Question

Which system of equations has only one solution? 4 x + 2 y = 8. Negative 4 x minus 2 y = 3. Negative 5 x + y = 6. 5 x minus y = negative 6. Negative 3 x + 4 y = 2. 3 x minus 4 y = 0. 2 x + 4 y = 6. 3 x minus 4 y = 9.

Answer 1

Answer:

Following system of equations have exactly one solution:

$2 x + 4 y = 6\\ 3 x - 4 y = 9$

Step-by-step explanation:

Given 4 system of equations:

1st  

$4 x + 2 y = 8\\ -4 x -2 y = 3$

2nd

$-5 x + y = 6\\ 5 x - y = - 6$

3rd

$- 3 x + 4 y = 2\\ 3 x - 4 y = 0$

4th

$2 x + 4 y = 6\\ 3 x - 4 y = 9$

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:

$a_1x+b_1y+ c_1=0$ and  $a_2x+b_2y+ c_2=0$

1. No solution:

There exists no solution if:

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$

2. Infinite solutions:

There exist infinitely many solutions if:

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

3. One solution:

There exists one solution if:

$\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$

Now, let us consider the given equations one by one.

1st system of equations:

The ratio is:

$\dfrac{4}{-4} , \dfrac{2}{-2} , \dfrac{8}{3}\\-1 =-1\neq \dfrac{8}{3}$

So, no solution.

2nd system of equations:

The ratio:

$\dfrac{-5}{5} , \dfrac{-1}{1} , \dfrac{6}{-6}\\-1 =-1=-1$

So, infinitely many solutions.

3rd system of equations:

$\dfrac{3}{-3} , \dfrac{-4}{4} , \dfrac{0}{2}\\-1 =-1\ne0$

So, no solution.

4th system of equations:

$\dfrac{2}{3} , \dfrac{4}{-4} , \dfrac{6}{9}\\\dfrac{2}{3} \ne-1$

Hence, only one solution.

So, the answer is:

Following system of equations have only one solution:

$2 x + 4 y = 6\\ 3 x - 4 y = 9$