Question 849344: Hi, can you please help me with this question? I'm stuck on this problem:"Solve this simultaneous equation: y-2=3x and (x-2)^2+y-30=0"
I thought by expanding (x-2)^2 into x^2-4x+4 would work, but it doesn't seem to. Thank you so much for your help in advanced!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the 2 equations are:
y - 2 = 3x
and:
(x-2)^2 + y - 30 = 0
solve for y in terms of x in the first equation.
you will get:
y = 3x + 2
substitute for y in the second equation to get:
(x-2)^2 + (3x + 2) - 30 = 0
simplify the equation to get:
(x-2)^2 + 3x - 28 = 0
now you want to expand the equation to get:
x^2 - 4x + 4 + 3x - 28 = 0
simplify this equation to get:
x^2 - x - 24 = 0
factor this equation by use of the quadratic formula to get:
x = (1 + square root of (97)) / 2
or:
x = (1 - square root of (97)) / 2
i confirmed the first solution is correct.
if x = (1 + square root of (97)) / 2, then:
y = 3 * (1 + square root of (97)) / 2 + 2
in decimal form, these answers become:
x = 5.42443 rounded to 5 decimal places.
y = 18.27329 rounded to 5 decimal places.
replacing x and y in the original equations confirms that these solutions are good.
you can do the same with x = (1 - square root of (97)) / 2
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