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3 years ago

The pair of linear equations 2x-3y = 14 and 3x - 2y = 4 has solution(s).answer plz....

Mathematics

1 answer:

Lesechka [4]3 years ago

8 0

Answer:

2x -3y =1 and 3x-2y= 4 has

Step-by-step explanation:

The pair of linear equations 2x-3y=1 and 3x-2y=4 has one unique solution. , then it has a unique solution other wise not. Since , , it means the pair of linear equations has only one unique solution. Hence, the pair of linear equations 2x-3y=1 and 3x-2y=4 has only one unique solution.

a1/a2= 2/3

b1/b2= 3/-2 (-) cancle so it remains

c1 / c2 = 1/4

a1/a2 = b1/b2 not equal to c1/c2 as no solution

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2 years ago

Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)

JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by:

$\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9$

Plugging in the values in formula, we get,

$\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}$

Hence, the midpoint of the segment is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

6 0

3 years ago

The pair of linear equations 2x-3y = 14 and 3x - 2y = 4 has solution(s).answer plz....​

The pair of linear equations 2x-3y = 14 and 3x - 2y = 4 has solution(s).answer plz....