Solving Systems

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• December 29th 2009, 03:16 PM
Vuong
Solving Systems
How do I solve this using any algebraic method?

3x-2y=10
5x+3y=15

2x-4y=-6
-x+2y=3
• December 29th 2009, 03:31 PM
integral
solve for x on one equation and substitute it in the other
• December 29th 2009, 03:32 PM
skeeter
Quote:

Originally Posted by Vuong
How do I solve this using any algebraic method?

3(3x-2y=10)
2(5x+3y=15)

2x-4y=-6
2(-x+2y=3)

solve using elimination ... distribute, add the equations and solve for the remaining variable. back substitute to find the other variable.
• December 29th 2009, 03:33 PM
Bacterius
$3x-2y=10$
$5x+3y=15$

From the second equation you know that :

$5x = 15 - 3y$

Thus $15x = 45 - 9y$

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

Substituting back this result into the first equation gives :

$9 - 1.8y - 2y = 10$

$9 - 3.8y = 10$

$3.8y = -1$

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

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

Can you do it for the second system now ? :)
• December 29th 2009, 03:36 PM
pickslides
Quote:

Originally Posted by Vuong

3x-2y=10
5x+3y=15

You can also use the elimination method

make

$3x-2y=10$ ...(1)
$5x+3y=15$ ...(2)

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

gives

$15x-10y=50$ ...(3)
$15x+9y=45$ ...(4)

Now taking (4) form (3) gives

$-19y = 5$

Can you take it from here?
• December 29th 2009, 03:41 PM
Vuong
Quote:

Originally Posted by Bacterius
$3x-2y=10$
$5x+3y=15$

From the second equation you know that :

$5x = 15 - 3y$

Thus $15x = 45 - 9y$

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

Substituting back this result into the first equation gives :

$9 - 1.8y - 2y = 10$

$9 - 3.8y = 10$

$3.8y = -1$

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

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

Can you do it for the second system now ? :)

How did you get the 9 for $9 - 1.8y - 2y = 10$?
• December 29th 2009, 03:43 PM
Vuong
Quote:

Originally Posted by pickslides
You can also use the elimination method

make

$3x-2y=10$ ...(1)
$5x+3y=15$ ...(2)

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

gives

$15x-10y=50$ ...(3)
$15x+9y=45$ ...(4)

Now taking (4) form (3) gives

$-19y = 5$

Can you take it from here?

y = -3.8 then plug it in?
• December 29th 2009, 03:46 PM
Bacterius
Quote:

Originally Posted by Vuong
How did you get the 9 for $9 - 1.8y - 2y = 10$?

Comes from the substitution of $3x = 9 - 1.8y$.
• December 29th 2009, 03:46 PM
pickslides
Quote:

Originally Posted by Vuong
y = -3.8 then plug it in?

(Nod)
To either (1),(2),(3) or (4) to find x.
• December 29th 2009, 03:57 PM
Vuong
Quote:

Originally Posted by pickslides
(Nod)
To either (1),(2),(3) or (4) to find x.

Thanks, I got 5.86
• December 29th 2009, 04:01 PM
Bacterius
That's not right, you should find something around $x \approx 3.15$ :(
• December 29th 2009, 04:02 PM
Vuong
Quote:

Originally Posted by Bacterius
That's not right, you should find something around $x \approx 3.15$ :(

I plugged -3.8 into 3x-2y=10 x_x
• December 29th 2009, 04:03 PM
pickslides
Quote:

Originally Posted by Vuong
y = -3.8 then plug it in?

Are you sure this is correct?

$-19y = 5 \implies y = -\frac{5}{19} \approx -0.263$
• December 29th 2009, 04:06 PM
Vuong
Quote:

Originally Posted by pickslides
Are you sure this is correct?

$-19y = 5 \implies y = -\frac{5}{19} \approx -0.263$

So I plug -0.263 into one of them?
• December 29th 2009, 04:08 PM
Bacterius
Quote:

Originally Posted by Vuong
So I plug -0.263 into one of them?

(Nod)
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