Problem:
Solve
2
x
+
5
y
=
−
3
;
3
x
−
y
=
21
Steps:
I will try to solve your system of equations.
3
x
−
y
=
21
;
2
x
+
5
y
=
−
3
Step: Solve
3
x
−
y
=
21
for y:
3
x
−
y
+
−
3
x
=
21
+
−
3
x
(Add -3x to both sides)
−
y
=
−
3
x
+
21
−
y
−
1
=
−
3
x
+
21
−
1
(Divide both sides by -1)
y
=
3
x
−
21
Step: Substitute
3
x
−
21
for
y
in
2
x
+
5
y
=
−
3
:
2
x
+
5
y
=
−
3
2
x
+
5
(
3
x
−
21
)
=
−
3
17
x
−
105
=
−
3
(Simplify both sides of the equation)
17
x
−
105
+
105
=
−
3
+
105
(Add 105 to both sides)
17
x
=
102
17
x
17
=
102
17
(Divide both sides by 17)
x
=
6
Step: Substitute
6
for
x
in
y
=
3
x
−
21
:
y
=
3
x
−
21
y
=
(
3
)
(
6
)
−
21
y
=
−
3
(Simplify both sides of the equation)
Answer:
y
=
−
3
and
x
=
6
Answer: y > x, y > 2
The horizontal line goes through 2 on the y axis. This boundary line is represented by the equation y = 2, since every point on this line has a y coord of 2. The shading above it means that the inequality is y > 2. Every point in the shaded region of y > 2 has a y coord that is larger than 2.
The other inequality is y > x because we shade above the dashed boundary line y = x, which is that slanted dashed line.
Combining the two regions of y > 2 and y > x leads to what is shown.
Answer:
where the choices?
Step-by-step explanation:
Answer:
ercgtbvyuh7
Step-by-step explanation:
vhcrc6vthby7uj7vt6crgtvy
Answer:
<em>-26</em>
Step-by-step explanation:
Given the sequence:
16, 9, 2, –5, ...,
To find:
7th measurement, if the above sequence continues:
Solution:
Let us examine the given sequence first:
First term is 16
Second term = 9
Third term = 2
Fourth term = -5
Difference between 2nd and 1st term = 9 - 16 = -7
Difference between 3rd and 2nd term = 2 - 9 = -7
Difference between 4th and 3rd term = -5 - 2 = -7
We can see that there is a common difference of -7 between each term.
That means, the sequence is in Arithmetic Progression.
whose first term, 
Common difference, 
To find:
7th term i.e. 
Solution:
Formula for 
term of an Arithmetic Progression is given as:
Let us put 
7th measurement will be <em>-26.</em>
