Solve: {y=1/2x-3; 4x-2y=16}

4x-2(1/2x-3)=16
4x-x+6=16
3x=10
x=10/3

y=
1/2(10/3)-3(3/3)=
5/3-9/3=
-4/3
Solution:
(10/3, -4/3)

Solve: 4m-5n=8, 2m+3n=10

1) 4m-5n=8
2) 2m+3n=10 <-m2 


and after we swap their positions... 


4m+6n=20
4m-5n=8 


then subtract to get 
11n=12
n=12/11 


from 1)... 


4m-5(12/11)=8
4m=8+60/11
4m=148/11
m=37/11

find the values of a,b and c of the quadratic equation y=ax^2+bx+c if it passes through the points (1,15), (-1,7), and (0,10).

y=ax^2+bx+c 
(0,10) => c=10 
(1,15) => a+b+10=15 or a+b=5 
(-1,7) => a-b+10=7 or a-b=-3 


adding the two last equations yields 2a=2 so that a=1 
and from the second equation a+b=5 we conclude b=4 


 

:) 

Solve: 2a+3b=7, 5a-4b=6

m4: 2a+3b=7
m3: 5a-4b=6 


8a+12a=28
15a-12a=18 
23a=46
a=2 

2(2)+3b=7
3b=3
b=1 

:)

Solve the system of Equations

the sum of a,b and c is 14. if 9a + 8b + 6c = 101, what is the value of 2a + b -c?

9a + 8b + 6c = 101 and a+b+c=14 so if we multiply every term by 7 we get: 7a+7b+7c=98 
and if we subtract the result from the first equation we get 2a+b-c=3 

so it's 3 

:)

Types of solutions

9) Chose A, B, or C from the list below. 4x+2y=8 Y=4-2x A) Inconsistent B) Consistent & Dependent C) Consistent & Independent

If we add 2x to both sides of the lower equation then double each of it's terms we'll get the upper equation. These lines both have the same solution, so the system is consistent and dependent.

:)

Linear systems of equations

7) Chose A, B, or C from the list below. Y=3x-5 Y=3x+6 A) Inconsistent B) Consistent & Dependent C) Consistent & Independent

Your Answer:

both equations have the same slope, but different y-intercepts, so they will never intersect. This system is inconsistent.


:)

Solve by elimination method 3x-5y=1 6x-10y=4

Solve by elimination method:
3x-5y=1
6x-10y=4

Solution:

 

First we solve for x in the first equation
 
 
The second equation,

becomes... 

which we simplify as follows... 


 
this clearly a contradiction. This contradiction tells us that our system has no solution. (Whenever we arrive at a contradiction after following the solution method correctly we know that the system has no solution. If we arrive at an identity we know that the lines are coincident)

John has a collection of nickels,dimes, and quarters. Altogether, there are 241 coins worth $26.20. There are 4 more nickels than dimes. How many of each type of coin are in the collection.

 John has a collection of nickels,dimes, and quarters. Altogether, there are 241 coins worth $26.20. There are 4 more nickels than dimes. How many of each type of coin are in the collection.


Solution:

Solve the linear system of equations