SOLUTION: the difference of two positive numbers is 3 and the sum of their square is 45. Find the numbers.
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Question 248047: the difference of two positive numbers is 3 and the sum of their square is 45. Find the numbers.
Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
the difference of two positive numbers is 3 and the sum of their squares is 45. Find the numbers.
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Equations:
x - y = 3
x^2 + y^2 = 45
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x = y+3
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Substitute for "x" and solve for "y":
(y+3)^2 + y^2 = 45
y^2 + 6y + 9 + y^2 = 45
2y^2 + 6y - 36 = 0
y^2 + 3y - 18 = 0
(y+6)(y-3) = 0
y = 3 or y = -6
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If y = 3, x-3 = 3, x = 6
If y = -6, x --6 = 3, x = -3
------------------------------------
Checking x^2 + y^2 = 45
3^2 + 6^2 = 9 + 36 = 45
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Solutions:
x=6 and y = 3
OR
x = -3 and y = -6
=============================
Cheers,
Stan H.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
The difference of two positive numbers is 3.
x - y = 3
.
x^2 + y^2 = 45
.
We can use the first equation to solve the second.
x - y = 3
.
Add y to both sides
x = y + 3
.
Substitute in the second equation and solve.
x^2 + y^2 = 45
(y+3)^2 + y^2 = 45
(y^ + 6y + 9) + y^2 = 45
.
Collect terms
2y^2 + 6y + 9 = 45
.
Subtract 45 from both sides.
2y^2 + 6y - 36 = 0
.
Divide by 2
.
y^2 + 3y - 18 = 0
.
Can we factor 18 into two terms that are 3 apart? Yes, 3 & 6.
.
(y + 6)(y - 3) = 0
.
So we have two candidate answers: y = -6 and y = 3.
.
The problem says the numbers are positive, so we assume y = 3 is one answer.
.
Given x^2 + y^2 = 45, we know y^9 = 9.
x^2 + 9 = 45
.
Subtract 9 from both sides
x^2 = 36
.
Take the square root.
x = 6.
.
So our second candidate answer is x = 6.
.
Checking our work.
.
What does x^2 + y^2 equal 45?
3^2 + 6^2 = 9 + 36 = 45
.
Does x - y = 3?
6-3 = 3
.
OK
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