Answer:
Following system of equations have exactly one solution:
[tex]2 x + 4 y = 6\\ 3 x - 4 y = 9[/tex]
Step-by-step explanation:
Given 4 system of equations:
1st
[tex]4 x + 2 y = 8\\ -4 x -2 y = 3[/tex]
2nd
[tex]-5 x + y = 6\\ 5 x - y = - 6[/tex]
3rd
[tex]- 3 x + 4 y = 2\\ 3 x - 4 y = 0[/tex]
4th
[tex]2 x + 4 y = 6\\ 3 x - 4 y = 9[/tex]
To find: Which system of equations has only one solution?
Solution:
First of all, let us learn the concept.
Let the equation of lines be:
[tex]a_1x+b_1y+ c_1=0[/tex] and [tex]a_2x+b_2y+ c_2=0[/tex]
1. No solution:
There exists no solution if:
[tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}[/tex]
2. Infinite solutions:
There exist infinitely many solutions if:
[tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}[/tex]
3. One solution:
There exists one solution if:
[tex]\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}[/tex]
Now, let us consider the given equations one by one.
1st system of equations:
The ratio is:
[tex]\dfrac{4}{-4} , \dfrac{2}{-2} , \dfrac{8}{3}\\-1 =-1\neq \dfrac{8}{3}[/tex]
So, no solution.
2nd system of equations:
The ratio:
[tex]\dfrac{-5}{5} , \dfrac{-1}{1} , \dfrac{6}{-6}\\-1 =-1=-1[/tex]
So, infinitely many solutions.
3rd system of equations:
[tex]\dfrac{3}{-3} , \dfrac{-4}{4} , \dfrac{0}{2}\\-1 =-1\ne0[/tex]
So, no solution.
4th system of equations:
[tex]\dfrac{2}{3} , \dfrac{4}{-4} , \dfrac{6}{9}\\\dfrac{2}{3} \ne-1[/tex]
Hence, only one solution.
So, the answer is:
Following system of equations have only one solution:
[tex]2 x + 4 y = 6\\ 3 x - 4 y = 9[/tex]
