Look at the picture.

The perimeter of the triangle:
Let's solve your equation step-by-step.
(−
1
/5
)(c)+2+c=
1
/5 (10c−20)
Step 1: Simplify both sides of the equation.
(−
1
/5
)(c)+2+c=
1
/5
(10c−20)
(−
1
/5
)(c)+2+c=(
1
/5
)(10c)+(
1
/5
)(−20) (Distribute)
−1
/5
c+2+c=2c+−4
(
−1
/5
c+c)+(2)=2c−4 (Combine Like Terms)
4
/5
c+2=2c−4
4
/5
c+2=2c−4
Step 2: Subtract 2c from both sides.
4/5
c+2−2c=2c−4−2c
−6
/5
c+2=−4
Step 3: Subtract 2 from both sides.
−6
/5
c+2−2=−4−2
−6
/5
c=−6
Step 4: Multiply both sides by 5/(-6).
(
5
/−6
)*(
−6
/5
c)=(
5
/−6
)*(−6)
c=5
Answer:
c=5
Answer:
The north campus had
while the south campus had
.
Step-by-step explanation:
The question stated that there are
students in the merged campus. That's the sum of the number of students in the north and the south campus before the merger.
Let
denote the number of students in the north campus before the merger. The other
students would all belong to the south campus before the merger.
Before the merger,
of the students of the north campus are of music majors. In terms of
, that's
students.
On the other hand,
of the students of the south campus are of music majors. That corresponds to
students.
With a similar logic, the number of music students in the merged east campus will be
.
The question implies that the sum of students in the two campuses should be equal to the number of music students in the east campus. That is:
.
Solve for
:
.
In other words, there are
students in the north campus and
students in the south campus before the merger.
Answer:
One solution graph is when the graphs of the equations intersect, then there is one solution that is true for both equations.
One solution = lines intersect
No solution graph is when the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.
No solution: lines are parallel
Infinitive solution graph is when the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.
Infinitive solutions: One line