a+b)({{a}^{2}}-2ab-{{b}^{2}})}=\frac{-y}{-(a-b)({{a}^{2}}+{{b}^{2}})+(a+b)({{a}^{2}}-2ab-{{b}^{2}})}=\frac{1}{(a-b)(a+b)-{{(a+b)}^{2}}} \) \(\Rightarrow \frac{x}{(a+b)[-({{a}^{2}}+{{b}^{2}})+(a+b)({{a}^{2}}-2ab-{{b}^{2}})]}=\frac{-y}{(a+b)({{a}^{2}}-2ab-{{b}^{2}})-(a-b)({{a}^{2}}+{{b}^{2}})}=\frac{1}{(a+b)(a-b-a-b)} \) \(\Rightarrow \frac{x}{(a+b)(-2ab-2{{b}^{2}})}=\frac{-y}{{{a}^{3}}-{{a}^{2}}b-3a{{b}^{2}}-{{b}^{3}}-{{a}^{3}}-a{{b}^{2}}+{{a}^{2}}b+{{b}^{3}}}=\frac{1}{(a+b)(-2b)} \) \(\Rightarrow \frac{x}{-(a+b)(2a+2b)b}=\frac{-y}{-4a{{b}^{2}}}=\frac{1}{-2b(a+b)} \) \(\Rightarrow \frac{x}{-2(a+b)(a+b)b}=\frac{1}{-2b(a+b)}\Rightarrow x=a+b \) \(and\text{ }\frac{-y}{-4a{{b}^{2}}}=\frac{1}{-2b(a+b)}\Rightarrow y=\frac{2ab}{a+b} \) Hence, the solution of the given system of equations is x = a + b, \(y=\frac{2ab}{a+b} \)
Example 9: Solve the following system of equations by cross-multiplications method. a(x + y) + b (x – y) = a2 – ab + b2 a(x + y) – b (x – y) = a2 + ab + b2 Sol: The given system of equations can be rewritten as ax + bx + ay – by – ( a2 – ab + b2) = 0 ⇒ (a + b) x + (a – b) y – (a2 – ab + b2) = 0 ….(1) And ax – bx + ay + by – (a2 + ab + b2) = 0 ⇒ (a – b) x + (a + b) y – (a2 + ab + b2) = 0 …(2) Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{x}{(a-b)\times \{-({{a}^{2}}+ab+{{b}^{2}})\}-(a+b)\times \{-({{a}^{2}}-ab+{{b}^{2}})\}}=\frac{-y}{(a+b)\times \{-({{a}^{2}}+ab+{{b}^{2}})\}-(a-b)\times \{-({{a}^{2}}-ab+{{b}^{2}})\}}=\frac{1}{(a+b)\times (a+b)-(a-b)(a-b)} \) \(\Rightarrow \frac{x}{-(a-b)({{a}^{2}}+ab+{{b}^{2}})+(a+b)({{a}^{2}}-ab+{{b}^{2}})}=\frac{-y}{-(a+b)({{a}^{2}}+ab+{{b}^{2}})+(a-b)({{a}^{2}}-ab+{{b}^{2}})}=\frac{1}{{{(a+b)}^{2}}-{{(a-b)}^{2}}}\) \(\Rightarrow \frac{x}{-({{a}^{3}}-{{b}^{3}})+({{a}^{3}}+{{b}^{2}})}=\frac{-y}{-{{a}^{3}}-2{{a}^{2}}b-2a{{b}^{2}}-{{b}^{3}}+{{a}^{3}}-2{{a}^{2}}b+2a{{b}^{2}}-{{b}^{3}}}=\frac{1}{{{a}^{2}}+2ab+{{b}^{2}}-{{a}^{2}}+2ab-{{b}^{2}}}\) \(\Rightarrow \frac{x}{2{{b}^{3}}}=\frac{-y}{-4{{a}^{2}}b-2{{b}^{3}}}=\frac{1}{4ab} \) \(\Rightarrow \frac{x}{2{{b}^{3}}}=\frac{-y}{-2b(2{{a}^{2}}+{{b}^{2}})}=\frac{1}{4ab}\) \(\Rightarrow \frac{x}{2{{b}^{3}}}=\frac{1}{4ab}\Rightarrow x=\frac{{{b}^{2}}}{2a}\) \(and\text{ }\frac{-y}{-2b(2{{a}^{2}}+{{b}^{2}})}=\frac{1}{4ab}\Rightarrow y=\frac{2{{a}^{2}}+{{b}^{2}}}{2a}\) Hence, the solution is \(x=\frac{{{b}^{2}}}{2a},y=\frac{2{{a}^{2}}+{{b}^{2}}}{2a}\)
Example 10: Solve the following system of equations by the method of cross-multiplication. \(\frac{a}{x}-\frac{b}{y}=0;\text{ }\frac{a{{b}^{2}}}{x}+\frac{{{a}^{2}}b}{y}={{a}^{2}}+{{b}^{2}};\) Where x ≠ 0, y ≠ 0 Sol: The given system of equations is \(\frac{a}{x}-\frac{b}{y}=0\) ………(1) \(\frac{a{{b}^{2}}}{x}+\frac{{{a}^{2}}b}{y}-\left( {{a}^{2}}+{{b}^{2}} \right)=0\) ………(2) Putting \(\frac { a }{ x }=u\) and \(\frac { b }{ y }=v\) in equatinos (1) and (2) the system of equations reduces to u – v + 0 = 0 b2u + a2v – (a2 + b2) = 0 Using formula for cross multiplication method: So, from equation (1) and (2) we can write the value of a,b and c. \(\Rightarrow \frac{u}{{{a}^{2}}+{{b}^{2}}-{{a}^{2}}\times 0}=\frac{-v}{-({{a}^{2}}+{{b}^{2}})-{{b}^{2}}\times 0}=\frac{1}{{{a}^{2}}-(-{{b}^{2}})}\) \(\Rightarrow \frac{u}{{{a}^{2}}+{{b}^{2}}}=\frac{-v}{-({{a}^{2}}+{{b}^{2}})}=\frac{1}{{{a}^{2}}+{{b}^{2}}} \) \(\Rightarrow \frac{u}{{{a}^{2}}+{{b}^{2}}}=\frac{1}{{{a}^{2}}+{{b}^{2}}}\Rightarrow u=1 \) \(and\text{ }\frac{-v}{-({{a}^{2}}+{{b}^{2}})}=\frac{1}{{{a}^{2}}+{{b}^{2}}}\Rightarrow v=1 and\text{ u}=\frac{a}{x}=1\Rightarrow x=a \) \(v=\frac{b}{y}=1\Rightarrow y=b \) Hence, the solution of the given system of equations is x = a, y = b.
Equations Reducible To A Pair Of Linear Equations Examples
Example 1: Solve the following system of equations \(\frac { 1 }{ 2x }\) – \(\frac { 1 }{ y }\) = – 1; \(\frac { 1 }{ x }\) + \(\frac { 1 }{ 2y }\) = 8 Sol. We have, \(\frac { 1 }{ 2x }\) – \(\frac { 1 }{ y }\) = – 1 ….(1) \(\frac { 1 }{ x }\) + \(\frac { 1 }{ 2y }\) = 8 ….(2) Let us consider 1/x = u and 1/y = v. Putting 1/x = u and 1/y = v in the above equations, we get; \(\frac { u }{ 2 }\) – v = – 1 ….(3) u + \(\frac { v }{ 2 }\) = 8 ….(4) Let us eliminate v from the system of equations. So, multiplying equation (3) with 1/2 and (4) with 1, we get \(\frac { u }{ 4 }\) – \(\frac { v }{ 2 }\) = \(-\frac { 1 }{ 2 }\) ….(5) u + \(\frac { v }{ 2 }\) = 8 ….(6) Adding equation (5) and (6), we get ; \(\frac { u }{ 4 }\) + u = \(-\frac { 1 }{ 2 }\) + 8 ⇒ \(\frac { 5u }{ 4 }\) = \(\frac { 15 }{ 2 }\) ⇒ u = \(\frac { 15 }{ 2 }\) × \(\frac { 4 }{ 5 }\) ⇒ u = 6 We know, \(\frac { 1 }{ x }\) = u ⇒ \(\frac { 1 }{ x }\) = 6 ⇒ x = \(\frac { 1 }{ 6 }\) Putting 1/x = 6 in equation (2), we get ; 6 + \(\frac { 1 }{ 2y }\) = 8 ⇒ \(\frac { 1 }{ 2y }\) = 2 ⇒ \(\frac { 1 }{ y }\) = 4 ⇒ y = \(\frac { 1 }{ 4 }\) Hence, the solution of the system is, x = \(\frac { 1 }{ 6 }\) , y = \(\frac { 1 }{ 4 }\)
Example 2: Solve \(\frac { 2 }{ x }\) + \(\frac { 1 }{ 3y }\) = \(\frac { 1 }{ 5 }\); \(\frac { 3 }{ x }\) + \(\frac { 2 }{ 3y }\) = 2 and also find ‘a’ for which y = ax – 2. Sol. Considering 1/x = u and 1/y = v, the given system of equations becomes 2u + \(\frac { v }{ 3 }\) = \(\frac { 1 }{ 5 }\) ⇒ \(\frac { 6u+v }{ 3 }\) = \(\frac { 1 }{ 5 }\) 30u + 5v = 3 ….(1) 3u + \(\frac { 2v }{ 3 }\) = 2 ⇒ 9u + 2v = 6 ….(2) Multiplying equation (1) with 2 and equation (2) with 5, we get 60u + 10v = 6 ….(3) 45u + 10v = 30 ….(4) Subtracting equation (4) from equation (3), we get 15u = – 24 u = \(-\frac { 24 }{ 15 }\) = \(-\frac { 8 }{ 5 }\) Putting u = \(-\frac { 8 }{ 5 }\) in equation (2), we get; 9 × \(\frac { -8 }{ 5 }\) + 2v = 6 ⇒ \(\frac { -72 }{ 5 }\) + 2v = 6 ⇒ 2v = 6 + \(\frac { 72 }{ 5 }\) = \(\frac { 102 }{ 5 }\) ⇒ v = \(\frac { 51 }{ 5 }\) Here \(\frac { 1 }{ x }\) = u = \(\frac { -8 }{ 5 }\) ⇒ x = \(\frac { -5 }{ 8 }\) And, \(\frac { 1 }{ y }\) = v = \(\frac { 51 }{ 5 }\) ⇒ y = ⇒ \(\frac { 5 }{ 51 }\) Putting x = \(\frac { -5 }{ 8 }\) and y = \(\frac { 5 }{ 51 }\) in y = ax – 2, we get; \(\frac { 5 }{ 51 }\) = \(\frac { -5a }{ 8 }\) – 2 \(\frac { 5a }{ 8 }\) = – 2 – \(\frac { 5 }{ 51 }\) = \(\frac { -102-5 }{ 51 }\) = \(\frac { -107 }{ 51 }\) a = \(\frac { -107 }{ 51 }\) × \(\frac { 8 }{ 5 }\) = \(\frac { -856 }{ 255 }\) a = \(\frac { -856 }{ 255 }\)
Example 3: Solve \(\frac{2}{x+2y}+\frac{6}{2x-y}=4\text{ ; }\frac{5}{2\left( x+2y \right)}+\frac{1}{3\left( 2x-y \right)}=1\) where, x + 2y ≠ 0 and 2x – y ≠ 0 Sol. Taking \(\frac { 1 }{ x+2y }\) = u and \(\frac { 1 }{ 2x-y }\) = v, the above system of equations becomes 2u + 6v = 4 ….(1) \(\frac { 5u }{ 2 }\) + \(\frac { v }{ 3 }\) = 1 ….(2) Multiplying equation (2) by 18, we have; 45u + 6v = 18 ….(3) Now, subtracting equation (3) from equation (1), we get ; –43u = – 14 ⇒ u = \(\frac { 14 }{ 43 }\) Putting u = 14/43 in equation (1), we get 2 × \(\frac { 14 }{ 43 }\) + 6v = 4 ⇒ 6v = 4 – \(\frac { 28 }{ 43 }\) = \(\frac { 172-28 }{ 43 }\) ⇒ v = \(\frac { 144 }{ 43 }\) Now, u = \(\frac { 14 }{ 43 }\) = \(\frac { 1 }{ x+2y }\) ⇒ 14x + 28y = 43 ….(4) And, v = \(\frac { 144 }{ 43 }\) = \(\frac { 1 }{ 2x-y }\) ⇒ 288x – 144y = 43 ….(5) Multiplying equation (4) by 288 and (5) by 14, the system of equations becomes 288 × 14x + 28y × 288 = 43 × 288 288x × 14 – 144y × 14 = 43 × 4 ⇒ 4022x + 8064y = 12384 ….(6) 4022x – 2016y = 602 ….(7) Subtracting equation (7) from (6), we get 10080y = 11782 ⇒ y = 1.6(approx) Now, putting 1.6 in (4), we get, 14x + 28 × 1.6 = 63 ⇒ 14x + 44.8 = 63 ⇒ 14x = 18.2 ⇒ x = 1.3 (approx) Thus, solution of the given system of equation is x = 1.3 (approx), y = 1.6 (approx).
Example 4: Solve \(\frac{1}{x+y}+\frac{2}{x-y}=2\text{ and }\frac{2}{x+y}-\frac{1}{x-y}=3 \) where, x + y ≠ 0 and x – y ≠ 0 Taking \(\frac { 1 }{ x+y }\) = u and \(\frac { 1 }{ x-y }\) = v the above system of equations becomes u + 2v = 2 ….(1) 2u – v = 3 ….(2) Multiplying equation (1) by 2, and (2) by 1, we get; 2u + 4v = 4 ….(3) 2u – v = 3 ….(4) Subtracting equation (4) from (3), we get
5v = 1 ⇒ v = \(\frac { 1 }{ 5 }\) Putting v = 1/5 in equation (1), we get; u + 2 × \(\frac { 1 }{ 5 }\) = 2 ⇒ u = 2 – \(\frac { 2 }{ 5 }\) = \(\frac { 8 }{ 5 }\) Here, u = \(\frac { 8 }{ 5 }\) = \(\frac { 1 }{ x+y }\) ⇒ 8x + 8y = 5 ….(5) And, v = \(\frac { 1 }{ 5 }\) = \(\frac { 1 }{ x-y }\) ⇒ x – y = 5 ….(6) Multiplying equation (5) with 1, and (6) with 8, we get; 8x + 8y = 5 ….(7) 8x – 8y = 40 ….(8) Adding equation (7) and (8), we get; 16x = 45 ⇒ x = \(\frac { 45 }{ 16 }\) Now, putting the above value of x in equation (6), we get; \(\frac { 45 }{ 16 }\) – y = 5 ⇒ y = \(\frac { 45 }{ 16 }\) – 5 = \(\frac { -35 }{ 16 }\) Hence, solution of the system of the given equations is ; x = \(\frac { 45 }{ 16 }\) , y = \(\frac { -35 }{ 16 }\)
