10th Class Mathematics Linear Equation in Two and Three 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

Answer: (a)

 

 

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

 

Answer: (d)

 

 

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

 

Answer: (b)

 

 

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

 

Answer: (d)

 

 

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

 

Answer: (a)

 

 

 

 

*         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

 

Answer: (a)

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

 

Answer: (c)

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

 

Answer: (a)

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

 

Answer: (b)

Here we have,

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

Thus, \[\frac{{{a}_{1}}}{{{a}_{2}}}\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\]

Hence, the given system of equations has no solution and hence is inconsistent.



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