Geometry B Period / Mr. Hansen
4/28/2004

Name: __________KEY___________

Unscheduled Quiz on Linear Systems of Equations

 

Instructions: Solve each system. Give answer as a solution set. Show work, including check of solution.

 

 

1.

 

 

 

I.      3x + 4y = 7
II.    6x + 8y = 12
Ia.    –6x – 8y = –14

 

 

 

Ia + II Þ 0 = –2 (®¬)

 

 

 

S = Æ

 

 

 

[Explanation by Mr. Hansen: Multiply eqn. I by –2 to get eqn. Ia. This is permitted because of the mult. prop. of =. Then add eqns. Ia and II to get 0 = –2. However, this is a contradiction (indicated by colliding arrows ®¬), which means that the solution set is the null set.]

 

 

2.

 

 

 

I.      3x + 4y = 7
II.    7x – 5y = 2

 

 

 

Ia.    21x + 28y = 49 [we multiplied through by 7]
IIa.  21x – 15y = 6 [we multiplied through by 3]

 

 

 

Ia – IIa Þ 43y = 43 Þ y = 1

 

 

 

Back-substitute into eqn. I to get 3x + 4(1) = 7 Þ 3x = 3 Þ x = 1

 

 

 

S = {(1, 1)}

 

 

 

Checks:
I.      3(1) + 4(1) = 7 ü
II.    7(1) – 5(1) = 2 ü

 

 

 

[Note: You must check both equations to make sure that your solution is valid.]

 

 

3.

 

 

 

First eqn. Þ –y = 11 – 2x Þ y = 2x – 11.
Now substitute y into second eqn.: –4x + 2(2x – 11) = –22 Û –4x + 4x – 22 = –22 Û 0 = 0, always true.

 

 

 

S = {(x, y): 2x – y = 11}

 

 

 

[Explanation by Mr. Hansen: We showed that the first eqn. is satisfied iff the second eqn. is satisfied. Therefore, any solution of the first is a solution of the second, and conversely. The solution set of the system is therefore the set of ordered pairs that satisfy the first.]