10th Class Mathematics Pair of Linear Equations in Two Variables Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables

Category : 10th Class

 

Pair of Linear Equations in Two Variables

 

Pair of Linear Equations in Two Variables

 

  1. Linear equation in two variables: An equation which can be put in the form \[ax\text{ }+\text{ }by\text{ }+\text{ }c\text{ }=\text{ }0\], where a, b, c are real numbers \[(a~\,\,\ne \,\,0,\text{ }b\,\,\ne \,\,~0)\] is called a linear equation in two variables x and y.

 

  1. Simultaneous linear equations in two variables: A pair of linear equations in two variables is said to form a          system of simultaneous linear equation.

 

  1. Solution of a given system of two simultaneous equations: A pair of value of the variable x and y satisfying each of the equations in a given system of two simultaneous equations in x and y is called a solution of the system.

 

  1. Consistent system: A system of simultaneous linear equations. Is said to be consistent if it has at least one solution.

 

  1. Inconsistent system: A system of simultaneous linear equations is said to be inconsistent if it has no solution.

 

  1. If a pair of linear equations is given by \[{{a}_{1}}x\text{ }+\text{ }{{b}_{1}}y\text{ }+\text{ }{{c}_{1}}=\text{ }0\] and \[{{a}_{2}}x\text{ }+\text{ }{{b}_{2}}y\text{ }+\text{ }{{c}_{2}}=\text{ }0\] then the following situations can arise:

(i)  if \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\], the pair of linear equations is consistent.

(ii) if \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\] the pair of linear equations is inconsistent.

(iii) if \[\frac{{{a}_{1}}}{{{a}_{2}}}\,\,=\,\,\frac{{{b}_{1}}}{{{b}_{2}}}\,\,=\,\,\frac{{{c}_{1}}}{{{c}_{2}}}\] the pair of linear equations is dependent and consistent.

 

Snap Test

 

 

  1. Find the value of x and y from the following equations.

\[\mathbf{x+}\frac{\mathbf{6}}{\mathbf{y}}\mathbf{=6;}\,\,\,\,\mathbf{3x-}\frac{\mathbf{8}}{\mathbf{y}}\mathbf{=5}\]

            (a) \[x\text{ }=\text{ }3,\text{ }y\text{ }=\text{ }2\]       

(b) \[x\text{ }=\text{ }2,\text{ }y\text{ }=\text{ }5\]

            (c) \[x\text{ }=\text{ }7,\text{ }y\text{ }=\text{ }3\]       

(d) \[x\,\,\text{=}\,\,4,\text{ }y\,\,=\,\,6\]

            (e) None of these

Ans.     (a)

            Explanation: Given equations are

            \[x+\frac{6}{y}=6\]      ..... (i)        and  \[3x-\frac{8}{y}=5\]         ..... (ii)

            Putting \[\frac{1}{y}\,\,=\,\,z\] in (i) and (ii), we get:

                        \[x\text{ }+\text{ }6z\text{ }=\text{ }6\]                                     ..... (iii)

                        \[3x\text{ }-\text{ }8z\text{ }=\text{ }5\]                         ….. (iv)

 

            Multiplying (iii) by 3 and subtracting (iv) from it we get:

            \[26z\,\,=\,\,13~~~\Rightarrow \,\,~~~z=\frac{1}{2}\]

            \[\therefore \,\,\,\,z=\frac{1}{y}\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,y\,\,=\,\,\frac{1}{z}\,\,=\,\,2\]

            Substituting \[y\text{ }=\text{ }2\] in (i) we get: \[x\text{ }=\text{ }3\]

 

  1. Find the value of k for which the system of equations:

\[\mathbf{kx}\text{ }-\text{ }\mathbf{4y}\text{ }=\text{ }\mathbf{3};\text{ }\mathbf{6x}\text{ }-\text{ }\mathbf{12y}\text{ }=\text{ }\mathbf{9}\] has an infinite number of solutions.

            (a) \[k\text{ }=\text{ }5\]                       

(b) \[k\text{ }=\text{ }6\]

            (c) \[k\text{ }=\text{ }2\]                       

(d) \[k\text{ }=\text{ }4\]

            (e) None of these

Ans.     (c)

Explanation: From the given equation: \[{{a}_{1}}=\text{ }k,\text{ }{{b}_{1}}=\text{ }-\text{ }4,\text{ }{{c}_{1}}=\text{ }-\text{ }3\] and \[{{a}_{2}}=\text{ }6,\text{ }{{b}_{2}}=\text{ }-\text{ }12,\text{ }{{c}_{2}}=\text{ }-9\]

 

We know that for the infinite number of solutions we must have \[\frac{{{a}_{1}}}{{{a}_{2}}}\,\,=\,\,\frac{{{b}_{1}}}{{{b}_{2}}}\,\,=\,\,\frac{{{c}_{1}}}{{{c}_{2}}}\]

            \[ie.\,\,\,\,\,\,\,\,\,\frac{k}{6}=\frac{1}{3}=\frac{1}{3}\,\,\,\,\,or\,\,k\,\,=\,\,2\]

 

  1. Solve for x and y:

\[\mathbf{47x}\text{ }+\text{ }\mathbf{31y}\text{ }=\text{ }\mathbf{63};\text{ }\mathbf{31x}\text{ }+\text{ }\mathbf{47y}\text{ }=\text{ }\mathbf{15}\]

            (a) \[x\text{ }=\text{ }2,\text{ }y\text{ }=-1\]                 

(b) \[x\text{ }=\text{ }1,\text{ }y\text{ }=-2\]

            (c) \[x\text{ }=\text{ }4,\text{ }y\text{ }=+1\]  

(d) \[x\text{ }=\text{ }2,\text{ }y\text{ }=+1\]

            (e) None of these

Ans.     (a)

            We have the equations:

                \[47x\text{ }+\text{ }31y\text{ }=\text{ }63\]                                                     .…….. (i)

                \[31x\text{ }+\text{ }47y\text{ }=\text{ }15\]                                                     …….. (ii)

            Adding (i) and (ii) we get: \[78\left( x\text{ }+\text{ }y \right)\text{ }=\text{ }78~~~~~~\]             \[\Rightarrow \]    \[x\text{ }+\text{ }y\text{ }=\text{ }1\]  …..... (iii)

Subtracting (ii) from (i) we get: \[16\left( x\text{ }-\text{ }y \right)\text{ }=\text{ }48~\]         \[\Rightarrow \]    \[x\text{ }-\text{ }y\text{ }=\text{ }3\]                …….. (iv)

Adding (iii) and (iv) we get: \[2x\text{ }=\text{ }4\]                        \[\Rightarrow \]    \[x\text{ }=\text{ }2.\]

            Substituting \[x\text{ }=\text{ }2\] in (i) we get: \[y\text{ }=\text{ }-\text{ }1\]

 

  1. Find the value of x and y from the following equations:

 

\[\frac{\mathbf{ax}}{\mathbf{b}}\,\,\mathbf{-}\,\,\frac{\mathbf{by}}{\mathbf{a}}\mathbf{=a}\,\,\mathbf{+}\,\,\mathbf{b;}\,\,\mathbf{ax}\,\,\mathbf{-}\,\,\mathbf{by}\,\,\mathbf{=}\,\,\mathbf{2ab}\]

 

            (a) \[x\,\,=\,\,a,~\,\,y=-\text{ }b\]          

(b) \[x\text{ }=\text{ }b,\text{ }y\text{ }=-a\]

            (c) \[x\text{ }=\text{ }b,\text{ }y\text{ }=\text{ }a\]      

(d) \[x\text{ }=\text{ }-\text{ }b,\text{ }y\text{ }=\text{ }-\text{ }a\]

            (e) None of these

Ans.     (b)

            Explanation: The given equations can be written as:

                        \[{{a}^{2}}\times {{b}^{2}}y\,\,=\,\,ab\left( a+b \right)\]                                     …….. (i)

                        \[ax-by=2ab\]                            …….. (ii)

            Multiplying (ii) by b and subtracting from (i) we get:

                        \[\left( {{a}^{2}}\text{ }ab \right)x=ab\left( a\text{ }+\text{ }b \right)-2a{{b}^{2}}\]

            \[\Rightarrow ~~a\left( a-b \right)x=\text{ }ab\left( a-b \right)\text{ }\Rightarrow \text{ }x\text{ }=\text{ }b\]

Subtracting \[x\text{ }=\text{ }b\] in (ii) we get: \[-\,by\text{ }=\text{ }ab~\,\,\,\,\Rightarrow \,\,\,\,~y\text{ }=-\text{ }a\]

 

  1. Find the value of x and y from the following equations.

 

\[\mathbf{ }\frac{\mathbf{x}}{\mathbf{a}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{b}}\mathbf{=2;}\,\,\mathbf{ax}\,\,\mathbf{-}\,\,\mathbf{by}\,\,\mathbf{=}\,\,{{\mathbf{a}}^{\mathbf{2}}}\,\,\mathbf{-}\,\,{{\mathbf{b}}^{\mathbf{2}}}\]

 

            (a) \[x=-\text{ }a,\text{ }\,\,y\text{ }=\text{ }b\]             

(b) \[x\text{ }=\text{ }a,\,\,\text{ }y=-b\]

            (c) \[x\text{ }=\text{ }a,\text{ }\,\,y\text{ }=\text{ }b\]               

(d) \[x=-\text{ }a,\,\,\text{ }y=-\text{ }b\]

            (e) None of these

Ans.     (c)

            Explanation: The given equations can be written as:

            \[bx\text{ }+\text{ }ay\text{ }=\text{ }2ab\]                               ….. (i)

            \[ac\text{ }\text{ }by\text{ }=\text{ }{{a}^{2}}\text{ }-\text{ }{{b}^{2}}\]                   …... (ii)

            Multiplying (i) by b and (ii) by a and adding we get:

\[\left( {{a}^{2}}+\text{ }{{b}^{2}} \right)x\text{ }=\text{ }2a{{b}^{2}}+\text{ }a\left( {{a}^{2}}-\text{ }{{b}^{2}} \right)\,\,\,\,\,\Rightarrow \,\,\,\,~~\left( {{a}^{2}}+\text{ }{{b}^{2}} \right)x\text{ }=\text{ }a\left( {{a}^{2}}+\text{ }{{b}^{2}} \right)~~\Rightarrow ~\,\,\,\,\,~x\text{ }=\text{ }a\]

            Substituting \[x\text{ }=\text{ }a\] in (i) we get: \[y\text{ }=\text{ }b\]


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