How do I solve this using any algebraic method?

3x-2y=10

5x+3y=15

2x-4y=-6

-x+2y=3

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- Dec 29th 2009, 03:16 PMVuongSolving Systems
How do I solve this using any algebraic method?

3x-2y=10

5x+3y=15

2x-4y=-6

-x+2y=3 - Dec 29th 2009, 03:31 PMintegral
solve for x on one equation and substitute it in the other

- Dec 29th 2009, 03:32 PMskeeter
- Dec 29th 2009, 03:33 PMBacterius
$\displaystyle 3x-2y=10$

$\displaystyle 5x+3y=15$

From the second equation you know that :

$\displaystyle 5x = 15 - 3y$

Thus $\displaystyle 15x = 45 - 9y$

And so, $\displaystyle 3x = 9 - 1.8y$.

Substituting back this result into the first equation gives :

$\displaystyle 9 - 1.8y - 2y = 10$

$\displaystyle 9 - 3.8y = 10$

$\displaystyle 3.8y = -1$

$\displaystyle y = \frac{-1}{3.8}$

Now substitute the value of $\displaystyle y$ found into any of the equations to retrieve $\displaystyle x$.

Can you do it for the second system now ? :) - Dec 29th 2009, 03:36 PMpickslides

You can also use the elimination method

make

$\displaystyle 3x-2y=10$ ...(1)

$\displaystyle 5x+3y=15$ ...(2)

To eliminate x take (1) $\displaystyle \times 5 $ and (2) $\displaystyle \times 3

$

gives

$\displaystyle 15x-10y=50$ ...(3)

$\displaystyle 15x+9y=45$ ...(4)

Now taking (4) form (3) gives

$\displaystyle -19y = 5$

Can you take it from here? - Dec 29th 2009, 03:41 PMVuong
- Dec 29th 2009, 03:43 PMVuong
- Dec 29th 2009, 03:46 PMBacterius
- Dec 29th 2009, 03:46 PMpickslides
- Dec 29th 2009, 03:57 PMVuong
- Dec 29th 2009, 04:01 PMBacterius
That's not right, you should find something around $\displaystyle x \approx 3.15$ :(

- Dec 29th 2009, 04:02 PMVuong
- Dec 29th 2009, 04:03 PMpickslides
- Dec 29th 2009, 04:06 PMVuong
- Dec 29th 2009, 04:08 PMBacterius