Pair of Linear Equations in Two Variables
Category : 10th Class
Pair of Linear Equations in Two Variables
Pair of Linear Equations in Two Variables
(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
\[\mathbf{x+}\frac{\mathbf{6}}{\mathbf{y}}\mathbf{=6;}\,\,\,\,\mathbf{3x-}\frac{\mathbf{8}}{\mathbf{y}}\mathbf{=5}\]
(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: \[x\text{ }=\text{ }3\]
\[\mathbf{kx}\text{ }-\text{ }\mathbf{4y}\text{ }=\text{ }\mathbf{3};\text{ }\mathbf{6x}\text{ }-\text{ }\mathbf{12y}\text{ }=\text{ }\mathbf{9}\] 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 \[\frac{{{a}_{1}}}{{{a}_{2}}}\,\,=\,\,\frac{{{b}_{1}}}{{{b}_{2}}}\,\,=\,\,\frac{{{c}_{1}}}{{{c}_{2}}}\]
\[ie.\,\,\,\,\,\,\,\,\,\frac{k}{6}=\frac{1}{3}=\frac{1}{3}\,\,\,\,\,or\,\,k\,\,=\,\,2\]
\[\mathbf{47x}\text{ }+\text{ }\mathbf{31y}\text{ }=\text{ }\mathbf{63};\text{ }\mathbf{31x}\text{ }+\text{ }\mathbf{47y}\text{ }=\text{ }\mathbf{15}\]
(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: \[y\text{ }=\text{ }-\text{ }1\]
\[\frac{\mathbf{ax}}{\mathbf{b}}\,\,\mathbf{-}\,\,\frac{\mathbf{by}}{\mathbf{a}}\mathbf{=a}\,\,\mathbf{+}\,\,\mathbf{b;}\,\,\mathbf{ax}\,\,\mathbf{-}\,\,\mathbf{by}\,\,\mathbf{=}\,\,\mathbf{2ab}\]
(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: \[-\,by\text{ }=\text{ }ab~\,\,\,\,\Rightarrow \,\,\,\,~y\text{ }=-\text{ }a\]
\[\mathbf{ }\frac{\mathbf{x}}{\mathbf{a}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{b}}\mathbf{=2;}\,\,\mathbf{ax}\,\,\mathbf{-}\,\,\mathbf{by}\,\,\mathbf{=}\,\,{{\mathbf{a}}^{\mathbf{2}}}\,\,\mathbf{-}\,\,{{\mathbf{b}}^{\mathbf{2}}}\]
(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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