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The equations x^2 + y^2 = 8 and x - y = 4 have to be solved by substitution.
x - y = 4
=> x = 4 + y
Substitute in x^2 + y^2 = 8
=> (4 + y)^2 + y^2 = 8
=> 16 + y^2 + 8y + y^2 = 8
=> y^2 + 4y + 4 = 0
=> (y + 2)^2 = 0
=> y = -2
x = 4 + y = 2
The solution of the given equations is x = 2 and y = -2
Knowing those two equations, and knowing that substitution is the way to solve it, we should start by deciding what to substitute. Since the first equation has squares in it, isolating and then using the value of either x or y would be playing with square roots, which could potentially get a bit messy :D So, we'll start with the second equation:
x-y=4, so by isolating x we get x=y+4
We can now plug that in as the 'x' of the first equation, i.e.
Now, we can expand that bracket and simplify:
Minus 8 from both sides to get equation =0
Dividing whole equation by 2, we get y^2+4y+4=0 which can be factorised to give (y+2)^2=0.
This gives the value y=-2, and by substituting this value into x-y=4, we get:
x-y= x-(-2) = x+2 =4
Woohoo! Problem = solved :D
The solution of the system of equations x^2 + y^2 = 8 and x - y = 4 is the point of contact of the circle x^2 + y^2 = 8 and the line x - y = 4.
From the equation of the line x - y = 4 we can express x in terms of y
x - y = 4
x = y + 4
Substitute x = y + 4 in the equation x^2 + y^2 = 8
(y+4)^2 + y^2 = 8
Use the expansion (a + b)^2 = a^2 + 2ab + b^2
y^2 + 8y + 16 + y^2 = 8
Subtract 8 from both the sides and simplify the polynomial.
2y^2 + 8y + 8 = 0
y^2 + 4y + 4 = 0
(y + 2)^2 = 0
y = -2
At y = -2, x = y + 4 = -2 + 4 = 2
The required solution of the equations x^2 + y^2 = 8 and x - y = 4 is x = 2 and y = -2
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