The solution to the given system of linear equations lies in which quadrant?
ex-3y = 6
1 x+y=2
quadrant iii
quadrant !
- 5
4
-3 -2
1
2
3
4
5
x
quadrant iii
| quadrant iv
? 17

The solution to the given systems of linear equation is {3, -1} lies in IV Quadrant.

Step-by-step explanation:

Given:

x-3y = 6  ........( 1 )

x+y=2    ........( 2 )

To find :

x = ?

y = ?

Solution:

Let solve the  above equation to get the solutions.

Eliminate X by subtracting the two equations equation 1 minus equation 2

Now substitute for y = -1 in equation 2 we get

Therefore, solution set is {3, -1}

In

I Quadrant {x,y} = {+,+}......both the coordinates are positive.

II Quadrant {x,y} = {-,+}.....X coordinate is negative and Y coordinate is positive.

III Quadrant {x,y} = {-,-}......both the coordinates are negative.

IV Quadrant {x,y} = {+,-}.....X coordinate is positive and Y coordinate is negative. {3,-1}

Point A on the graph is the solution.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

95.03 is the answer. hope this !