# Simultaneous equations

• Jan 18th 2008, 02:59 AM
duckegg911
Simultaneous equations
After some much apreciated help on here by several people i would like to display this question complete with working out and soloution, could people please pick up on any mistakes i may have made

Many Thanks

Question 2

2x + y = -1
x + 2y = 4

Multiply the first equation by 2

Thus gives 4x + 2y = -2

Multiply the second equation by 4

Thus gives 4x + 8y = 16

Now subtract equation 2 from equation 1

Thus 0 + -6y = -18

Thus y = 3

Enter the value for y back into the first equation which gives…

2x + 3 = -1

Thus x must equal -2

The solutions are..

Y= 3
X= -2
• Jan 18th 2008, 03:24 AM
mr fantastic
Quote:

Originally Posted by duckegg911
After some much apreciated help on here by several people i would like to display this question complete with working out and soloution, could people please pick up on any mistakes i may have made

Many Thanks

Question 2

2x + y = -1
x + 2y = 4

Multiply the first equation by 2

Thus gives 4x + 2y = -2

Multiply the second equation by 4

Thus gives 4x + 8y = 16

Now subtract equation 2 from equation 1

Thus 0 + -6y = -18

Thus y = 3

Enter the value for y back into the first equation which gives…

2x + 3 = -1

Thus x must equal -2

The solutions are..

Y= 3
X= -2

I would however make two suggestions:

1. Check your final answer by substituting it into the given equations.

2. Multiplying the first equation by 2 and the second by 4 to eliminate x is a bit inefficient. I'd suggest just multiplying the second by 2 .... you'll achieve the same result with less work and therefore more quickly and with less chance of error:

Quote:

Originally Posted by duckegg but edited (in red) by Mr F
[snip]
2x + y = -1
x + 2y = 4

[snip]

Multiply the second equation by 2

Thus gives 2x + 4y = 8

Now subtract equation 1 from equation 2

Thus 0 + 3y = 8 - -1 = 9

Thus y = 3