In which quadrant does the solution of the system fall? x+y=4 2x-y=2 a.I b.II c.III d.IV
do you know how to find the solution of the system?
would you like to use elimination or substitution methods?
@Quxxn
substitution
ok do you know how?
no /:
ok start by solving one of the two equations for x or for y, you choose the equation and what variable to solve for
the first equation looks easiest to solve for either x or y
is it x=4-y?
yep, ok now since you know x = 4 - y, look at the other equation and whereever u see an x, replace it with 4-y
2x-y=2 2(4-y)=2 8-2y=2 -2y=2-8 -2y=-6 y=3
very close but the second line you are missing a -y 2(4-y)=2 should be 2(4-y) - y=2
ohh hold on let me try again
k
2(4-y)-y=2 8-2y-y=2 8+ -3y=2 -3y=-6 y=2
yep perfect, so now you know y=2, you can replace y with 2 in either equation to solve for x, i think the first is easiest though..
so whats x?
would it be 2x-2=2? Like is that what i use to solve for x?
yeah that will work,
i would have used the first equation to get x + 2 = 4, but that will work too
2x-2=2 2x=2 x=2/4 x=1/2 x+2=4 x=2 ??
yeah x = 2, the first one should look like this 2x - 2 = 2 (add 2 to both sides) 2x = 4 (added 2 to both sides) x = 4/2 x = 2
oh oops lol
but yeah either way you get 2, so now you know x = 2 and y = 2, so you have the point (2,2), that is where both of the lines intersect, also called the solution of the system (the equations are equations of lines by the way)
ok so for the point (2,2), you have a positive x coordinate and a positive y coordinate, what quadrant would that be in?
Quadrant II right?
no, they are labeled starting with 1 in the upper right, and going counter clockwise, 2 in the upper left, 3 in the lower left and 4 in the lower right
oh i forgot so then its the first quadrant?
yep, x values are positive and y values are positive in the first quadrant, Quadrant I
okay thank you!
yw
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