Simultaneous Equation difficulty

J

jonnburton

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Hi

I have been unable to solve the following simultaneous equation and was wondering whether anyone could point me in the right direction/show what I'm doing wrong in my working.

A) 5x + 3y = 4

B) 5x^2 - 3y^2 = 8

Rearranging A:

C) 3y = 4-5x


so B is:

5x^2 - (4-5x)^2 = 8

5x^2 - (16-40x+25) = 8

-20x^2 +40x -24 = 0


And this is where I reach a dead end as the final quadratic equation is not factorizable (b^2 - 4ac = 3520).

Any information would be greatly appreciated!
 
Hello, jonnburton!

\(\displaystyle \begin{Bmatrix}(A)& 5x + 3y &=& 4 \\ (B)& 5x^2 - 3y^2 &=& 8 \end{Bmatrix}\)

Rearranging (A): .\(\displaystyle 3y \:=\: 4-5x\)
so B is: .\(\displaystyle 5x^2 - (4-5x)^2 \:=\: 8\) . This is incorrect.
Click to expand...

Rearranging (A): .\(\displaystyle y \:=\:\frac{4-5x}{3}\)
Substitute into (B): .\(\displaystyle 5x^2 - 3\left(\frac{4-5x}{3}\right)^2 \:=\:8\)
Got it?


By the way, most quadratics are not factorable.
That's why they invented the Quadratic Formula.
 
Hi Jon

Soroban has given you your answer, but I want to point out HOW you made your mistake.

To substitute for y in equation B you must find y. You did not. You skipped a step. You found 3y and saw 3y^2 in equation B and thought you were saving time by substituting 3y for 3y^2. But of course 3y = 3y^2 only if y = 0 or y = 1. I suspect that 75% of the mistakes made in math come from skipping steps.

You made several other mistakes. You came up with the equation:

\(\displaystyle -20x^2 + 40x - 24 = 0.\) You made your arithmetic much harder by skipping the step of simplification.

\(\displaystyle -20x^2 + 40x - 24 = 0 \implies 5x^2 - 8x + 6 = 0.\)

Much easier arithmetic. Because you did not simplify you made an arithmetic mistake.

\(\displaystyle b^2 - 4ac \implies (40)^2 - 4(-20)(-24) = 1600 - 4(+480) = 1600 - 1920 = - 320 \ne + 3520.\)

MORAL: DONT SKIP STEPS
 
Hi Soroban and Jeff, many thanks to both of you for your help. I can see where I went wrong with this now - and thanks for the advice on not skipping steps! I need to pay attention to this.
 
I have been trying to get further with these simultaneous equations but still can't solve them!

This is what I have done:

\(\displaystyle 5x + 3y = 4\)

\(\displaystyle 5x^2 + 3y^2 = 8\)

\(\displaystyle 5x^2 + 3(\dfrac{4-5x}{3}) - 8 = 0 \)

Multiply throughout, to get rid of the fraction:

\(\displaystyle 15x^2-3(4-5x)^2 - 24 =0\)

\(\displaystyle 15x^2-3(16-40x+25x^2) - 24 =0\)

\(\displaystyle 15x^2-48 +120x-75x^2 - 24 =0\)

\(\displaystyle -60x^2+120x -72 =0\)

Simplifies to:

\(\displaystyle -5x^2+10x -6 =0\)

According to the answer in the book, the solution is:
\(\displaystyle x=2\)
\(\displaystyle y=-2\)

But when using the quadratic formula, I get:

\(\displaystyle x = \dfrac {-10+/-\sqrt-100-(4*-5*-6)}{-10}\)

\(\displaystyle x = \dfrac {-10+/-\sqrt-20}{-10}\)

Which I don't think makes any sense because a negative number doesn't have a square root.
 
jonnburton said:
I have been trying to get further with these simultaneous equations but still can't solve them!

This is what I have done:

\(\displaystyle 5x + 3y = 4\)

\(\displaystyle 5x^2 + 3y^2 = 8\) ← You have careless typo here

\(\displaystyle 5x^2 + 3(\dfrac{4-5x}{3}) - 8 = 0 \) ← You have careless typo here

Multiply throughout, to get rid of the fraction:

\(\displaystyle 15x^2-3(4-5x)^2 - 24 =0\) ← This should be \(\displaystyle 15x^2 - 1* (4-5x)^2 - 24 =0\)

\(\displaystyle 15x^2-3(16-40x+25x^2) - 24 =0\)

\(\displaystyle 15x^2-48 +120x-75x^2 - 24 =0\)

\(\displaystyle -60x^2+120x -72 =0\)

Simplifies to:

\(\displaystyle -5x^2+10x -6 =0\)

According to the answer in the book, the solution is:
\(\displaystyle x=2\)
\(\displaystyle y=-2\)

But when using the quadratic formula, I get:

\(\displaystyle x = \dfrac {-10+/-\sqrt-100-(4*-5*-6)}{-10}\)

\(\displaystyle x = \dfrac {-10+/-\sqrt-20}{-10}\)

Which I don't think makes any sense because a negative number doesn't have a square root.
Click to expand...


.
 
jonnburton said:
I have been trying to get further with these simultaneous equations but still can't solve them!

This is what I have done:

\(\displaystyle 5x + 3y = 4\) \(\displaystyle \implies 3y = 4 - 5x \implies y = \dfrac{4 - 5x}{3}.\) Still skipping steps.


\(\displaystyle 5x^2 + 3y^2 = 8\) How did a minus sign become a plus sign. Be careful.

\(\displaystyle 5x^2 + 3(\dfrac{4-5x}{3}) - 8 = 0 \) No \(\displaystyle 5x^2 - 3\left(\dfrac{4 - 5x}{3}\right)^2 - 8 = 0.\) Denominator is \(\displaystyle 3^2 = 9.\)

By this time you are so far off track that there is no point in checking further.

Multiply throughout, to get rid of the fraction:

\(\displaystyle 15x^2-3(4-5x)^2 - 24 =0\)

\(\displaystyle 15x^2-3(16-40x+25x^2) - 24 =0\)

\(\displaystyle 15x^2-48 +120x-75x^2 - 24 =0\)

\(\displaystyle -60x^2+120x -72 =0\)

Simplifies to:

\(\displaystyle -5x^2+10x -6 =0\)

According to the answer in the book, the solution is:
\(\displaystyle x=2\)
\(\displaystyle y=-2\)

But when using the quadratic formula, I get:

\(\displaystyle x = \dfrac {-10+/-\sqrt-100-(4*-5*-6)}{-10}\)

\(\displaystyle x = \dfrac {-10+/-\sqrt-20}{-10}\)

Which I don't think makes any sense because a negative number doesn't have a square root.
Click to expand...
.
 
I think I might have got closer to solving this.

\(\displaystyle 5x + 3y = 4\)

\(\displaystyle 5x^2 - 3y^2 = 8\)


\(\displaystyle 3y = 4 - 5x\)

\(\displaystyle y = \dfrac {4 - 5x}{3}\)



\(\displaystyle 5x^2 - 3(\dfrac{4-5x}{3})^2 - 8 = 0 \)

\(\displaystyle (\dfrac{4-5x}{3})^2 = (\dfrac {16-40x+25x^2}{9})[\tex]


so,
\(\displaystyle 5x^2 - 3(\dfrac{16-40x+25x^2}{9}) - 8 = 0 \)

Here I multiplied everything by 9, to get rid of the fraction:

\(\displaystyle 45x^2 - 3(16-40x+25x^2) - 72 = 0 \)

Simplifies to:

\(\displaystyle 15x^2 - (16-40x+25x^2) - 24 = 0 \)

\(\displaystyle -10x^2+40x-40 = 0 \)

Simplifies to:
\(\displaystyle -x^2+4x-4 = 0 \)

This won't factorize as

\(\displaystyle b^2-4ac = 16-(-4*-4)= 0\)


So, using the quadratic equation:

\(\displaystyle x =\dfrac{-b +/- \sqrt b^2-4ac}{2a}\)

\(\displaystyle x = \dfrac{-4 +/- \sqrt 0}{-2} = 2\)

Substituting into the first equation gives:

\(\displaystyle 10 + 3y = 4\)
\(\displaystyle 3y = 4 - 10\)
\(\displaystyle 3y = -6\)
\(\displaystyle y = -2\)

Many thanks for all of your guidance, I do appreciate it a lot!\)
 
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jonnburton said:
I think I might have got closer to solving this.

\(\displaystyle 5x + 3y = 4\)

\(\displaystyle 5x^2 - 3y^2 = 8\)


\(\displaystyle 3y = 4 - 5x\)

\(\displaystyle y = \dfrac {4 - 5x}{3}\)



\(\displaystyle 5x^2 - 3(\dfrac{4-5x}{3})^2 - 8 = 0 \)

\(\displaystyle (\dfrac{4-5x}{3})^2 = (\dfrac {16-40x+25x^2}{9})[\tex]


so,
\(\displaystyle 5x^2 - 3(\dfrac{16-40x+25x^2}{9}) - 8 = 0 \)

Here I multiplied everything by 9, to get rid of the fraction:

\(\displaystyle 45x^2 - 3(16-40x+25x^2) - 72 = 0 \)

Simplifies to:

\(\displaystyle 15x^2 - (16-40x+25x^2) - 24 = 0 \)

\(\displaystyle -10x^2+40x-40 = 0 \)

Simplifies to:
\(\displaystyle -x^2+4x-4 = 0 \)

This won't factorize as =====> WRONG. It does factor.

\(\displaystyle b^2-4ac = 16-(-4*-4)= 0\) ====> In fact, the discriminant = 0, so not only does it factor, but it has only one solution (see below).


So, using the quadratic equation:

\(\displaystyle x =\dfrac{-b +/- \sqrt b^2-4ac}{2a}\)

\(\displaystyle x = \dfrac{-4 +/- \sqrt 0}{-2} = 2\)

Substituting into the first equation gives:

\(\displaystyle 10 + 3y = 4\)
\(\displaystyle 3y = 4 - 10\)
\(\displaystyle 3y = -6\)
\(\displaystyle y = -2\)

Many thanks for all of your guidance, I do appreciate it a lot!\)
Click to expand...
\(\displaystyle

\(\displaystyle -x^2+4x-4=0\)

I think it is easiest to factor when the leading coefficient is not negative, so divide through by the leading coefficient, in this case -1:

\(\displaystyle x^2-4x+4=0\)

\(\displaystyle (x-2)(x-2)=0 \text{ or } (x-2)^2=0\)

\(\displaystyle x-2=0\)

\(\displaystyle x=2\)\)
 
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