Answer:
Second option: (-2,-3) and (1,0)
Step-by-step explanation:
Given the system of equations
, you can rewrite them in this form:
![x^2 + 2x-3= x - 1](https://tex.z-dn.net/?f=x%5E2%20%2B%202x-3%3D%20x%20-%201)
Simplify:
Factor the quadratic equation. Choose two number whose sum be 1 and whose product be -2. These are: 2 and -1, then:
![(x+2)(x-1)=0\\\\x_1=-2\\\\x_2=1](https://tex.z-dn.net/?f=%28x%2B2%29%28x-1%29%3D0%5C%5C%5C%5Cx_1%3D-2%5C%5C%5C%5Cx_2%3D1)
Substitute each value of "x" into any of the original equation to find the values of "y":
![y_1= (-2) - 1=-3\\\\y_2=(1)-1=0](https://tex.z-dn.net/?f=y_1%3D%20%28-2%29%20-%201%3D-3%5C%5C%5C%5Cy_2%3D%281%29-1%3D0)
Then, the solutions are:
(-2,-3) and (1,0)
<span>2√(x-5) = 2
</span><span>√(x-5) = 1
</span>x-5 = 1
x=6(I don't know what extraneous solution means though)
You would have to solve for x the plug in the answer for c into the next equation to find your answer
Answer:
c- (2x+3)(x+1)
Step-by-step explanation:
Let me know if you want step by step
the length is given and i got the width as 8 so you just add up all the sides and i got 40