# 10th Class Mathematics Pair of Linear Equations in Two Variables Condition for Consistency

Condition for Consistency

Category : 10th Class

### Condition for Consistency

For the system of linear equation${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}\,and\,{{a}_{2}}x+{{b}_{2}}\,y={{c}_{2}}$,

1. If $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$, then the system of equation has unique solution.
2. If $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$, then the system of equation has no solution.
3. If $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$ then the system of equation has infinitely many solution.

The types of solution the pair of linear equation $3x+4y=7\, and \,4x-3y=7$ have?

(a) Unique solution

(b) No solution

(c) Infinitely many solution

(d) All of these

(e) None of these

Which one of the following is the condition for infinitely many solution?

(a) ${{a}_{1}}{{a}_{2}}={{b}_{1}}{{b}_{2}}$

(b) $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$

(c) $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$

(d) $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$

(e) None of these

The value of m for which the given system of equation $mx-5y=10\, and \,8x=9y=24$ has no solution.

(a) $\left( m\ne \frac{40}{9}\And m=\frac{10}{3} \right)$

(b) $\left( m=\frac{40}{9}\And m\ne \frac{10}{3} \right)$

(c) $\left( m=\frac{40}{9}\And m=\frac{10}{3} \right)$

(d) $\left( m\ne \frac{40}{9}\And m\ne \frac{10}{3} \right)$

(e) None of these

Find the relation between m and n for which the system of equation $4x+6y=14\, and \,(m+n)x+(2m-n)y=21$, has unique solution.

(a) 2m = 3n

(b) m = 5n

(c) $2m\ne 3n$

(d) $m\ne 5n$

(e) None of these

The ratio of income of Mack and Jacob is 3 : 4 and the ratio of their expenditure is 1: 2. If their individual saving is Rs. 2000, then their monthly income is:

(a) (Rs. 3000 & Rs. 4000)

(b) (Rs. 2000 & Rs. 3000)

(c) (Rs. 4000 & Rs. 6000)

(d) (Rs. 1000 & Rs. 4000)

(e) None of these

Cross Multiplication Method

Solve the system of equations $2x+3y=17,3x-2y=6$ by the method of cross multiplication.

(a) $X=4\And y=-3$

(b) $X=2\And y=-5$

(c) $X=-4\And y=1$

(d) $X=-5\And y=7$

(e) None of these

Explanation

By cross multiplication, we have

$\therefore \,\,\frac{x}{\left[ 3\times (-6)-(-2)\times (-17) \right]}=\frac{y}{\left[ (-17)\times 3-(-6)\times 2 \right]}=\frac{1}{\left[ 2\times (-2)-3\times 3 \right]}$
$\Rightarrow \,\,\,\frac{x}{(-18-34)}=\frac{y}{(-51+12)}=\frac{1}{(-4-9)}$

$\Rightarrow \,\,\,\frac{x}{(-52)}=\frac{y}{(-39)}=\frac{1}{(-13)}=3$

$\Rightarrow \,\,x=\,\frac{-52}{-13}=4,y=\frac{-39}{13}=3$

Hence, $x=4\And y=3$ is the required solution

Solve the system of equation by cross multiplication method.

$4x-7y+28=0$

$5y-7x+9=0$

(a) $x=6\And y=3$

(b) $x=2\And y=8$

(c) $x=7\And y=8$

(d) $x=7\And y=7$

(e) None of these

By cross multiplication, we have

$\therefore \,\,\,\,\frac{x}{\left[ (-7)\times 9-5\times 28 \right]}=\frac{y}{\left[ 28\times (-7)-9\times 4 \right]}=\frac{1}{\left[ 4\times 5-(-7)\times (-7) \right]}$$\Rightarrow \,\,\frac{x}{(-63-140)}=\frac{y}{(-196-36)}=\frac{1}{20-49}$

$\Rightarrow \,\,\frac{x}{-203}=\frac{y}{-232}=\frac{1}{-29}$

$\Rightarrow \,\,x=\left( \frac{-203}{-29} \right)=7\And y=\left( \frac{-232}{-29} \right)=8$
Hence, x = 7 & y = 8 is the required solution.

Find the nature of solution of the given system of equation.

$2x-5y=17$

$5x-3y=14$

(a) Unique solution

(b) No solution

(c) Infinitely many solution

(d) All of these

(e) None of these

Explanation

$\Rightarrow \,\,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{5},\frac{{{b}_{1}}}{{{b}_{2}}}\frac{5}{3},\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-17}{-14}=\frac{17}{14}$

Thus $\Rightarrow \,\,\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$

Hence, the given system of equations has a unique solution.

Find the nature of solution of the given system of equations.

$3x-5y=11$

$6x-10y=7$

(a) Unique solution

(b) No solution

(c) Infinitely many solution

(d) All of these

(e) None of these

$\Rightarrow \,\,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{6}=\frac{1}{2},\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-5}{-10}=\frac{1}{2}\And \frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-11}{-7}=\frac{11}{7}$
Thus, $\frac{{{a}_{1}}}{{{a}_{2}}}\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$