2x-3y = -7
4x -y = -14
what I keep on getting is no solution. when I do substitution it gives me -14 = -14.
i also get no solution for the following:
5x+2y= -3
15x+6y = 15
2007-08-20 13:59:12 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
well are you trying to find the absolute value? if you are then
-14=-14 is right...
2007-08-20 14:06:30 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Seems easier to do this $4x-y=-14$, in this eq. $y=4x+14$, then substitute in ec. 1, $5x+2(4x+14)=-7$ now it is $5x+8x+28=-7$, $13x=-7-28$, $13x=-35$, $x=-35/13$. Substitution of $x$, $y=4(-35/13)+14$.
For the second system, there is no solution, if you see carefully, the second equation in the system is exactly twice the first. In other words, both equations are the same, and of course you need (at least) two equations to know $x$ and $y$.
2007-08-20 21:20:27 · answer #2 · answered by Pablo C 1 · 0⤊ 0⤋
In the second equation, add y to both sides and add 14 to both getting 4x + 14 = y. Replace the other y with (4x + 14) getting
2x - 3(4x + 14) = -7
Distribute the -3: 2x - 12x - 42 = -7
That's -10x - 42 = -7, or -10x = 35, or x = -3.5. Go from there.
The other one actually does have no solution.
2007-08-20 21:08:31 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋
solve second eqn for y
y = 14 + 4x
sub in the first eqn
2x-3*(14+4x) = -7
use distributive property.
2x-42-12x = -7
-10x=35
x= -3.5
sub in x to find y (either eqn)
y=4(-3.5)+14
y = -14+14 = 0
use same method for other set :)
2007-08-20 21:10:39 · answer #4 · answered by chicabonita 4 · 0⤊ 0⤋
1st set ( -3.5, 0 ) Just graph it ! and set x and y to 0
2nd system is parallel see 15x + 6y = 15 or 5x + 2y = 5 ( factor out the 3 from both sides )
2007-08-20 21:40:34 · answer #5 · answered by Will 4 · 0⤊ 0⤋
-14 = -14 accually means 1 not no solution. Knowing one variable plug it in and find for the second one.
2007-08-20 21:06:07 · answer #6 · answered by dudas_91 4 · 0⤊ 0⤋