SOLUTION: I am studing the substitution Method.. and for some reason i can not seem to figure out how to do it. My teacher has been trying to work with me, but yet when I look at the problem

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons  -> Linear Equations Lesson -> SOLUTION: I am studing the substitution Method.. and for some reason i can not seem to figure out how to do it. My teacher has been trying to work with me, but yet when I look at the problem      Log On


   



Question 2732: I am studing the substitution Method.. and for some reason i can not seem to figure out how to do it. My teacher has been trying to work with me, but yet when I look at the problem I go blank on what to do.
I have an equation y = 2x + 7
y = 10x - 1 I dont understand how to do this...

Found 2 solutions by xcentaur, MathTherapy:
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
y = 2x + 7 .... [1]
y = 10x - 1 .... [2]


y=2x+7
2x=y-7
x=(y-7)/2 ...... [3]


Substituting value of [3] in [2],
y=10x-1
y=(10(y-7)/2)-1
y=5(y-7)-1
y=5y-35-1
y-5y=-35-1
-4y=-36
y=-36/-4
y=9


Also,putting y=9 in equation [1] we get,
y=2x+7
9=2x+7
9-7=2x
2x=2
x=1


ANS:
x=1,y=9


Hope this helps,
best of luck.


Answer by MathTherapy(10791) About Me  (Show Source):
You can put this solution on YOUR website!
I am studing the substitution Method.. and for some reason i can not seem to figure out how to do it.
My teacher has been trying to work with me, but yet when I look at the problem I go blank on what to do. 
I have an equation y = 2x + 7
                   y = 10x - 1   I dont understand how to do this...
********************************************************************

I have no idea why the person who responded did this: y=2x+7
2x=y-7
x=(y-7)/2 ...... [3]

What could possibly make someone solve for "x," and create a 3rd equation, WITH  a FRACTION,
and making things MUCH MORE COMPLEX, when the given equations are: y = 2x + 7
                                                                                                                                y = 10x - 1   
Isn't that the MOST SIMPLEST a system can ever be, READY to be solved using the SUBSTITUTION method? 
y = 2x + 7 ---- eq (i)
y = 10x - 1 --- eq (ii)

The above system MEANS that 2x + 7 = 10x - 1
                                                  2x - 10x = - 1 - 7  
                                                          - 8x = - 8
                                                              x = %28-+8%29%2F%28-+8%29+=+1

                                                              y = 2x + 7
                                                              y = 2(1) + 7 ----- Substituting 1 for x in eq (i)
                                                              y = 2 + 7 = 9

                                     Solution set: system%28x+=+1%2C+y+=+9%29