Question

How do I solve this with elimination?
2x+3y=4
x+2y=-5

Answer 1

Thank you :)

Answer 2

2x+3y=4
x+2y=-5

y = (4-2x)/3
y = (-5-x)/2

Set y = y and cross multiply:

2(4-2x) = 3(-5-x)

8-4x = -15-3x

x = 23
y = -14

Answer 3

Sorry, I have to disagree with you on that

Answer 4

both of our answers work checked with the calculator

Answer 5

Weird, but okay

Answer 6

I believe mine is the correct, correct solution though

Answer 7

I'm so sorry, its not supposed to be two additions, the equation is;

2x+3y=4
x-2y=-5

Answer 8

so its it supposed to be;

x=-5-2y?

Answer 9

c'mon man. You have to post the right equations. I did all that hard work for nothing

Answer 10

i said sorry

Answer 11

You're forgiven

Answer 12

:)

Answer 13

moneybird, sorry but i didnt get where you got 2x-4y=-10 from?

Answer 14

2x+3y=4
x-2y=-5

y = 4-2x)/3
y = (-5-x)/-2

Set y = y and cross multiply

-2(4-2x) = 3(-5-x)

-8+4x = -15 -3x

7x = -7
x = -1
y = 2

Answer 15

Moneybird, whatever your approach is with these... It isn't working

Answer 16

There's only one method that works everytime.

Answer 17

By the way, moneybird...your solutions are not "true" solutions. They're the result of a flawed substitution method

Answer 18

2x+3y=4
x-2y=-5

2x - 4y = -10
- 2x + 3y = 4

-7y = -14
y = 2

2x + 6 = 4
2x = -2
x = -1

Answer 19

elimination gets us to the cramer rule

\begin{array}l
ax+by=n\\
cx+dy=m
\end{array}

\begin{array}r
*-c)&ax+by=n\\
*a)&cx+dy=m
\end{array}

\begin{array}l
-acx-cby&=-cn\\
\ \ \ acx+ady&=\ \ am\\
-----&----\\
\ \ \ \ y(ad-cb)&=am-cn
\end{array}
\[y=\frac{am-cn}{ad-cb}\]

and x is similar to construct

Answer 20

amistre...lol

Answer 21

I saw my first mistake now sry

Answer 22

Amistre always ends up doing some kind of extra cirriccular activity.
Good catch money

Answer 23

thank you guys for the help :)