# System of Equations with different coefficients

#### skooter

##### New member
Hi it's me again with yet another problem that I cant solve

I'm currently solving systems of equations which until now has been extremely simplistic. Now this book is changing coefficients on me and I'm completely lost :lol:

I solved the first one just fine

5a + 2b = -16
a + 3b = 15

so:

5(a + 3b) = 5(15)
or 5a + 15b = 75

then:

5a + 2b = -16
subtract
5a + 15b = 75

to get:

-13b = -91

b = 7

then solve for x

a + 3(7) = 15

a = -6

then I substituted x and y into the original problem and it worked...I thought I had the method down.

So I tried it for the second problem:

5d - 3e = 5
d - e = -1

so:

5(d - e) = 5(-1)
or:
5d - 5e = -5

now subtract

5d - 3e = 5
subtract
5d - 5e = -5

to get -2e = 10

e = -5

then:

d - -5 = -1

d = -6

#### stapel

##### Super Moderator
Staff member
skooter said:
5d - 3e = 5
subtract
5d - 5e = -5

to get -2e = 10
Not quite:

. . . . .5d - 5e = -5
. . . . .5d - 3e = 5

. . . . .5d - 5d = 0

. . . . .-5e - (-3e) = -5e + 3e = -2e

. . . . .-5 - (5) = -10

. . . . .-2e = -10
. . . . .e = 5

Try it from there.

Eliz.

#### skooter

##### New member
Thank you for the response (I get how you found the answer)

how do you know when to put one equation on top of the other? In the example my book gives it places the equations in their original order when subtracting the two. I assumed it should always be this way.

#### stapel

##### Super Moderator
Staff member
skooter said:
In the example my book gives it places the equations in their original order when subtracting the two. I assumed it should always be this way.
Heck, no! :wink: You can rearrange to your heart's content!

Eliz.

#### skooter

##### New member
So pretty much the only way to tell when to rearrange the equations is if it's wrong the first time? :x