# 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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