Set? Are you missing part of the question
System of Linear Equations entered :
[1] y - 2x/3 = -1
[2] y + x = 4
// To remove fractions, multiply equations by their respective LCD
Multiply equation [1] by 3
// Equations now take the shape:
[1] 3y - 2x = -3
[2] y + x = 4
Graphic Representation of the Equations :
-2x + 3y = -3 x + y = 4
Solve by Substitution :
// Solve equation [2] for the variable x
[2] x = -y + 4
// Plug this in for variable x in equation [1]
[1] 3y - 2•(-y +4) = -3
[1] 5y = 5
// Solve equation [1] for the variable y
[1] 5y = 5
[1] y = 1
// By now we know this much :
y = 1
x = -y+4
// Use the y value to solve for x
x = -(1)+4 = 3
I hope this help you
Answer:
See below.
Step-by-step explanation:
Draw segment OB.
In triangle OBC, points R and S are the midpoints of sides OC and BC, respectively. That makes RS parallel to OB.
In triangle OBA, points P and Q are the midpoints of sides OA and BA, respectively. That makes PQ parallel to OB.
Since segments RS and PQ are parallel to segment OB, then RS and PQ are parallel to each other.
A. 6x-y=1 can go to 6x-y-1=0 and then 6x-1 = y
Answer:
D
Step-by-step explanation:
Hope this helped!
