Condition for Consistency
Category : 10th Class
For the system of linear equation\[{{a}_{1}}x+{{b}_{1}}y={{c}_{1}}\,and\,{{a}_{2}}x+{{b}_{2}}\,y={{c}_{2}}\],
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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