Question 348952
solve the following system of equations:
x + y + {{{sqrt(xy)}}} = 28
x^2 + y^2 + xy = 336
:
Simplify the 1st equation to use for elimination
x + y = {{{28 - sqrt(xy)}}}
square both sides
{{{x^2 + 2xy + y^2}}} = {{{784 - 56sqrt(xy) + xy}}}
subtract xy from both sides
{{{x^2 + 2xy - xy+ y^2}}} = {{{784 - 56sqrt(xy)}}}
{{{x^2 + xy+ y^2}}} = {{{784 - 56sqrt(xy)}}}
:
Use for elimination with 2nd equation
{{{x^2 + y^2+ xy}}} = {{{784 - 56sqrt(xy)}}}
{{{x^2 + y^2 + xy = 336}}}
---------------------------Subtraction eliminates everything on the left, we have:
{{{0 = 448 - 56sqrt(xy)}}}
{{{56sqrt(xy) = 448}}}
divide both sides by 56, results:
{{{sqrt(xy) = 8}}}
square both sides
xy = 64
y = {{{64/x}}}
:
Substitute {{{64/x}}} for y in the 1st equation
x + y + {{{sqrt(xy)}}} = 28
x + {{{64/x}}} + {{{sqrt(x(64/x))}}} = 28
x's inside the radical cancel
x + {{{64/x}}} + {{{sqrt(64))}}} = 28
x + {{{64/x}}} + 8 = 28
x + {{{64/x}}} = 28 - 8
x + {{{64/x}}} = 20
multiply by x
x^2 + 64 = 20x
A quadratic equation
x^2 - 20x + 64 = 0
Factors to
(x-16)(x-4)
x = 16
x = 4
:
Find y:
When x=16: y = 64/16 = 4
When x=4: y = 64/4 = 16
:
:
Check solution in the 1st equation
16 + 4 + {{{sqrt(16*4)}}} = 28
20 + {{{sqrt(64)}}} = 28; confirms our solutions; x=16, y-4 and x=4, y=16