### Extra Review for Unit 1 Test

```Review for Unit 1 Test
1. 1 and 2 are a linear pair.
1 = 5x + 25
2 = 4x + 20
Find m2
5x + 25 + 4x + 20 = 180
9x + 45 = 180
9x = 135
x = 15
2 = 4(15) + 20 = 80
Y is=the
midpoint
of RL
x
+
4y
–
15
+
2x
+
y
–
10
6x
–
5y
2.
M
YI
bisects
LYE
-3x + 10y = 25
x + 4y
R – 15 = 2x +Y y – 10 L
-x + 3y = 5 2(x + 4y – 15) = 6x – 5y
E 2x + 8yI – 30 = 6x – 5y
-4x + 13y = 30
mEYL
2(2x + y=–(6x
10)– =5y)
6x – 5y
Write
of equations to find
4x +and
2ysolve
– 20a=system
6x – 5y
the values
and
-2x +of7yx =
20y.
x = 25
y = 10
3. Six times an angle’s supplement is 5
less than the angle. Find the measure of
the angle’s supplement.
6(180 – x) = x – 5
1080 – 6x = x – 5
1085 = 7x
155 = x
Supplement = 180 – 155 = 25
4. 1 and 2 form a right angle.
1 = 10x + 12
2 = 3x
Find m1
10x + 12 + 3x = 90
13x + 12 = 90
13x = 78
x=6
1 = 10(6) + 12= 72
5. 1 and 2 are complementary.
1 is 15 more than twice 2
Find m2
2 = x; 1 = 90 – x
90 – x = 2x + 15
75 = 3x
25 = x
6. 1 and 2 are vertical angles.
1 = 3x2 – 2x + 6
2 = x2 – 11x + 2
Find the value of x.
3x2 – 2x + 6 = x2 – 11x + 2
2x2 + 9x + 4 = 0
(2x + 1)(x + 4) = 0
x = -½ x = -4
(both work!)
7. The sum of an angle’s supplement and its
complement is equal to 8 times the angle.
Find the measure of the angle.
180 – x + 90 – x = 8x
270 – 2x = 8x
270 = 10x
27 = x
8. JM intersects GW at point J
GJM = 2x + 5
MJW = x + 25
Draw and label a diagram.
W
J
G
M
9. Can you assume:
M
Y is the midpoint of RL
YI bisects LYE
R
Y
E
L
I
d)
e)
Angle
R, Y,
L
LYI
are==collinear
a)
c)MY
SO
OU
LYE
SU
isangle
right EYI
RY
YL
f)b)
=+=YI
MYR
d)
e)
yes,
you’re
BISECTOR
c)no!
yes
no! b/c
b)
MIDPOINT
f)a)
e)
yes,
b/c they
aretold
vertical
angles
10. An angle is twice it’s complement.
Find the measure of the angle’s
supplement.
x = 2(90 – x)
x = 180 – 2x
3x = 180
x = 60
Supplement = 180 – 60 = 120
11. Solve the system using either
substitution or elimination.
3x + 2y = 11
-6x + y = -32
x=5
y = -2
```