Solving Systems

**Vuong** · December 29th 2009, 03:16 PM

How do I solve this using any algebraic method?

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

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

**integral** · December 29th 2009, 03:31 PM

solve for x on one equation and substitute it in the other

**skeeter** · December 29th 2009, 03:32 PM

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.

**Bacterius** · December 29th 2009, 03:33 PM

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

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 ?

**pickslides** · December 29th 2009, 03:36 PM

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) $\times 3<br />$

gives

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

Now taking (4) form (3) gives

$-19y = 5$

Can you take it from here?

**Vuong** · December 29th 2009, 03:41 PM

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 $9 - 1.8y - 2y = 10$ ?

**Vuong** · December 29th 2009, 03:43 PM

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) $\times 3<br />$

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?

**Bacterius** · December 29th 2009, 03:46 PM

Originally Posted by Vuong

How did you get the 9 for $9 - 1.8y - 2y = 10$ ?

Comes from the substitution of $3x = 9 - 1.8y$ .

**pickslides** · December 29th 2009, 03:46 PM

Originally Posted by Vuong

y = -3.8 then plug it in?

To either (1),(2),(3) or (4) to find x.

**Vuong** · December 29th 2009, 03:57 PM

Originally Posted by pickslides

To either (1),(2),(3) or (4) to find x.

Thanks, I got 5.86

**Bacterius** · December 29th 2009, 04:01 PM

That's not right, you should find something around $x \approx 3.15$

**Vuong** · December 29th 2009, 04:02 PM

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

**pickslides** · December 29th 2009, 04:03 PM

Originally Posted by Vuong

y = -3.8 then plug it in?

Are you sure this is correct?

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

**Vuong** · December 29th 2009, 04:06 PM

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?

**Bacterius** · December 29th 2009, 04:08 PM

Originally Posted by Vuong

So I plug -0.263 into one of them?

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