Linear and non-linear equations

• Nov 8th 2011, 07:34 AM
spirit79
Linear and non-linear equations
I can't find the right answer to this question... Is it just me, or are the solutions given wrong?

x^2 + 3x + 5y = 20
x + 3y = 1

Given solutions are x = 3 or x = -9/2

Where did I go wrong?

X = 1 – 3y ---- (1)
X^2 + 3x + 5y = 20 --- (2)
Sub (1) into (2), (1 – 3y)^2 + 3(1 – 3y) + 5y = 20
1 – 6y + 9y^2 + 3 – 9y + 5y = 20
9y^2 – 10y – 16 = 0
(9y + 8 ) (y – 2) = 0
y = -8/9 or y = 2

Sub y = -8/9 in (1),
x = 1 - 3(-8/9)
= 11/3

Sub y = 2 in (1),
x = 1 - 3(2)
= -5

Are the given solutions correct? And are what I worked the right answers? If not, please tell me where I made a mistake. I'm not too sure if I substituted the y values correctly...
• Nov 8th 2011, 07:56 AM
e^(i*pi)
Re: Linear and non-linear equations
Quote:

Originally Posted by spirit79
I can't find the right answer to this question... Is it just me, or are the solutions given wrong?

x^2 + 3x + 5y = 20
x + 3y = 1

Given solutions are x = 3 or x = -9/2

Where did I go wrong?

X = 1 – 3y ---- (1)
X^2 + 3x + 5y = 20 --- (2)
Sub (1) into (2), (1 – 3y)^2 + 3(1 – 3y) + 5y = 20
1 – 6y + 9y^2 + 3 – 9y + 5y = 20
9y^2 – 10y – 16 = 0
(9y + 8 ) (y – 2) = 0
y = -8/9 or y = 2

Sub y = -8/9 in (1),
x = 1 - 3(-8/9)
= 11/3

Sub y = 2 in (1),
x = 1 - 3(2)
= -5

Are the given solutions correct? And are what I worked the right answers? If not, please tell me where I made a mistake. I'm not too sure if I substituted the y values correctly...

Your answer is correct, you can also test your results by subbing them into the original equation.

Spoiler:
For (-5,2) : $\displaystyle (-5)^2 + 3(-5) + 5(2) = 20 \Leftrightarrow 25 - 15+10 = 20$ which is correct as is $\displaystyle (-5) + 3(2) = 1$ so (-5,2) is a valid solution

For (11/3 , -8/9): $\displaystyle \left(\dfrac{11}{3}\right)^2 + 3\left(\dfrac{11}{3}\right) + 5\left(\dfrac{-8}{9}\right) = 20$ so (11/3 , 8/9) is a valid solution.
• Nov 8th 2011, 09:17 AM
spirit79
Re: Linear and non-linear equations
Quote:

Originally Posted by e^(i*pi)
Your answer is correct, you can also test your results by subbing them into the original equation.

Spoiler:
For (-5,2) : $\displaystyle (-5)^2 + 3(-5) + 5(2) = 20 \Leftrightarrow 25 - 15+10 = 20$ which is correct as is $\displaystyle (-5) + 3(2) = 1$ so (-5,2) is a valid solution

For (11/3 , -8/9): $\displaystyle \left(\dfrac{11}{3}\right)^2 + 3\left(\dfrac{11}{3}\right) + 5\left(\dfrac{-8}{9}\right) = 20$ so (11/3 , 8/9) is a valid solution.

I have another question, but I'm not too keen to start another thread topic on it seeing that the questions are related...

Solve:

5x + 3y + 7 = 0
x^2 - 4y + 3 = 3y^2

I tried, but got the wrong answers...?

5x = -3y – 7
25x^2 = 9y^2 – 42y + 49 --- (1)
X^2 – 4y + 3 = 3y^2
25x^2 – 100y + 75 = 75y^2 --- (2)
Sub (1) into (2),
9y^2 – 42y + 49 – 100y + 75 = 75y^2
9y^2 – 42y – 100y – 75y^2 + 49 + 75 = 0
-142y – 75y^2 + 124 = 0
66y^2 + 142y – 124 = 0
2 (33y^2 + 71y – 62) = 0
(11y + 31) (3y – 2) = 0
Y = -31/11 or y = 2/3
When y = -31/11,
5x = -3 (-31/11) – 7
X = 16/55
When y = 2/3,
5x = -3 (2/3) – 7
X = -9/5

The answer key states y = -1 29/33 or y = 1, x = 3/11 or x = -2.