How many solutions does the system have?

Mathematics

1 answer:

umka21 [38]3 years ago

3 0

One solution

<h2>Explanation:</h2>

For a system of linear equations in two variables, we could have three possible cases:

<h3>Case 1. No solution.</h3>

This happens when the lines are parallel and have different y-intercepts.

<h3>Case 2. One solution.</h3>

This happens when the lines intersect at a single point.

<h3>Case 1. Infinitely many solutions</h3>

This happens when the lines are basically the same having the same slope and y-intercept.

So, let's rewrite our lines in Slope-intercept form $y=mx+b$ :

$Line \ 1: \\ \\ 4x-2y = 8 \\ \\ 2y=4x-8 \\ \\ y=\frac{4}{2}x-\frac{8}{2} \\ \\ y=2x-4 \\ \\ \\ Line \ 2: \\ \\ 2x + y = 2 \\ \\ y=-2x+2$

As you can see, they have different slopes and y-intercepts. So they will intersect at a single point which is the solution of the system. By using graphing tool we get that this point is (1.5, -1) as indicated in the figure below.

<h2>Learn more:</h2>

Parametric equations: brainly.com/question/10022596

#LearnWithBrainly

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3 years ago

You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4,

Kitty [74]

Answer:

D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

Let the coordinate of the fourth vertex be D(x,y).

We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.

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Midpoint of BD = Midpoint of AC

$\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3$

The coordinates of the fourth vertex is D(4,-3)

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3 years ago

Question 1

Vikki [24]

Answer:

(B)15,36,39

Step-by-step explanation:

To determine which of the line segments could create a Right Triangle, we check if it satisfies the Pythagoras Theorem, taking the longest part to be the Hypotenuse in all cases.

By Pythagoras Theorem: $Hypotenuse^2=Opposite^2+Adjacent^2$