Kenty,
How about
4x  5y = 12
x + 10y = 3.
In the elimination method you want to manipulate the equations so that the coefficients of x match or the coefficients of y match. For this pair of equations I see two ways to accomplish this.
 Multiply the second equation by 4. This would result in the term 4x in each equation.
 Multiply the first equation by 2. This would result in the term 10y in the first equation and 10y in the second equation.
Lets do it both ways.
 Multiply the second equation by 4. This results in
4x  5y = 12
4x + 40y = 12
Subtract the two equations to get
45y = 0 or y = 0. If you substitute y = 0 into either of the two initial equations you get x = 3. Thus the answer is x = 3, y = 0.
 Multiply the first equation by 2. This results in
8x  10y = 24
x + 10y = 3
This time add the two equations to get
9x = 27 or x = 27/9 = 3. Now substitution of x = 3 into either of the initial equations gives y = o.
In some systems you may need to manipulate both equations to produce a system where two of the coefficients match.
Try your system and write back if you still need help.
Penny
