1. ## Solving Systems

How do I solve this using any algebraic method?

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

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

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

3. 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.

4. $\displaystyle 3x-2y=10$
$\displaystyle 5x+3y=15$

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 ?

5. Originally Posted by Vuong

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

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?

6. Originally Posted by Bacterius
$\displaystyle 3x-2y=10$
$\displaystyle 5x+3y=15$

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

7. Originally Posted by pickslides
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?
y = -3.8 then plug it in?

8. Originally Posted by Vuong
How did you get the 9 for $\displaystyle 9 - 1.8y - 2y = 10$?
Comes from the substitution of $\displaystyle 3x = 9 - 1.8y$.

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

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

10. Originally Posted by pickslides

To either (1),(2),(3) or (4) to find x.
Thanks, I got 5.86

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

12. Originally Posted by Bacterius
That's not right, you should find something around $\displaystyle x \approx 3.15$
I plugged -3.8 into 3x-2y=10 x_x

13. Originally Posted by Vuong
y = -3.8 then plug it in?
Are you sure this is correct?

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

14. Originally Posted by pickslides
Are you sure this is correct?

$\displaystyle -19y = 5 \implies y = -\frac{5}{19} \approx -0.263$
So I plug -0.263 into one of them?

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

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