To enable the completion of the proof that line <em>l</em> is parallel to line <em>m</em>, a
diagram showing the lines and their common transversal is attached.
The completed two column proof is presented as follows;
Statement 
Reason
1. ∠1 and ∠2 are supplementary angles 1. Given
2. m∠1 + m∠2 = 180° 2. <u>Definition of supplementary ∠s</u>
3. ∠1 and ∠3 are supplementary angles 3. Exterior sides in opposite rays
4. <u>m∠1 + m∠3 = 180° </u> 4. <u>Definition of supplementary ∠s</u>
5. m∠1 + m∠2 = m∠1 + m∠3 5. <u>Transitive property of equality</u>
6. <u>m∠2 = m∠3 </u> 6. <u>Subtraction property of equality</u>
7. l ║ m 7. <u>Converse of alternate interior </u>
<u>angles postulate</u>
Reasons:
- Reason for statement 2: Supplementary angles are defined as two angles that sum up to 180°
- Reason for statement 3: Two angles are supplementary if the exterior sides that form each angle are opposite rays (rays that are drawn out infinitely in opposite direction but have the same endpoint)
- Statement 4: Mathematical expression of the sum of ∠1 and ∠3; Reason for statement 4 is the definition of supplementary angles
- Reason for statement 5: Transitive property of equality describes the property that if a number <em>x</em> = <em>y</em>, and <em>z </em>= <em>y</em>, then <em>x</em> = <em>z</em>.
- Statement 6: Subtracting m∠1 from both sides of the equation in statement 5. gives; m∠1 + m∠2 - m∠1 = m∠1 + m∠3 - m∠1 ⇒ m∠2 = m∠3. Reason for statement 6 is the subtraction property of equality
- Reason for statement 7: The converse of the alternate interior angles postulate states that if the alternate interior angles formed between two lines and a common transversal are congruent, the two lines are parallel.
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Answer: left one is no solution and right one is infinite solutions
explanation: the left one the lines don’t touch at all which has no solution and the right one means that both lines are on top of each other
Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)
Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)
Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)
Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)
Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)
Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Collect results:
Answer: {x = 1, y = 1, z = 0
Answer: 
Step-by-step explanation:
Hope this helps, have a BLESSED AND WONDERFUL DAY! As well as a great Black History Month and Valentine's day! :-)
- Cutiepatutie ☺❀❤
An expression that is equivalent to the statement Subtract a from the quotient of 6 and B is 6/b - a
<h3>How to write equivalent expression</h3>
Given statement
Subtract a from the quotient of 6 and B
- The quotient of 6 and B can be written as 6/B
- a subtracted from 6/B is written as
6/B - a
Therefore, the correct option equivalent to the statement is option A) 6/B - a
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