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

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.

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

3. 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.

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

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

6. 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: \[x\text{ }=\text{ }3\]

2. 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 \[\frac{{{a}_{1}}}{{{a}_{2}}}\,\,=\,\,\frac{{{b}_{1}}}{{{b}_{2}}}\,\,=\,\,\frac{{{c}_{1}}}{{{c}_{2}}}\]

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

3. 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: \[y\text{ }=\text{ }-\text{ }1\]

4. 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: \[-\,by\text{ }=\text{ }ab~\,\,\,\,\Rightarrow \,\,\,\,~y\text{ }=-\text{ }a\]

5. 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: \[y\text{ }=\text{ }b\]