Solve the following system of equations graphically on the set of axes below. y=x+8 y=−2x−1
1 answer:
This makes sense because every point on the gold line is a solution to the equation
y
=
1
2
x
+
3
y=
2
1
x+3start color #e07d10, y, equals, start fraction, 1, divided by, 2, end fraction, x, plus, 3, end color #e07d10, and every point on the green line is a solution to
y
=
x
+
1
y=x+1start color #0d923f, y, equals, x, plus, 1, end color. So, the only point that's a solution to both equations is the point of intersection
y
=
1
2
x
+
3
y=
2
1
x+3start color #e07d10, y, equals, start fraction, 1, divided by, 2, end fraction, x, plus, 3, end color #e07d10, and every point on the green line is a solution to
y
=
x
+
1
y=x+1start color #0d923f, y, equals, x, plus, 1, end color. So, the only point that's a solution to both equations is the point of intersection
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Answer:
The variable, y is 11°
Step-by-step explanation:
The given parameters are;
in triangle ΔABC; in triangle ΔFGH;
Segment = 14
Segment
= 14
Segment = 27
Segment
= 19
Segment = 19
Segment
= 2·y + 5
∡A = 32° ∡G = 32°
∡A = ∠BAC which is the angle formed by segments = 14 and
= 19
Therefore, segment = 27, is the segment opposite to ∡A = 32°
Similarly, ∡G = ∠FGH which is the angle formed by segments = 14 and
= 19
Therefore, segment = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;
≅
by Congruent Parts of Congruent Triangles are Congruent (CPCTC)
∴ =
= 27° y definition of congruency
= 2·y + 5 = 27° by transitive property
∴ 2·y + 5 = 27°
2·y = 27° - 5° = 22°
y = 22°/2 = 11°
The variable, y = 11°