1. ## Solving two equations

I have two equations;

(-1/3 x + 14/3) = (2x + 2)

I require to find the value of "x" to show where two lines cross simultaneously on a graph which I have plotted.

On the graph the two lines intersect at approximately x- = 1.1.

I have tried to transpose the above equations to solve for "x", this is what I have concluded;

( - 1/3x + 14/3) = (2x + 2)

( - 1/3x) = (2x + 2 - 14/3)

( - 1/3x + 2x) = (2 - 14/3)

( - 1/3 + 2x^2) = (2 - 14/3)

(2x^2) = (2 - 14/3 + 1/3)

(x^2) = (4/1 - 14/3 + 1/3)

(x^2) = (4/9 - 14/9 + 1/9)

(x^2) = -1

x = 1

Now I don't profess to say that the above transposition is correct, but the answer appears to be very close to the intersection of the graph, where the graph is visually about x- = 1.1

Please advise if there is a more precise way to solve for "x"

Thanks

David

2. ## Re: Solving two equations

Originally Posted by David Green
I have two equations;

(-1/3 x + 14/3) = (2x + 2)

I require to find the value of "x" to show where two lines cross simultaneously on a graph which I have plotted.

On the graph the two lines intersect at approximately x- = 1.1.

I have tried to transpose the above equations to solve for "x", this is what I have concluded;

( - 1/3x + 14/3) = (2x + 2)

( - 1/3x) = (2x + 2 - 14/3)

Fine to here, then errors consistently creep in.

( - 1/3x - 2x) = (2 - 14/3)

( - 1/3 - 2x^2) = (2 - 14/3)?? If you multiply the left side by $x$, then you have to remember to multiply the right side by $x$ too!

(2x^2) = (2 - 14/3 + 1/3)

(x^2) = (4/1 - 14/3 + 1/3)

(x^2) = (4/9 - 14/9 + 1/9)

(x^2) = -1

x = 1 Please review this. 1^2\neq{-1}. $x^2=-1$ has no real solutions for $x$.

Now I don't profess to say that the above transposition is correct, but the answer appears to be very close to the intersection of the graph, where the graph is visually about x- = 1.1

Please advise if there is a more precise way to solve for "x"

Thanks

David
We have:

$\frac{1}{3x}+\frac{14}{3}=2x+2$

Subtracting $\frac{1}{3x}$ from both sides:

$\frac{14}{3}=2x-\frac{1}{3x}+2$

Subtracting $2$ from both sides:

$\frac{14}{3}-2=2x-\frac{1}{3x}$

We can simplify the left hand side by writing $2$ as $\frac{6}{3}$, giving:

$\frac{14}{3}-\frac{6}{3}=2x-\frac{1}{3x}$

$\frac{8}{3}=2x-\frac{1}{3x}$

Now, to solve this, we could, as you've suggested, multiply through by x, as long as we remember that $x\neq{0}$.

$\frac{8x}{3}=2x^2-\frac{1}{3}$

I'd then multiply through by 3 for simplicity, giving:

$8x=6x^2-1$

$6x^2-8x-1=0$

Go for it!

Edit: Removed an error from my calculations.

3. ## Re: Solving two equations

Originally Posted by Quacky
We have:

$\frac{1}{3x}+\frac{14}{3}=2x+2$

Subtracting $\frac{1}{3x}$ from both sides:

$\frac{14}{3}=2x-\frac{1}{3x}+2$

Subtracting $2$ from both sides:

$\frac{14}{3}-2=2x-\frac{1}{3x}$

We can simplify the left hand side by writing $2$ as $\frac{6}{3}$, giving:

$\frac{14}{3}-\frac{6}{3}=2x-\frac{1}{3x}$

$\frac{8}{3}=2x-\frac{1}{3x}$

Now, to solve this, we could, as you've suggested, multiply through by x, as long as we remember that $x\neq{0}$.

$\frac{8x}{3}=2x^2-\frac{1}{3}$

I'd then multiply through by 3 for simplicity, giving:

$8x=6x^2-1$

$6x^2-8x-1=0$

Go for it!
Before I bring the quadratics into it to solve for "x", can I just clarify that you did understand my bad presentations on here?

- 1 / 3 (x)

This means that the fraction - 1/3 is multiplied by "x"

are we on the same wavelength with this?

4. ## Re: Solving two equations

No we weren't on the same wavelength; this was my mistake.

Revised solution:

We have:

$\frac{-1}{3}x+\frac{14}{3}=2x+2$

Immediately, I'd remove the fractions by multiplying through by $3$ to give:

$-x+14=6x+6$

We can then add $x$ to both sides, and subtract $6$ from both sides, to obtain:

$8=7x$

5. ## Re: Solving two equations

Originally Posted by Quacky
No we weren't on the same wavelength; this was my mistake.

Revised solution:

We have:

$\frac{-1}{3}x+\frac{14}{3}=2x+2$

Immediately, I'd remove the fractions by multiplying through by $3$ to give:

$-x+14=6x+6$

We can then add $x$ to both sides, and subtract $6$ from both sides, to obtain:

$8=7x$
Thank you for your advice, I have carried out the following algebra to solve for "x".

- 1/3x + 14/3 = 2x + 2

-1/3x + 14 = 2x + 6

-x + 14 = 6x + 6

-x = 6x + 6 - 14

-x - 6x = - 8

- 7x = - 8

x = 1.1

Thank you

David

6. ## Re: Solving two equations

Just to be sure you know, in your solution you took this step
Originally Posted by David Green

- 1/3x + 14/3 = 2x + 2

-1/3x + 14 = 2x + 6 <----- This line is not true.

-x + 14 = 6x + 6
If you were just using the above line as an intermediate step to get to the 3rd line thats fine. As long as you know you can't stop there if you wanted.

In algebra addition and subtraction tend to work on separate things. That is you can only add up Xs together, or only add up numbers together.

Multiplication and division are different. When you multiply by 3 on each side you have to do it to every term on both sides. I wanted to make sure you didn't think you were multiplying by 3 just the xs, and then multiplying by 3 the regular numbers. Those 2 lines are just 1 step not 2. and the last line is true, while the middle line is incorrect.

7. ## Re: Solving two equations

Originally Posted by takatok
Just to be sure you know, in your solution you took this step

If you were just using the above line as an intermediate step to get to the 3rd line thats fine. As long as you know you can't stop there if you wanted.

In algebra addition and subtraction tend to work on separate things. That is you can only add up Xs together, or only add up numbers together.

Multiplication and division are different. When you multiply by 3 on each side you have to do it to every term on both sides. I wanted to make sure you didn't think you were multiplying by 3 just the xs, and then multiplying by 3 the regular numbers. Those 2 lines are just 1 step not 2. and the last line is true, while the middle line is incorrect.
Hi takatok,

Thank you for your advice, I appreciate what you point out, however that as you also mentioned is just one step towards the final solution. I am not 100% why you point out that it is incorrect, although I don't doubt that there will be other ways to perform the algebra, but to me, and I am no expert, the algebra does seem to follow through to the correct solution, which is backed up by the intersection of the graph.

If you feel that there is a better way to carryout the algebra, please do point me in the right direction as this is a learning curve for me.

Thanks

David

8. ## Re: Solving two equations

When 2 sides of an equation are equal, to solve them you do the exact same thing to both sides. Thus they remain equal. Simple example:
5*2 = 10
5*2-5=10-5.

All the way through the equation both sides remain equal.
If you plug in x= $\frac{8}{7}$ into everyone of your equations you will see that each side equals the other EXCEPT the line i pointed out.
what you did in that line while it got you to the correct result in the next line was totally "illegal" as far as algebra was concerned.

I was just making you sure you realized that those 2 lines together are ONE step.. and the illegal line wasn't a step all on its own. Nor something you could do by itself in other problems.

Normally every line in a solution is taken to be equivalent to every other line in the solution. However if you divide just a few terms by 3, then write them out. Then divide the rest by 3 and write them out. Your first line is NOT equivalent to the rest.. though you 2nd line is and you can proceed from the 2nd line to the final solution. A more normal way to do it, is just divide all the terms by 3, and write out that line.