cutiecomittee123
  • cutiecomittee123
Solve this system of conic section equations 7y^2+x^2=64 x+y=4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
First solve the equation x+y = 4 for y \[\Large x+y=4\] \[\Large y=4-x\] Now plug this into the other equation so you boil things down to one variable only \[\Large 7y^2+x^2=64\] \[\Large 7(4-x)^2+x^2=64\] \[\Large 7(16-8x+x^2)+x^2=64\] \[\Large 112-56x+7x^2+x^2=64\] \[\Large 112-56x+8x^2=64\] \[\Large 112-56x+8x^2-64=0\] \[\Large 8x^2-56x+48=0\] From this point, you need to solve for x. To do so, I recommend using the quadratic formula. I'll let you do this part. Tell me what x values you get as the solutions.
cutiecomittee123
  • cutiecomittee123
I get x=1 and x=6
jim_thompson5910
  • jim_thompson5910
me too

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jim_thompson5910
  • jim_thompson5910
if x = 1, then y = ???
cutiecomittee123
  • cutiecomittee123
sweet and then just plug that into on of the original equations to solve for y
jim_thompson5910
  • jim_thompson5910
yes or plug into y = 4-x since that already has y isolated
cutiecomittee123
  • cutiecomittee123
so like y=4-1 and y=4-6
cutiecomittee123
  • cutiecomittee123
y=3 and y=-2
jim_thompson5910
  • jim_thompson5910
correct, so the solutions are these two ordered pairs (1,3) and (6,-2) and this confirms it http://www.wolframalpha.com/input/?i=7y^2%2Bx^2%3D64%2Cx%2By%3D4
cutiecomittee123
  • cutiecomittee123
wanna help me with another one?
jim_thompson5910
  • jim_thompson5910
sure
jim_thompson5910
  • jim_thompson5910
what's your question?
cutiecomittee123
  • cutiecomittee123
x^2+y^2+2x+2y=0 x^2+y^2+4x+6y+12=0 solve this system of equations
jim_thompson5910
  • jim_thompson5910
ok at first this looks really complicated, but we can eliminate quite a bit here notice how each equation has x^2+y^2 in it so we can subtract the equations (either equation1 - equation2 or equation2-equation1) to eliminate the x^2+y^2 terms what do you get when you subtract?
cutiecomittee123
  • cutiecomittee123
2x+4y+12=0
jim_thompson5910
  • jim_thompson5910
Now let's solve 2x+4y+12=0 for x 2x+4y+12=0 2x+4y+12-12=0-12 2x+4y = -12 2x+4y-4y = -12 - 4y 2x = -4y - 12 2x/2 = (-4y-12)/2 x = -2y - 6
cutiecomittee123
  • cutiecomittee123
now plug that into the other equation
jim_thompson5910
  • jim_thompson5910
Next, plug x = -2y - 6 into either original equation. I'll pick the first equation x^2+y^2+2x+2y=0 (-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6 (4y^2+24y+36)+y^2+2(-2y-6)+2y=0 4y^2+24y+36+y^2-4y+12+2y=0 5y^2+22y+48=0 solve that for y (use the quadratic formula). Tell me what you get
jim_thompson5910
  • jim_thompson5910
yes correct
jim_thompson5910
  • jim_thompson5910
hmm I messed up, let me fix
jim_thompson5910
  • jim_thompson5910
x^2+y^2+2x+2y=0 (-2y-6)^2+y^2+2(-2y-6)+2y=0 ... replace x with -2y-6 (4y^2+24y+36)+y^2+2(-2y-6)+2y=0 4y^2+24y+36+y^2-4y-12+2y=0 ... it should be -12, not +12 5y^2+22y+24=0
cutiecomittee123
  • cutiecomittee123
well I tried and I got the b^2-4(a)(c) = -476 you cant take a square root of a negative number
jim_thompson5910
  • jim_thompson5910
I fixed my mistake and got 5y^2+22y+24=0
cutiecomittee123
  • cutiecomittee123
oh gotcha let me try that
cutiecomittee123
  • cutiecomittee123
y=-2 and y=-2.4
jim_thompson5910
  • jim_thompson5910
-2.4 or -12/5
jim_thompson5910
  • jim_thompson5910
now use each y value to find the corresponding x value
cutiecomittee123
  • cutiecomittee123
-12/5? how did you get that?
cutiecomittee123
  • cutiecomittee123
-22+ or =- the sqrt of 4
cutiecomittee123
  • cutiecomittee123
over 10
jim_thompson5910
  • jim_thompson5910
I used the quadratic formula or you can convert -2.4 to fraction form -2.4 = -2.4*(10/10) = -24/10 = -12/5
cutiecomittee123
  • cutiecomittee123
makes sense
cutiecomittee123
  • cutiecomittee123
so now we plug in the y values to get x
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
you can use x = -2y - 6
cutiecomittee123
  • cutiecomittee123
x=-10.8 and x=-10
jim_thompson5910
  • jim_thompson5910
incorrect on both
jim_thompson5910
  • jim_thompson5910
if y = -2, then x = -2y - 6 x = -2(-2) - 6 x = 4 - 6 x = -2
jim_thompson5910
  • jim_thompson5910
if y = -2.4, then x = -2y - 6 x = -2(-2.4) - 6 x = 4.8 - 6 x = -1.2
cutiecomittee123
  • cutiecomittee123
so the solutions are (-2,-2) and (-1.2, -12/
jim_thompson5910
  • jim_thompson5910
correct and the graph confirms it
1 Attachment
jim_thompson5910
  • jim_thompson5910
I think you meant to say -12/5
jim_thompson5910
  • jim_thompson5910
if you use -12/5 then convert -1.2 to fraction form as well -1.2 = -1.2*(10/10) = -12/10 = -6/5
cutiecomittee123
  • cutiecomittee123
Yeah I did mean -12/5 thanks:))
jim_thompson5910
  • jim_thompson5910
np

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