Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

solve the system of equations
y=2x^2-3
y=3x-1

Sagot :

y= 2x^2-3
x= 2y^2-3
x+3=2y^2
(x+3)/2= y^2
y= squareroot [(x+3)/2)] --> inverse equation
Domain: (-3, infinity)
Range: (0, infinity) 

y= 3x-1

Let x=0
y=3(0)-1
y=-1, Therefore when x=0,y=-1, so point of graph will be (0.-1)

Let x=1
y= 3(1)-1
y= 2, therefore when x=1, y=2 ...point on graph will be (1,2)

Let x be-1
y= 3(-1-)-1
y=-4, therefore when x=-1, y=-4...points on graph will be (-1,-4)

Hope this helps!!!






artasticdrawing

Answer:

The answer would be C) (-1/2,-5/2),(2,5) I believe. :D

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.