SOLUTION: I am SO confused on this one, i already tried to solve this but when i checked by substittuting, they weren't equal. Solving Systems by Substitution: a = 2/b - 3 a = 2b -

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I am SO confused on this one, i already tried to solve this but when i checked by substittuting, they weren't equal. Solving Systems by Substitution: a = 2/b - 3 a = 2b -      Log On


   

Question 127894: I am SO confused on this one, i already tried to solve this but when i checked by substittuting, they weren't equal.
Solving Systems by Substitution:
a = 2/b - 3
a = 2b - 18
I'm supposed to find the solution as a point like: [2,-4] but i have no idea if the a or b is x. Please help me with this!
I tried this: 2/b-3=2b-18 thats all i know that im pretty sure is right then times 2 on both sides -3=4b-18 then add 18 and -15=4b divide by 4 and thats 3.75 but when i put that into the other equation like 2*3.75 thats -10.5 and the final was -10.5 = 2/3.75 - 3 and thats -10.5 = .5 - 3 which is -10.5 does not equal -3.5 Im so confused and don't know what i did wrong!
THANK YOU SO MUCH
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Solving Systems by Substitution:
a = 2/b - 3
:
a = 2b - 18
:
I see you substituted (2/b - 3), from the 1st equation, for a in the 2nd equation
2%2Fb+-+3 = 2b - 18
:
get rid of the denominator by multiplying equation by b; (You multiplied by 2) then solve for b
2 - 3b = 2b^2 - 18b
:
0 = 2b^2 - 18b + 3b - 2; arrange as a quadratic on the right
OR
2B^2 - 15B - 2 = 0; the normal way a quadratic equation is writte
:
This is a quadratic equation that has to be solved using the quadratic formula
a = 2; b = -15; c = -2
:
The two solutions from that will be: b =-.131 and b = 7.631
:
Let's use b = 7.631 in the second equation to find a
a = 2(7.631) - 18
a = 14.262 - 18
a = -2.738
:
So one pair of solutions will be a = -2.738, b = 7.631
:
We can assume a = y and b = x from the way the equations are written
(the dependent variable is y and the left)
The x/y pair would be 7.631, -2.738
:
Check for equality in the 1st equation
-2.738 = 2%2F7.631+-+3
-2.738 = .262 - 3
-2.738 = -2.738; confirms this set of solutions
:
But the quadratic equation has another solution; b = -.131
:
Find a using this value for b in the 2nd equation
a = 2(-.131) - 18
a = -.262 - 18
a = -18.262
:
Our other pair will be a = -18.262, b = -.131 (x/y -.131, -18.262)
:
Check to see if this will have equality in the 1st equation
-18.262 = 2%2F-.131+-+3
-18.262 = -15.267 - 3
-18.262 = -15.267; so this pair of solutions check out too
:
Did this help? I assume you know quadratic formula so I did not go thru that.