anonymous
  • anonymous
Help Please!!! Solve the following system of equations and show all work. y = x2 + 3 y = x + 5
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@mathmale
anonymous
  • anonymous
Or anyone that can help me
anonymous
  • anonymous
what do you mean by "solve?"

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I'm not sure thats all they told us.
anonymous
  • anonymous
you would write it: x+5=x2+3 and then solve from there. the reason that is corrret is because youre just repacing y with what y is equal to- y
anonymous
  • anonymous
Okay thank you
anonymous
  • anonymous
Do I solve for x?
anonymous
  • anonymous
do the substitution method where you multiple one of the equations by -1
NotTim
  • NotTim
Do not do what ew.sorry said. it may work, but @arek 's method is much, much safer.
anonymous
  • anonymous
Okay I will. For the substitution method do I make y -1? Sorry I'm not that great at math.
whpalmer4
  • whpalmer4
\[y=x^2+3\]\[y=x+5\] Those are two equations to solve. A couple of routes you can take: substitution setting the two right hand sides equal to each other substitution means solving one equation for one variable in terms of the other. Then you substitute for that variable in the other equation. In this problem, this would a fine way to go because the first step has already been done for you: you have \(y=x+5\) as one of the equations, and that gives you \(y\) in terms of \(x\). Take the other equation \((y = x^2+3)\) and replace \(y\) with \(x+5\), then solve for the values of \(x\). There will be two solutions thanks to the \(x^2\) term, although in some cases they may turn out to be the same value. another approach (which is really substitution as well) is to notice that we have two equations with the same thing on one side. Here, both equations have \(y\) by itself on one side of the equals sign. We can simply set the other two sides equal to each other, because they are equal to the same thing. After that, you solve just as before. In both approaches, after you find the value(s) of \(x\), you plug them into one of the original equations to find the corresponding values of \(y\).
whpalmer4
  • whpalmer4
Another approach would be to graph the two equations and observe where they intersect. \[y=x^2+3\]is a parabola, and \[y=x+5\]is a line. The line crosses the parabola in two locations, which are the solutions to this system.
whpalmer4
  • whpalmer4
And yet another approach, if you had the luxury of a multiple-choice problem where you could know that the answers are provided and you merely need to identify the correct answers, would be to plug each set of answers into all of the equations and select the answer choice where all of the equations work. Note well that you can have "answers" that work for some equations but all...the technical term for them is "wrong" :-)
anonymous
  • anonymous
Wow! Thank you so much!
anonymous
  • anonymous
|dw:1433532657818:dw|
anonymous
  • anonymous
I'm stuck at this point. Did I do it wrong?
whpalmer4
  • whpalmer4
\[x^2+3=x+5\] collect everything on one side, what do you get?
anonymous
  • anonymous
0 = -x^2 +x + 2
whpalmer4
  • whpalmer4
Yes, or \[x^2-x-2 = 0 \] Do you know how to factor that equation, or solve it with the quadratic formula, or complete the square?
whpalmer4
  • whpalmer4
@torvia you still here?
anonymous
  • anonymous
Yes sorry my dad had to use the computer
anonymous
  • anonymous
I know how to solve using quadratic formula
anonymous
  • anonymous
|dw:1433535440928:dw|
anonymous
  • anonymous
did i do that right?
whpalmer4
  • whpalmer4
close, but not close enough. I'm intrigued by how you went from \[\sqrt{1+8}\] to \[\sqrt{4}\]in particular :-)
whpalmer4
  • whpalmer4
I think you probably just didn't read your handwriting...
anonymous
  • anonymous
Yeah sorry it was really messy.
whpalmer4
  • whpalmer4
correct answer is...
anonymous
  • anonymous
My solutions then would be 2 and -2?
whpalmer4
  • whpalmer4
let's try them out! \[x^2-x-2=0\]\[(2)^2-(2)-2=0\]\[4-2-2=0\checkmark\] you do the other one (x=-2)
whpalmer4
  • whpalmer4
we are just verifying that we solved for \(x\) correctly here...
anonymous
  • anonymous
\[x^2-x-2 = 0\] \[-2^2-(-2)-2 =0\] 4+2-2 = 0 4 = 0
whpalmer4
  • whpalmer4
Hmm. maybe one of those solutions is incorrect :-)
anonymous
  • anonymous
-2 won't be a solution
whpalmer4
  • whpalmer4
by the way, you really need to write \[(-2)^2\]instead of \[-2^2\] The first one is \((-2)*(-2) = 4\) and the second one is \(-(2)(2) = -4\)
anonymous
  • anonymous
Oh, sorry.
whpalmer4
  • whpalmer4
I'll speed things along a little bit: \[\frac{1\pm\sqrt{9}}{2} = \frac{1+3}{2},\frac{1-3}{2} = \frac{4}{2},\frac{-2}{2} = 2,-1\]
anonymous
  • anonymous
Thanks a million!
whpalmer4
  • whpalmer4
and if we try \(-1\) in the equation: \[(-1)^2 -(-1)-2=0\]\[1+1-2=0\checkmark\] Now, you can plug those two values of \(x\) into one of the equations and find the corresponding values of \(y\).
anonymous
  • anonymous
okay so then y will equal 7 and -4
anonymous
  • anonymous
(2,7) (-1, -4)
whpalmer4
  • whpalmer4
Yep! And if you plug them all in to all the equations, they should all work.
anonymous
  • anonymous
Great! Thank you :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.