By looking at the point 1,1 and 0,3 you rise two and over one creating the ratio 2/1 which reduces down to 2
:D
<h2>Given</h2>
- A quadrilateral ABCD with opposite pairs of sides being congruent.
<h2>To prove</h2>
- Opposite sides are parallel.
<h2>Solution</h2>
We'll prove first the triangles formed by diagonal are congruent then we'll prove the angles on either side of diagonal are congruent.
<u>The steps are:</u>
- Step ⇒ Statement ⇒ Reason
=============================================================
- 1 ⇒ AB ≅ CD and BC ≅ AD ⇒ Given
- 2 ⇒ AC = CA ⇒ Common side
- 3 ⇒ ΔABC ≅ ΔCDA ⇒ Side-side-side congruence
- 4 ⇒ ∠BAC ≅ ∠DCA ⇒ Corresponding angles of congruent triangles
- 5 ⇒ ∠BAC and ∠DCA are alternate interior angles ⇒ Definition of alternate interior angles
- 6 ⇒ AB║CD ⇒ The converse of alternate interior angles theorem
- 7 ⇒ ∠BCA ≅ ∠DAC ⇒ Corresponding angles of congruent triangles
- 8 ⇒ ∠BCA & ∠DAC are alternate interior angles ⇒ Definition of alternate interior angles
- 9 ⇒ BC║AD ⇒ The converse of alternate interior angles theorem
You cannot eliminate x and end up with two different equations.
5x+3y=9.. B
2x-4=40...A
Solve A for x.
2x - 4 = 40
2x = 40 + 4
2x = 44
x = 44/2
x = 22
The only thing I can think of doing is to plug 22 into B.
5(22) + 2y = 9
110 + 2y = 9...This leads to an answer for y.
No two NEW EQUATIONS is possible.
2y = 9 - 110
2y = -101
y = -101/2
Check your question again.
