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.
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.
I get x=1 and x=6
if x = 1, then y = ???
sweet and then just plug that into on of the original equations to solve for y
yes or plug into y = 4-x since that already has y isolated
so like y=4-1 and y=4-6
y=3 and y=-2
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
wanna help me with another one?
what's your question?
x^2+y^2+2x+2y=0 x^2+y^2+4x+6y+12=0 solve this system of equations
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?
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
now plug that into the other equation
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
hmm I messed up, let me fix
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
well I tried and I got the b^2-4(a)(c) = -476 you cant take a square root of a negative number
I fixed my mistake and got 5y^2+22y+24=0
oh gotcha let me try that
y=-2 and y=-2.4
-2.4 or -12/5
now use each y value to find the corresponding x value
-12/5? how did you get that?
-22+ or =- the sqrt of 4
I used the quadratic formula or you can convert -2.4 to fraction form -2.4 = -2.4*(10/10) = -24/10 = -12/5
so now we plug in the y values to get x
you can use x = -2y - 6
x=-10.8 and x=-10
incorrect on both
if y = -2, then x = -2y - 6 x = -2(-2) - 6 x = 4 - 6 x = -2
if y = -2.4, then x = -2y - 6 x = -2(-2.4) - 6 x = 4.8 - 6 x = -1.2
so the solutions are (-2,-2) and (-1.2, -12/
I think you meant to say -12/5
if you use -12/5 then convert -1.2 to fraction form as well -1.2 = -1.2*(10/10) = -12/10 = -6/5
Yeah I did mean -12/5 thanks:))