How do you solve the system 2x^2+2y^2=15 and x+2y=6?

1 Answer
Nallasivam V
Sep 23, 2016

x=0.72; y=2.64
x=1.68; y=2.16

Explanation:

Given -

2x^2+2y^2=15-----------(1)
x+2y=6 ---------------(2)

Solve equation (2) for x

x=6-2y

Substitute x=6-2y in equation (1)

2(6-2y)^2+2y^2=15
2(36-24y+4y^2)+2y^2=15
72-48y+8y^2+2y^2=15
10y^2-48y+72-15=0
10y^2-48y+57=0

y has two values

y=(-b+-sqrt(b^2-(4xxaxxc)))/(2xxa

y=(-(-48)+-sqrt((-48)^2-(4xx10xx57)))/(2xx10

y=(48+-sqrt(2304-2280))/(2xx10
y=(48+-sqrt(24))/(2xx10)

y=(48+-4.9)/(20)
y=(48+4.9)/(20)=52.9/20=2.64
y=2.64
y=(48-4.9)/(20)=43.1/20=2.16
y=2.16

x also has two values

Substitute y=2.64 in equation (1)

x+2(2.64)=6
x+5.28=6
x=6-5.28=0.72
x=0.72

Substitute y=2.16 in equation (1)

x+2(2.16)=6
x+4.32=6
x=6-4.32=1.68
x=1.68

x,y values are -
x=0.72; y=2.64
x=1.68; y=2.16

Look at the graph

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