Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. x − 3y = −5 4x + 3y = 10
a) try to solve it if you get to one single solution x=something y=something_else then it is a single solution. b) if you left with x=something*y+something_else that is infinite solutions c) if you left with x=something and x=something_else - this is a contradiction thus - no solutions
i put no solotions the answer was wrong
with the first equation you solve for x subtracting it on both sides
This has unique solution x-3y=-5 4x+3y=10. Multiplying first one by 4, 4x-12y=-20. Subtracting from second. 15y=30,y=2. x=1
lets do (a) solve it (from line 1) x - 3y = -5 thus x=3y-5 (from line 2) 4x + 3y = 10 thus 4(3y-5)+3y=10 thus 12y-20+3y=10 thus 15y=30 thus y = 2 back to x=3y-5 means x=3*2-5 x=1
giselle3214 you understand what i did?
ohkay so with this 2x − y = 5 and 3x+ y = −6 (first equation ) x = y-5 ( second equation ) ??
line two confused me bratner
the idea is to start with one line as if it was the only one and solve for one of the unknowns. i solved for x. then go to the second line and substitute x with what you found from the first line . you should be left with an equation with one unknown. solve it and you are through.
ohhhhhh you have to substitue for the second line i thought you solve for it again
here is example of "no solution" situation: x+y=10 2x+2y=30 tell me if you can't calculate and make sure there is actually no solution.
i get it know then
thank you!
now*
it equals out to the same thing
i specifically "cooked" the equations to have no solution. please write your calculations so i can see what is wrong.
with the first one x and y equals to one so that would make the second one equals itself
no this is incorrect. concentrate on the first line and solve for x. tell me what you get.
the first line is x+y = 10
you minus x on both sides. y=x-10
actually you should get y=10-x
it matters the order?
the sign infront of the x and 10 matters because => 15 = 20-5 but not 15 = 5-20
see my point?
ohkay so it does. i see your point
no go to the second line and substitute y for 10-x ( we found that y = 10-x) the second line was 2x+2y=30 . you will be left with equation with one unknown (x) - solve it and tell me what you get.
2x+2(10-x)=30
good. now you can solve it for x.
2x=20-2x=30
x=10, x= -15
not really. 2x+2(10-x)=30 means that 2x+20-2x=30
i don't understand how the second "equals" sign got into one equation.
now you have 2x+20-2x=30 and that leads us to 20=30 (2x-2x is 0) which is not true ("contradiction"). which means that the set of equations that i gave you is absurd and has no solution.
im sooooo confsued
2x − y = 5 3x + y = −6
would that be one solution?
yes.
start with the first line and solve for say... y because it is multiplied only by (minus sign). what you get?
y=2x+5
that is wrong. 2x-y=5 / -2x from both sides / you get -y=5-2x /multiply by -1 both sides you get y=-5+2x / change order you get y=+2x-5
-5?
yep . if you multiply 5 by -1 you get 5*(-1)=-5
i see you have trouble with multiplication by -1. it goes like this -y = 5 - 2x /multiply by -1 both sides means (-1)*(-y) = -1*(5-2x) /open paranthesis y = -1*5-1*(-2x) y = -5+2x y = 2x - 5
the whole thing is to get rid of that minus sign that we get stuck with in front of y
now go to the second line and substitute that y with 2x-5
3x+y+-6 3x+y(2x-5)
its an equation it should have an equal sign. and by substituting i mean "write 2x-5 instead of y"
the idea is that we "already know what y is! y is 2x-5" that is how we get rid of all y in the second line.
try again?
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