SOLUTION: I am struggling with some of these problems please help
Solve each system by substitution:
(b) y = 13x + 14
x = y - 1
John
Algebra.Com
Question 295209: I am struggling with some of these problems please help
Solve each system by substitution:
(b) y = 13x + 14
x = y - 1
John
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve each system by substitution:
(a) 5x + y = 10
3x – 3y = 6
---------------
1st: Solve the 1st equation for "y":
y = 10-5x
2nd:
Substitute for "y" in the 2nd equation and solve for "x":
3x - 3(10-5x) = 6
3x -30+15x = 6
18x = 36
x = 2
----
3rd: Since y = 10-5x, solve for "y":
y = 10-5*2 = 0
------
Solution: (2,0)
Check your answer in the two original equations.
=======================================================
(b) y = 13x + 14
x = y - 1
----------------
1st: The 2nd equation is already solved for x,
so substitute for "x" in the 1st equation:
---
y = 13(y-1) + 14
y = 13y-13+14
y = 13y +1
12y = -1
y = -1/12
---
2nd: Since x = y-1
x = (-1/12)-1
x = -13/12
------------------------
Cheers,
Stan H.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
john, some website rules
one problem at a time
no similar problems
limit 4 problems a day
(b) y = 13x + 14
x = y - 1
plug in y-1 for x in the first equation to get
y=13*(y-1)+14
y=13y-13+14
y=13y+1
-12y=1
y=-1/12
x=y-1
x=-1/12-1
x=-13/12
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