Solve the system below and identify the shape of the graph of the equation. 7y^2+x^2=64 x+y=4
x = - y + 4 substitute, it's a long process doing this stuff lol. You can tell the first one is a circle from its form and the second equation is a line.
The first one is close to a circle but not exactly.
what do i do with x = -y+4
Substitute it into the other equation for x. \[\huge 7y^2 + (4-y)^2 = 64\] and solve for y.
and then find the y (factor y)
first would I multiply (4-y)(4-y)??
yeah
I got 16+y^2, is that correct?
16 + y^2 - 8y
oh! I thought one was positive and one was negative!
(4-y)^2 = (4-y)(4-y)
so after i get that solution what do i plug it into?
\[\huge 7y^2 + 16 + y^2 - 8y = 64\] get everything on one side and combine the like terms and then use the quadratic formula or factor it. to get your y values.
if u confuse, try using this formula : \[\frac{ (-b)\pm \sqrt{b ^{2}-4ac} }{ 2a }\]
thanks! let me solve first using what shamil gave me :)
okay I got 1+- 40/2 which gave me 41/2 and -39/2. is that correct? i used the quadratic formula
did i do this write though if the question says "For questions 10 - 11, solve each system. In your work, identify the shape of the graph of each equation." ??
7y^2 +16 + y^2 - 8y = 64 8y^2 - 8y - 48 = 0 8(y^2 - y - 6 ) = 0 y^2 -y - 6 = 0 (y-2)(y+3) = 0 y = 2, y = -3
How did you end up with those values? o_O
tblue said to use quadratic formula ._. so no?
it just another way
(y-3) ( y+2) = 0 oops
y = 3, y = -2 *
i like using factoring where possible it's much easier.
hm, what would i do with those numbers then?
so u can substitute to x + y = 4
So anyways. y = 3, y = -2 x+y=4 x + 3 =4 x = 1, x -2 =4 x =6
So, your solutions are (1,3) and ( 6,-2)
he's right
thank you! so @shamil98 (1,3) and (6,-2) would be the equation to each system?
Those are the solutions to the system.
& when it says " identify the shape of the graph of each equation" that would be where you said the top eqation was close to being a circle and the bottom equation was a line?
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