Question 1196472: Given the simultaneous equations
lgx + 2lgy = 1, x-3y^2 = 13, show that x^2 -13x - 30 = 0.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Log Rules:
log(A)+log(B) = log(A*B)
log(A)-log(B) = log(A/B)
B*log(A) = log(A^B)
Use those rules to get:
log(x) + 2log(y) = 1
log(x) + log(y^2) = 1
log(xy^2) = 1
Assuming base 10 logs, then,
log(xy^2) = 1
xy^2 = 10^1
xy^2 = 10
y^2 = 10/x
Now let's use the second equation
x - 3y^2 = 13
x - 3(10/x) = 13
x - 30/x = 13
x^2 - 30 = 13x
x^2-13x-30 = 0
In the second to last step, I multiplied both sides by x to clear out the fraction.
This is the same as multiplying every term by x.
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