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3 years ago

Need help solving homework

Mathematics

2 answers:

bearhunter [10]3 years ago

4 0

8+9 and 10+7 are both 17
12-7 and 3+2 are both 5
19-2 and 2+15 are both 17
14-9 and 1+4 are both 5

8+9=10+7=19-2=2+15

12-7=3+2=14-9=1+4

Lilit [14]3 years ago

3 0

3 + 2 = 1 + 4
2 + 15 = 10 + 7
8 + 9 = 19 - 2
12 - 7 = 14 - 9

Hope this helped!

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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

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2nd

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

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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:

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

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