# Thread: Solving simultaneous eqautions by inspection

1. ## Solving simultaneous eqautions by inspection

Hi,

I have attached what I am trying to do hopefully it is clear any ideas?

Thanks

2. Subtract the 2nd equation from the first equation and you have

-2y = -2 so y = 1 then substitute y = 1 into 1st equation and you have

x + 8(1) = 4 so x = -4

3. Originally Posted by Caity
-2y = -2 so y = 1
Thanks for that but why does -2 become y=1, I would think logically it was y=-2

4. $\displaystyle x+8y=4$

$\displaystyle x+10y=6$

Subtract the two equations.

$\displaystyle -2y=-2$

Divide both sides of the equation by -2.

$\displaystyle \frac{-2y}{-2}=\frac{-2}{-2}$

$\displaystyle y=1$

Substitute y = 1 into the first equation and solve for x.

5. Thanks that really helped but I got stuck again on this one

$\displaystyle 5x-3y=47$ 1

$\displaystyle 7x-3y=67$ 2

So we ignore the 3y as $\displaystyle (-3y) - (-3y)$ is 0.

On the right: $\displaystyle 47-67 = -20$
On the left: $\displaystyle 5x-7x = -2x$

Therfore:
$\displaystyle \frac{-2x}{-20} = 10x$

Now we substitute $\displaystyle x = 10$ into formulae 1, because it has the lowest number on right

$\displaystyle 5 x 10 = 47$

$\displaystyle 50 = 47$

$\displaystyle 47 - 50$

$\displaystyle = -3$

$\displaystyle y = -3$

Although the answer says $\displaystyle y = -1$

Hope this is clear, and thank you.

6. You have it correct up to -3y = -3
then you divide both sides by the -3 because you want y by itself.
so y= 1 not -1. sometimes the answers in the book have typos

7. Hey thanks for that. Ive just got to the bit where the y or x's are more than 1 never knew u had to devide ... Thanks