Answer:
x = 2 and y = 5
Step-by-step explanation:
Given the simultaneous linear equations
y = 2x + 1 and x + y = 7
We are to find the value of x and y;
Substitute 1 into 2
From 2;
x+y = 7
x+2x+1 = 7
3x+1 = 7
3x = 7-1
3x = 6
x = 6/3
x = 2
Recall that y = 2x+1
y = 2(2) + 1
y = 4+1
y = 5
Hence the solution from the graph will be the point where the line cuts the x axis and this will be at x = 2 and y = 5
Answer:
1.
5
x
−
2
y
=
4
; (−1, 1)
2.
3
x
−
4
y
=
10
; (2, −1)
3.
−
3
x
+
y
=
−
6
; (4, 6)
4.
−
8
x
−
y
=
24
; (−2, −3)
5.
−
x
+
y
=
−
7
; (5, −2)
6.
9
x
−
3
y
=
6
; (0, −2)
7.
1
2
x
+
1
3
y
=
−
1
6
; (1, −2)
8.
3
4
x
−
1
2
y
=
−
1
; (2, 1)
9.
4
x
−
3
y
=
1
;
(
1
2
,
1
3
)
10.
−
10
x
+
2
y
=
−
9
5
;
(
1
5
,
1
10
)
11.
y
=
1
3
x
+
3
; (6, 3)
12.
y
=
−
4
x
+
1
; (−2, 9)
13.
y
=
2
3
x
−
3
; (0, −3)
14.
y
=
−
5
8
x
+
1
; (8, −5)
15.
y
=
−
1
2
x
+
3
4
;
(
−
1
2
,
1
)
16.
y
=
−
1
3
x
−
1
2
;
(
1
2
,
−
2
3
)
17.
y
=
2
; (−3, 2)
18.
y
=
4
; (4, −4)
19.
x
=
3
; (3, −3)
20.
x
=
0
; (1, 0)
Find the ordered pair solutions given the set of x-values.
21.
y
=
−
2
x
+
4
; {−2, 0, 2}
22.
y
=
1
2
x
−
3
; {−4, 0, 4}
23.
y
=
−
3
4
x
+
1
2
; {−2, 0, 2}
24.
y
=
−
3
x
+
1
; {−1/2, 0, 1/2}
25.
y
=
−
4
; {−3, 0, 3}
26.
y
=
1
2
x
+
3
4
; {−1/4, 0, 1/4}
27.
2
x
−
3
y
=
1
; {0, 1, 2}
28.
3
x
−
5
y
=
−
15
; {−5, 0, 5}
29.
–
x
+
y
=
3
; {−5, −1, 0}
30.
1
2
x
−
1
3
y
=
−
4
; {−4, −2, 0}
31.
3
5
x
+
1
10
y
=
2
; {−15, −10, −5}
32.
x
−
y
=
0
; {10, 20, 30}
Find the ordered pair solutions, given the set of y-values.
33.
y
=
1
2
x
−
1
; {−5, 0, 5}
34.
y
=
−
3
4
x
+
2
; {0, 2, 4}
35.
3
x
−
2
y
=
6
; {−3, −1, 0}
36.
−
x
+
3
y
=
4
; {−4, −2, 0}
37.
1
3
x
−
1
2
y
=
−
4
; {−1, 0, 1}
38.
3
5
x
+
1
10
y
=
2
; {−20, −10, −5}
Part B: Graphing Lines
Given the set of x-values {−2, −1, 0, 1, 2}, find the corresponding y-values and graph them.
39.
y
=
x
+
1
40.
y
=
−
x
+
1
41.
y
=
2
x
−
1
42.
y
=
−
3
x
+
2
43.
y
=
5
x
−
10
44.
5
x
+
y
=
15
45.
3
x
−
y
=
9
46.
6
x
−
3
y
=
9
47.
y
=
−
5
48.
y
=
3
Find at least five ordered pair solutions and graph.
49.
y
=
2
x
−
1
50.
y
=
−
5
x
+
3
51.
y
=
−
4
x
+
2
52.
y
=
10
x
−
20
53.
y
=
−
1
2
x
+
2
54.
y
=
1
3
x
−
1
55.
y
=
2
3
x
−
6
56.
y
=
−
2
3
x
+
2
57.
y
=
x
58.
y
=
−
x
59.
−
2
x
+
5
y
=
−
15
60.
x
+
5
y
=
5
61.
6
x
−
y
=
2
62.
4
x
+
y
=
12
63.
−
x
+
5
y
=
0
64.
x
+
2
y
=
0
65.
1
10
x
−
y
=
3
66.
3
2
x
+
5
y
=
30
Part C: Horizontal and Vertical Lines
Find at least five ordered pair solutions and graph them.
67.
y
=
4
68.
y
=
−
10
69.
x
=
4
70.
x
=
−
1
71.
y
=
0
72.
x
=
0
73.
y
=
3
4
74.
x
=
−
5
4
75. Graph the lines
y
=
−
4
and
x
=
2
on the same set of axes. Where do they intersect?
76. Graph the lines
y
=
5
and
x
=
−
5
on the same set of axes. Where do they intersect?
77. What is the equation that describes the x-axis?
78. What is the equation that describes the y-axis?
Part D: Mixed Practice
Graph by plotting points.
79.
y
=
−
3
5
x
+
6
80.
y
=
3
5
x
−
3
81.
y
=
−
3
82.
x
=
−
5
83.
3
x
−
2
y
=
6
84.
−
2
x
+
3
y
=
−
12
Step-by-step explanation:
Answer:
<h2>b = 15°</h2>
Step-by-step explanation:
If Pq = RQ then ΔPQR is the isosceles triangle. The angles QPR and PRQ have the same measures.
We know: The sum of the measures of the angeles in the triangle is equal 180°. Therefore we have the equation:
m∠QPR + m∠PRQ + m∠RQP = 180°
We have
m∠QPR = m∠PRQ and m∠RQP = 60°
Therefore
2(m∠QPR) + 60° = 180° <em>subtract 60° from both sides</em>
2(m∠QPR) = 120° <em>divide both sides by 2</em>
m∠QPR = 60° and m∠PRQ = 60°
Therefore ΔPRQ is equaliteral.
ΔPSR is isosceles. Therefore ∠SPR and ∠PRS are congruent. Therefore
m∠SPR = m∠PRS
In ΔAPS we have:
m∠SPR + m∠PRS + m∠RSP = 180°
2(m∠SPR) + 90° = 180° <em>subtract 90° from both sides</em>
2(m∠SPR) = 90° <em>divide both sides by 2</em>
m∠SPR = 45° and m∠PRS = 45°
m∠PRQ = m∠PRS + b
Susbtitute:
60° = 45° + b <em>subtract 45° from both sides</em>
15° = b
