Practice Problem(Prashanth Gowda)
Question- The m<A and m<B are supplementary. The m<A=x^2 + 5 while m<B=28x+10. What is x equal to and what is the measure of Angles A and B and which angle measure is greater?
Answer- x=5, m<A=30, m<B=150
m<B > m<A
Explanation- m<A + m<B = 180 Supplementary Angles
(x^2+5) + (28x+10) = 180 Substitution Property
x^2 + 28x + 15 = 180 Combining Like Terms
x^2 + 28x - 165 = 0 Subtraction Property of Equality
(x+33)(x-5) = 0 Factoring Trinomials
x+33=0 x-5=0
x= -33 and 5
X can't equal -33 because that would make <B negative and Angles can't have a negative measure.
x=5. Also angle A can't be greater than 360 degrees.
(x^2+5) + (28x+10) = 180
((5)^2+5) + (28(5)+10) = 180
30 + 150 = 180
180 = 180
m<A=30 and m<B=150
150 > 30
m<B > m<A
Paragraph- I first saw that the two angles were supplementary. That means the measure of the angles add up to 180. That sets up the equation m<A + m<B = 180. Then I substituted x^2 + 5 for m<A and 28x + 10 for m<B. This got me (x^2+5) + (28x+10) = 180. Then I combined like terms and got x^2 + 28x + 15 = 180. Then I subtracted 180 from both sides and got x^2 + 28x - 165 = 0. Then I factored the trinomial and got (x+33)(x-5). This helped me get that x = -33 and 5. In this case x couldn't equal -33 because that made one segment length greater than a 1000, and another length negative. Angle measures can't be greater than 360 degrees or negative. Now that I found that x was 5, I plugged in 5 for x and got that m<A=30 and m<B=150. This helps me answer the last part of the question which is which angle measure is greater. I got m<B is greater than m<A because 150 is greater than 30.
Question- Point A, B, and C are collinear and Point B is between Points A and C. Segment AB = 3x-8, Segment BC = 5x-10 and Segment AC = 30.
What's the measure of Segment AB, BC , and what's the value of x?
Answer- x=6, AB=10, and BC = 20
Explanation- AB + BC = AC Segment addition Postulate
(3x-8) + (5x-10) = 30 Substitution Property
8x - 18 = 30 Combining like terms
8x = 48 Addition Property
x = 6 Division Property
AB = 3x - 8
AB = 3(6) - 8
AB = 18 - 8
AB = 10
BC = 5x -10
BC = 5(6) - 10
BC = 30 - 10
BC = 20
Paragraph- I first saw that Point B is between Points A and C, so that means AB =BC = AC. Then I plugged in 3x - 8 for AB, 5x - 10 for BC and 30 for AC. Then I combined like terms and got 8x -18 = 30. Then I added 18 to both sides and got 8x = 48. I finally got x by dividing both sides by 8. this got me x = 6. Then I plugged in 6 for x and got that AB=10 and BC=20.
Question- The m<A and m<B are supplementary. The m<A=x^2 + 5 while m<B=28x+10. What is x equal to and what is the measure of Angles A and B and which angle measure is greater?
Answer- x=5, m<A=30, m<B=150
m<B > m<A
Explanation- m<A + m<B = 180 Supplementary Angles
(x^2+5) + (28x+10) = 180 Substitution Property
x^2 + 28x + 15 = 180 Combining Like Terms
x^2 + 28x - 165 = 0 Subtraction Property of Equality
(x+33)(x-5) = 0 Factoring Trinomials
x+33=0 x-5=0
x= -33 and 5
X can't equal -33 because that would make <B negative and Angles can't have a negative measure.
x=5. Also angle A can't be greater than 360 degrees.
(x^2+5) + (28x+10) = 180
((5)^2+5) + (28(5)+10) = 180
30 + 150 = 180
180 = 180
m<A=30 and m<B=150
150 > 30
m<B > m<A
Paragraph- I first saw that the two angles were supplementary. That means the measure of the angles add up to 180. That sets up the equation m<A + m<B = 180. Then I substituted x^2 + 5 for m<A and 28x + 10 for m<B. This got me (x^2+5) + (28x+10) = 180. Then I combined like terms and got x^2 + 28x + 15 = 180. Then I subtracted 180 from both sides and got x^2 + 28x - 165 = 0. Then I factored the trinomial and got (x+33)(x-5). This helped me get that x = -33 and 5. In this case x couldn't equal -33 because that made one segment length greater than a 1000, and another length negative. Angle measures can't be greater than 360 degrees or negative. Now that I found that x was 5, I plugged in 5 for x and got that m<A=30 and m<B=150. This helps me answer the last part of the question which is which angle measure is greater. I got m<B is greater than m<A because 150 is greater than 30.
Question- Point A, B, and C are collinear and Point B is between Points A and C. Segment AB = 3x-8, Segment BC = 5x-10 and Segment AC = 30.
What's the measure of Segment AB, BC , and what's the value of x?
Answer- x=6, AB=10, and BC = 20
Explanation- AB + BC = AC Segment addition Postulate
(3x-8) + (5x-10) = 30 Substitution Property
8x - 18 = 30 Combining like terms
8x = 48 Addition Property
x = 6 Division Property
AB = 3x - 8
AB = 3(6) - 8
AB = 18 - 8
AB = 10
BC = 5x -10
BC = 5(6) - 10
BC = 30 - 10
BC = 20
Paragraph- I first saw that Point B is between Points A and C, so that means AB =BC = AC. Then I plugged in 3x - 8 for AB, 5x - 10 for BC and 30 for AC. Then I combined like terms and got 8x -18 = 30. Then I added 18 to both sides and got 8x = 48. I finally got x by dividing both sides by 8. this got me x = 6. Then I plugged in 6 for x and got that AB=10 and BC=20.
Practice Problems- Patrick Flaherty
Question- D, E and F are collinear and DE = EF. If DE = x^2 -7x + 17 and EF= 3x - 8, find the value of x and DF.
Answer- x=5, DF = 14.
Explanation- DE = EF , DE = x^2 -7x + 17, and EF = 3x - 8 Given
x^2 - 7x + 17 = 3x - 8 Substitution
x^2 - 10x + 17 = -8 SPOE
x^2 - 10x + 25 = 0 APOE
a = 1 b = 10 c = 25 Quadratic Formula (a b and c)
a*c = 25, so find factors of 25 Factoring
1 and 25, 5 and 5. Factoring
(x - 5) (x - 5) = 0 Factoring
x = 5 Zero Property
EF = 3(5) - 8 Substitution
EF = 7 Simplifying
DE = 7 Transitive
DE + EF = DF Segment Addition Postulate
7 + 7 = DF Substitution
DF = 14 Simplifying
x=5, DF = 14
Paragraph- D, E and F are collinear and DE = EF. If DE = x^2 -7x + 17 and EF= 3x - 8, find the value of x and DF. Since DE = EF, DE = x^2 - 7x + 17, and EF = 3x - 8 are given, I marked them down. I need to solve for x, because DF is in terms of x. Because DE and EF are equal, x^2 - 7x + 17 = 3x - 8 through substitution. I decided to use the SPOE to move 3x to the other side, getting x^2 - 10x + 17 = - 8. Next, I added 8 to both sides to get x^2 - 10x + 25 = 0. Since I have a quadratic equation now, I can factor it if I know my ABCs. In this case, A = 1, B = 10, and C = 25. A times C equals 25, so we will factor that. 1 and 25, 5 and 5. -5 - 5 = B, so we will use that. (x - 5) (x - 5) = 0, after factoring. Since x can only be 5, it can be substituted for the DE or EF. EF is simpler, so (with substitution) EF = 3(5) - 8, which simplifies to EF = 7. We know that DE + EF = DF, and DE = EF, so, by substitution, 7 + 7 = DF. Simplified, DF = 14. Our answers are DF = 14, and x = 5.
Answer- x=5, DF = 14.
Explanation- DE = EF , DE = x^2 -7x + 17, and EF = 3x - 8 Given
x^2 - 7x + 17 = 3x - 8 Substitution
x^2 - 10x + 17 = -8 SPOE
x^2 - 10x + 25 = 0 APOE
a = 1 b = 10 c = 25 Quadratic Formula (a b and c)
a*c = 25, so find factors of 25 Factoring
1 and 25, 5 and 5. Factoring
(x - 5) (x - 5) = 0 Factoring
x = 5 Zero Property
EF = 3(5) - 8 Substitution
EF = 7 Simplifying
DE = 7 Transitive
DE + EF = DF Segment Addition Postulate
7 + 7 = DF Substitution
DF = 14 Simplifying
x=5, DF = 14
Paragraph- D, E and F are collinear and DE = EF. If DE = x^2 -7x + 17 and EF= 3x - 8, find the value of x and DF. Since DE = EF, DE = x^2 - 7x + 17, and EF = 3x - 8 are given, I marked them down. I need to solve for x, because DF is in terms of x. Because DE and EF are equal, x^2 - 7x + 17 = 3x - 8 through substitution. I decided to use the SPOE to move 3x to the other side, getting x^2 - 10x + 17 = - 8. Next, I added 8 to both sides to get x^2 - 10x + 25 = 0. Since I have a quadratic equation now, I can factor it if I know my ABCs. In this case, A = 1, B = 10, and C = 25. A times C equals 25, so we will factor that. 1 and 25, 5 and 5. -5 - 5 = B, so we will use that. (x - 5) (x - 5) = 0, after factoring. Since x can only be 5, it can be substituted for the DE or EF. EF is simpler, so (with substitution) EF = 3(5) - 8, which simplifies to EF = 7. We know that DE + EF = DF, and DE = EF, so, by substitution, 7 + 7 = DF. Simplified, DF = 14. Our answers are DF = 14, and x = 5.
Question- Base on the image, what is the value of x?
Answer- x=24
Explanation- 2y + 4 = 3y - 64 Vertical Angles
4 = y - 64 SPOE
y = 68 APOE
2x - 8 + 2(68) + 4 = 180 Linear Pair and Substitution
2x + 132 = 180 Simplifying
2x = 48 SPOE
x = 24 DPOE
Paragraph-
Based on the image, what is the value of x? Looking at the diagram, we can see that the lines form linear pairs and vertical angles. To solve for x, we must first find y, so we can use it as a supplement. Because of the definition of vertical angles, 2y + 4 = 3y - 64. So solve, I moved 2y to the other side using subtraction, getting 4 = y - 64. Next, I moved 64 over, using addition, to get y = 68. Since I have y, and I know (2x - 8) + (2y + 4) = 180 due to the definition of linear pairs, I substituted y for 68 in that equation. After simplification, it looked like 2x + 132 = 180. I needed to isolate x, so I subtracted 132 from both sides, to get 2x = 48. Finally, I got rid of the coefficient by dividing everything by 2, and got x = 24.
Answer- x=24
Explanation- 2y + 4 = 3y - 64 Vertical Angles
4 = y - 64 SPOE
y = 68 APOE
2x - 8 + 2(68) + 4 = 180 Linear Pair and Substitution
2x + 132 = 180 Simplifying
2x = 48 SPOE
x = 24 DPOE
Paragraph-
Based on the image, what is the value of x? Looking at the diagram, we can see that the lines form linear pairs and vertical angles. To solve for x, we must first find y, so we can use it as a supplement. Because of the definition of vertical angles, 2y + 4 = 3y - 64. So solve, I moved 2y to the other side using subtraction, getting 4 = y - 64. Next, I moved 64 over, using addition, to get y = 68. Since I have y, and I know (2x - 8) + (2y + 4) = 180 due to the definition of linear pairs, I substituted y for 68 in that equation. After simplification, it looked like 2x + 132 = 180. I needed to isolate x, so I subtracted 132 from both sides, to get 2x = 48. Finally, I got rid of the coefficient by dividing everything by 2, and got x = 24.