2x + 5y = 2
-3x - y =-3
-3x - y = -3
y = 3x - 3
substitute y = 3x - 3 into the first equation.
2x + 5y = 2
2x + 5(3x - 3) = 2
2x + 15x - 15 = 2
17x - 15 = 2
solve for x in 17x - 15 = 2
17x - 15 = 2
17x = 2 + 15
17x = 17
x = 1
substitute x = 1 into y = 3x - 3
y = 3x - 3
y = 3(1) - 3
y = 3 - 3
y = 0
(1, 0) << the answer
hope this helped, God bless!
Answer:
It's like solving a quadratic, but in reverse, and in this case you'll arrive at x2+x−12=0
Explanation:
We're going to go "backwards" with this problem - normally we're asked to take a quadratic equation and find the roots. So we'll do what we normally do, but in reverse:
Let's start with the roots:
x=3, x=−4
So let's move the constants over with the x terms to have equations equal to 0:
x−3=0, x+4=0
Now we can set up the equation, as:
(x−3)(x+4)=0
We can now distribute out the 2 quantities:
x2+x−12=0
Answer:
Hello! answer: 55
Step-by-step explanation:
55 + 35 = 90 HOPE THAT HELPS!
