SOLUTION: i have to write my own linear system problem for school and solve it by using elimination and substitution but when i tried solving it i got two different solutions the equations

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: i have to write my own linear system problem for school and solve it by using elimination and substitution but when i tried solving it i got two different solutions the equations       Log On


   

Question 412009: i have to write my own linear system problem for school and solve it by using elimination and substitution but when i tried solving it i got two different solutions
the equations are y=1500+20x and y=3000+10x
when i used substitution i got (150,4500)and with elimination i got (-300,-4500)
is that correct or wrong?
Found 2 solutions by nerdybill, ewatrrr:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
We'll do it both ways here...
Substitution method:
y=1500+20x
y=3000+10x
.
1500+20x=3000+10x
1500+10x=3000
10x=1500
x = 150
.
y=1500+20x
y=1500+20(150)
y=1500+3000
y=4500
.
.

Applying the elimination method to:
y=1500+20x
y=3000+10x
.
moving the x's to the left:
y-20x=1500
y-10x=3000
.
multiply the bottom equation by -1 to get add both together:
y-20x=1500
-y+10x=-3000
---------------
-10x = -1500
x = 150
.
substitute into top equation to find y:
y=1500+20x
y=1500+20(150)
y=1500+3000
y=4500

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Hm...the system of linear equations You have picked has only 'one' solution:

solving by substitution method

y=1500+20x and 

y=3000+10x

1500+2x = 3000+10x

1200 = 8x

 150 = x and y = 1500 + 3000 = 4500



solving by elimination method

y=1500+20x and 

y=3000+10x



 y= 1500+20x and 

-y=-3000-10x    |multiplying 2nd equation thru by -1 and adding so as to eliminate y

 0 = -1500 +10x

1500 = 10x

 150 = x  and y = 1500 + 3000 = 4500



CHECKING our Answer(s)***

4500 = 3000 + 1500

4500 = 1500 + 3000