Try substituting those values, and you will see they do not fit either equation.
looking at the 2nd equation, y = x+2, so
x+2 = x^2+5x-3
x^2 + 4x - 5 = 0
(x+5)(x-1) = 0
x = -5 or 1
so, y = -3 or 3
the solutions are thus (-5,-3)(1,3)
15. What is the solution of the linear-quadratic system of equations?
{y = x^2 + 5x - 3}
{y - x = 2}
13 answers
first things first...calm down. yes, its extremely hard. but we go simple approach. y-x=2 implies y=x+2.we substitute that in the first equation to get x+2 = x^2+5x-3 =x^2+4x-5=0.
We factor to get (x-1)(x+5)=0. Thus, x=1, or -5. PLugging in, we get y=-3,3. Thus, (1,3)(-5,-3) are the solutions
We factor to get (x-1)(x+5)=0. Thus, x=1, or -5. PLugging in, we get y=-3,3. Thus, (1,3)(-5,-3) are the solutions
makes no damn sense
what about the first equation?
Find the solutions of the system of equations braceleftze y = x2 + 3 y = x + 15 .
you guys are life savers thank you
does anyone else try to understand this but just can't so you end up cheating?
Yeah but it because algebra get confusing real quick for me.
@ight, no I cheat reguardless. I would have no idea where to start. lmao
ong tho i try to understand math sometimes but it's just so poorly described in class i feel weird trying to ask for extra help when everyone else already understands ._.
lmfao learning algebra 2 is like learning Japanese for me π it makes no sense to meπππ
yall r life savers
How do you know at the end of this equation that y=-3 and 3?
9 answers