Answer:
14
Step-by-step explanation:
Answer:
2 13/20
Step-by-step explanation:
Answer:
On occasions you will come across two or more unknown quantities, and two or more equations
relating them. These are called simultaneous equations and when asked to solve them you
must find values of the unknowns which satisfy all the given equations at the same time.
Step-by-step explanation:
1. The solution of a pair of simultaneous equations
The solution of the pair of simultaneous equations
3x + 2y = 36, and 5x + 4y = 64
is x = 8 and y = 6. This is easily verified by substituting these values into the left-hand sides
to obtain the values on the right. So x = 8, y = 6 satisfy the simultaneous equations.
2. Solving a pair of simultaneous equations
There are many ways of solving simultaneous equations. Perhaps the simplest way is elimination. This is a process which involves removing or eliminating one of the unknowns to leave a
single equation which involves the other unknown. The method is best illustrated by example.
Example
Solve the simultaneous equations 3x + 2y = 36 (1)
5x + 4y = 64 (2) .
Solution
Notice that if we multiply both sides of the first equation by 2 we obtain an equivalent equation
6x + 4y = 72 (3)
Now, if equation (2) is subtracted from equation (3) the terms involving y will be eliminated:
6x + 4y = 72 − (3)
5x + 4y = 64 (2)
x + 0y = 8
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
