SOLUTION: solve the system of equations by graphing, then classifying the system. Then classify the system. 4u+v=14 4u=v+26 What is the system of the solution?

Algebra ->  Expressions-with-variables -> SOLUTION: solve the system of equations by graphing, then classifying the system. Then classify the system. 4u+v=14 4u=v+26 What is the system of the solution?      Log On


   

Question 172878: solve the system of equations by graphing, then classifying the system. Then classify the system.
4u+v=14
4u=v+26
What is the system of the solution?
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!
4u%2Bv=14, EQN 1
4u=v%2B26, EQN 2
Let highlight%28u=y%29 & highlight%28v=x%29
Evaluating EQN 1:
4y%2Bx=14--->4y=-x%2B14
cross%284%29y%2Fcross%284%29=%28-x%2B14%29%2F4
y=%28-1%2F4%29x%2B%2814%2F4%29------>y=%28-1%2F4%29x%2B%287%2F2%29>>>>>> Line 1
Evaluating EQN 2:
4y=x%2B26--->cross%284%29y%2Fcross%284%29=%28x%2B26%29%2F4
y=%281%2F4%29x%2B%2826%2F4%29--->y=%281%2F4%29x%2B%2813%2F2%29>>>>>>>>> Line 2
We'll see the graph:
-----> RED,>>>> Line 1: GREEN,>>>> Line 2
Let' ssee the point of intersection:
In EQN 1,
v=14-4u},condition 1>>>> subst. in EQN 2
4u=highlight%2814-4u%29%2B26
4u%2B4u=30 ----->8u=30
cross%288%29u%2Fcross%288%29=cross%2840%295%2Fcross%288%29
u=highlight%28y=5%29
Subst. in condition 1:
v=14-4%285%29=14-20
v=highlight%28x=-6%29
Therefore, point of intersection @ (-6,5):

Let' see the intercepts of Line 1:
f%28y%29=0
%281%2F4%29x=%287%2F2%29 ----->cross%281%2F4%29x%2Fcross%281%2F4%29=%287%2F2%29%2F%281%2F4%29
x=%287%2F2%29%284%2F1%29=28%2F2=14
f%28x%29=0
y=7%2F2
Line 1 (RED) touches on points (14,0) & (0,7/2)
.
Let's see Line 2:
f%28y%29=0
%28-1%2F4%29x=13%2F2--->cross%28-1%2F4%29x%2Fcross%28-1%2F4%29=%2813%2F2%29%2F%28-1%2F4%29
x=%2813%2F2%29%28-4%2F1%29=52%2F2=-26
f%28x%29=0
y=13%2F2
Line 2 (GREEN) touches on points (-26,0) & (0,13/2)
We'll see the graph:

Thank you,
Jojo